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(1)

"!$#&%(')#&*,+

-(. -,/10 /

23

400 × 0, 4 = 4 × 100 × 4 × 0, 1 = 16 × 10 = 160

4 . -,/10 /

2 3

300 0, 3 = 3 × 100

3 × 0, 1 = 100

0, 1 = 100 × 10 = 1000

05.7698 0

2

/:-

;1<=>?

02/

12>>

-

<,

0

2@BA&C

-

DE>F 4 2HGI<=,

/ .

555

^C ^-

7 555 b 7

− 49

79 65

− 63 2

JKDL"')M&*N*2!$O%P=*,+

-,/10 /

2

-(.

(10 ² ) ⁴ × 10 ⁻⁵ = 10 ^2×4 × 10 ⁻⁵ = 10 ⁸ × 10 ⁻⁵ = 10 ⁸⁻⁵ = 10 ³

4 .

10 ⁴ + 10 ² = 10000 + 100 = 10100

05.

1 10 ⁻ ² = 10 ² = 100

Q;HRTSU!$%,VUMXWYO*,+[Z

0

1,\$

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] -,0

,>c12

0

4 / d\

/

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1>7;1<

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1 3 + ³ ₅ = ₁₅ ⁵ + ₁₅ ⁹ = ⁵⁺⁹ ₁₅ = ¹⁴ ₁₅

4 .

1 3 3 5

= ¹ ₃ × ⁵ ₃ = ⁵ ₉

05.

1 3

6 = ¹ ₃ × ¹ ₆ = ₁₈ ¹

e "Wgfih)!;Sj!Mk*2WgOmlPnOWgfpo)SUP=*,+

F_C

-

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-

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<

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2 3 > ³ ₅

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³ ₅ = ₁₅ ⁹

4 .

ln ⁴ ₇

< 0

^3T2>v ⁸

4 < 7

^,> ⁰

⁴ ₇ < 1

^\T,> ⁰

ln ⁴ ₇

< ln 1

^A^0N-

ln

² ⁰ ^- ^>(B ^/ ^2ECgs^]

0

.

$w

/

2xGyC;

0

2Dzq

ln 1 = 0

05.

√

3 ² + 4 ² < 7

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√

3 ² + 4 ² = √

9 + 16 = √

25 = 5

{KD|O%!$lSjP=fiP=OYV?lPiOWgfpo)SUP=*,+)} > 0

>;;~2^2K

/

x

y

^/^zq

1 6 x 6 3

2 6 y 6 5

⁶ ^>

E>>

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FB>qH

x + y

\>c2>

02-

2F_2>(H

x − y

^- >azqa`>n>

0N-

;2F_2>(H

1 y

-(.

1 + 2 6 x + y 6 3 + 5

^> ⁰

3 6 x + y 6 8

4

. -

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4

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0

− 2 > − y > − 5

^\ ⁰ ^`2@rGjrt

− 5 6 − y 6 − 2

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1 + ( − 5) 6 x + ( − y) 6 3 + ( − 2)

^\ ⁰ `2@rGjr1

− 4 6 x − y 6 1

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0

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-

>(

0

1

-

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T>F 4 2Cgs

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a

0

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5

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x

y

^/^x3

(2)

-(.

(x + y) ² = x ² + 2xy + y ² .

4 .

( x − y ) ² = x ² − 2 xy + y ² .

05.

( x − y )( x + y ) = x ² − y ²

HRT!$%,V5W SjMk*!;VUMXWYO+

-d0 ,2

/

27KC2,>1<

-

>(2x3

-(.

3x ² + 4x = x(3x + 4)

4 .

( x + 1) x − (2 + 3 x )( x + 1) = ( x + 1)[ x − (2 + 3 x )] = ( x + 1)[ x − 2 − 3 x ] = ( x + 1)( − 2 x − 2) =

− 2(x + 1)(x + 1) = − 2(x + 1) ² .

05.

49x ² − 4 = (7x) ² − 2 ² = (7x + 2)(7x − 2)

;a'!;VUMXWgO)* U+ ;E

-

>;

R

^/^27z= ^- 1>Ds<

-

>(2E3

-(.

5 x = 0 ⇐⇒ x = ⁰ ₅ = 0 .

4 .

− 3 x = 15 ⇐⇒ x = ₋ ¹⁵ ₃ = − 5 .

05.

4x + 3 − 2(1 − x) = 0 ⇐⇒ 4x + 3 − 2 + 2x = 0 ⇐⇒ 6x = − 1 ⇐⇒ x = − ¹ ₆ .

;a'!;VUMXWgO)*U+ ;E

-

>;

R

^/^27z= ^- 1>Ds<

-

>(2E3

-(.

x ² = 9 ⇐⇒ ( x = √

9 x = − √

9) ⇐⇒ ( x = 3

x = − 3) .

4 .

x ² = 2 x ⇐⇒ x ² − 2 x = 0 ⇐⇒ x ( x − 2) = 0 ⇐⇒ ( x = 0

x − 2 = 0) ⇐⇒ ( x = 0

x = 2) .

05.

x ² + 1 = 0 ⇐⇒ x ² = − 1

^q ^> ⁰ ^C ^- ^D ^/^{;>^2} ^//^7G ⁰ ^D2zq ^- ,>_C;z=;

/ 0N-

7a`>_>F

4

/

H,7Cgs

]

,c>q

/

;a'!;VUMXWgO)*U+ ;E

-

>;

R

³

-(.

2x ² − 3x + 1 = 0

^$ ⁰ ^1FB1> ^- ^>(

∆ = 1 > 0

^> ⁰ ^2K? ^/^;1>2 ^/1/³

x ₁ = ⁻ ^b ⁻ _2a ^√ ^∆ = ³⁻¹ _2×2 = ¹ ₂

x ₂ = ⁻ ^b+ _2a ^√ ^∆ = ³⁺¹ _2×2 = 1.

4 .

x ² − 3x = 0 ⇐⇒ x(x − 3) = 0 ⇐⇒ (x = 0

x − 3 = 0) ⇐⇒ (x = 0

x = 3).

05.

2 x + x ² − 3 = 0 ⇐⇒ x ² + 2 x − 3 = 0

⁾ ⁰ ^1FB1> ^- ^>(

∆ = 16 > 0

^> ⁰ ²⁼ⁱ ^/^K,>E2 ^//^2_3

x ₁ = ⁻²⁻⁴ ₂ _× ₁ = − 3

x ₂ = ⁻²⁺⁴ ₂ _× ₁ = 1 .

7O=a'!;VUMXWgO)*,+

;E

-

>

R

^/^27>;2zq ^- ^>71< ^- ^>(2x3

-(.

9x 6 0 ⇐⇒ x 6 ⁰

9 ⇐⇒ x 6 0,

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0$-,0

>n;27F_2D

/

`>

-,/

1C

-

7>c>F 4 Cgs

]

4 .

− 3 x > 15 ⇐⇒ x 6 ₋₃ ¹⁵ ⇐⇒ x 6 − 5 ,

^0N- ^> ^0$- ^> ^/^> ^/ ^`>;2 ^-,/¹ ^/z=[`b>F^

/

C

/

0$-,0

>

H2F_27;

/

`b1>2

-,/

sC

-

7>c>F 4 >2

-

]

05.

x ² − 3x + 2 6 0

^>F_

x ² − 3x + 2

^2@;c>;

a

⁰ `brGjr;Cg1

]

A&

0

a = 1

^. ^Cg

,;

x

^G ^/ ^`b ^K22;H;2 ^-,0 ^1>2 ^g} ⁰ ^x>FB ^- ^FBHCY;H; ⁰ ^1FB1> ^- ^>(

∆ = 1 > 0

Cg~2">

0

Kc

-,0

>7

//

2

x ₁ = ³⁻¹ _2×1 = 1

x ₂ = ³⁺¹ _2×1 = 2.

^- ⁰ ,>z=;2>(

x ² − 3x + 2 6 0 ⇐⇒ 1 6 x 6 2

UJK?qWgfi(V5SjMXPg+

-(.

`-

A

^a`>^- ^> ^/ H>>;2xC -

/:-^]

,FI

/

I3

A =

"!$#&%

×

^'

!$(*)&%+(,

2 ·

4

.7

`-

A

a`>z=;x;x

-

u>

R

H>>;2EC -

/:-"]

FI

/

I3

A = πR ² .

(3)

05.

} > 0

>~>

-

>

/

ABC

⁰ ^- ^> ^/^{^>}

B

⁶ ^KC1FB ^/ ⁰ ^,>q; ^/ ^`^- ^> ^/

BCA [

^G ^/ ^`^- ^{1^;2}

/

,>D

0

,Dc

-

>

/

I3

cos( BCA [ ) = BC AC ·

5Q;?SUWYMV5P=*,+

-(.

>>"

-

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>,F__; C

/:-

> /

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A

B

^;>( ^/ ⁰ K,>;>2I>(3

A(1; 2)

B(2; − 1)

^$ ^2FB1>2 ^/ ^`bz= ^- ^1> ^/:- ¹

(AB )

x _A 6 = x _B

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(AB)

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y = ax + b

F_FB

A

;>pCY1>(

/:-

;s,\ 2

0

=>>2 ;1<,2>(<,2$2

/

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-

> g ,>

0

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2 = a + b

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B

³

− 1 = 2 a + b

^$ ^/- ^CFB1~22 ^-,/^1>

b = 2 − a

^zq ^/ ^`b>

FBC

/:-,0 x

-

>

/-

0

>^2

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1CY;H 4 >

− 1 = 2a + (2 − a)

^g ^> ⁰

a = − 3

^F_FB

b = 2 − a

^\

>c 4 2>(

b = 2 − ( − 3) = 5

^`bzq ^- ^1> ^/:- ¹

( AB )

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y = − 3 x + 5

4 .

>>;

-

>D>n2Cg~2E@

,>F_C

/:-

>

/-

;s

(D)

^a`2zq ^- ^>

y = 2x + 3

^/- ^s

(∆)

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y = − 2 x + 1

x;2Ks>(rt /1/

2

02-

>(

0

>

]

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0

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0

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> C

- .

2

/

2F_2>(

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0

=

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0

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2

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0 /

0

K

0

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0

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8

22>( A

2 6 = − 2

^. ^,> ⁰ ^/^{^12x>(}

0N-

>(2

05.

2

/

u=~2F_"1<

-

>(x3

2x − 5y = − 4 3 x + y = 11

2

/

;>C

-

7 4 @1;1>

-

2KK1~2F_xzq

-

1>

]

>1

y = 11 − 3 x

^g} ^>c2F_C ^/-,0

y

^C ^-

11 − 3 x

-

>;

/:-

CFB1~2B2zq

-

>iCY;^ 4 2>1

2x − 5(11 − 3x) = − 4

^/- ^>>;

2x + 15x = − 4 + 55

^\

0

`2@rGjr1

17x = 51

^`

x = ⁵¹ ₁₇ = 3

^1> ^-,/^2F_2>(

y = 11 − 3x = 11 − 3 × 3 = 11 − 9 = 2

>

0/

>3$2

/T/

0 C

/

(3; 2)

^/;1>?u=@~2F_

e 7L !SjMV5 +)L=SjMXWTlMk%(MV5 +

-(.

,>>2HKKFBC

/

2D

]

>

0

1>7C

-

1x3

f(x) = x ²

g(x) = cos x

4 .

>;>2DKcK2F_C

/

2D

]

>

0

,>7F_C

-

123

f (x) = x ³

g(x) = sin x

05.

,>>2DKcKFBC

/

2D

]

>

0

1>

2π

rCg21K;zq23

f(x) = cos x

g(x) = sin x

U{KMkf MV5P=*,+

F_C

/

2

/

2

/

F_11<

-

>(E3

-(.

lim

x →−∞

x ³ = −∞ .

4 .

x lim →0

1 x ² = + ∞ .

05.

lim

x→+∞

x

1 + x ² = lim

x→+∞

x

x ² = lim

x→+∞

1 x = 0

^\ ^02- ^/:-/1FBsHa`>zqd>(;xCg

/

uK>FB72>

/

`>>[

2

-,/

EG

/:-/

F_1Hczq,12>(DH2F_2DEC

/

$-

K;2

5;!o)#XP=!' lP;!SjM&!KVUM&WgO*,+

,1

f

^;> ^] ^> ⁰ ^> $>s<

- 4 /

;

/

`b1>q2@<

-d//

[ − 5; 13]

^>( ^/

- 4 / -

<

-

,> 2@;>>

0 rx3

x − 5 1 2 8 13

f (x)

− 10 % − 3

& − 5 % 4

& − 6

} >;21>EC

-

C _f

^/:-I0 ⁴ ^EC2>( ^- ^s<x

f

(4)

-(.7

/

2

/

F

-

K1FIF ;

f

^/ `>(<

-,/1/

[ − 5; 13]

⁶ ^>czq ^/1/^< ^-,/^27rt^/T- ^>(

EF

-

K1FIF 2

4

^D^/ ^2@ ^- 2>(D>

x = 8

4 .

2FB1>2

/

>F 4 x

/

;1>\

-

CC

-

2>

-

>(G

/

`>(<

-,/1/

[ − 5; 13]

^\; ^/ ^`2zq ^- ^>

f (x) = − 2

`bzq

-

1>

-

F_7;2K

/

K,>2\

/

`>

-

CC

-

2>

-

>(HG

/

`>(2<

-,/1/

[2; 8]

^A ^0N-

− 5 6 − 2 6 4

^. ^/ ^`^- ^;

G /

`>(<

-,/1/

[8; 13]

^A ^0N-

− 6 6 − 2 6 4

^.

05.

} >

-

sDzq

f ⁰ ( − 5) = 1

^$ ^F_> ^/ ^`2zq ^- ^>? ^/- ^- >>(xG

C _f

^- cCg1>qD[`

- 4 0

1

x = − 5

`bzq

-

1>_

0

D

-

>,2>(H2@,>>2HC

-

y = f ⁰ ( − 5) × ( x − ( − 5)) + f ( − 5)

^- ^/ ⁰ ^D_ ^- ⁴ ^/ ^-

<

-

,>2\5>x<szq

f ( − 5) = − 10

^d ^> ^0/ ^`2zq ^- ^>x2@

y = 1 × (x+5) − 10

^\ ⁰ ^`2@rGjrt

y = x − 5

DRTWgO)%,VUMXWgO)*IV5SjM

WYOWgfi(V5SjMk['P=*,+

F_C

/

3

-(.

cos(0) = 1

cos( ^π ₂ ) = 0.

4 .

sin(0) = 0

sin( ^π ₂ ) = 1 .

05.

Cg;

x

² ^/ ³

cos ² x + sin ² x = 1 .

;W

!SjMVfiPOqh =SjMXP=O+

-(.7

/

2

/

`2>;2F 4 /

E$>s,>?

ln

^{^ `2@}

]0; + ∞ [

4 . F_C

/

3

lim

x → 0 ⁺ ln x = −∞ .

05.

1FBC

/22\

0

`2@rGjr1

/

`b

0

1

-

>

/

@uKF 4 /

ln

^\ ^{/:-^]} ^-d0 ^1>s< ^- ^>(3

ln 8

ln 2 = ln(2 ³ )

ln 2 = 3 ln(2) ln 2 = 3.

;W

!SjMVfiPOqh =SjMXP=O P(V MkO =a'!;VUMXWYO*,+

-(.

F_>H;>2

/

x

^< ^- ^>(

ln x = ln 2 + ln 3

} >

-

1z=;

ln a + ln b = ln(ab)

^> ⁰

ln 2 + ln 3 = ln 6

^$ ^`

x = 6

4 . 2,

/

`bz=

-

1>

ln x = 1

ln x = 1 ⇐⇒ e ^ln ^x = e ¹ ⇐⇒ x = e .

05.

2

/

`b1>2zq

-

,>

ln x > 0

-I]

>

0

> ;Cg>>(

//

-

>q

0

1

-

>qE

R

^\> ^-

ln x > 0 ⇐⇒ e ^lnx > e ⁰ ⇐⇒ x > 1.

J ;7|Th WgOP=OYVUMXPK#&#XPg+

-(.7

/

2

/

`2>;2F 4 /

E$>s,>?

exp

^{^ `2}

R

4 . F_C

/

3

lim

x →−∞ e ^x = 0

^A&HF ^2F_

0 ⁺

^.

05.76

KC1FB

e ^2x

^G ^/ ^`^-

e ^x

} >

-

1z=;

e ^a+b = e ^a × e ^b

^$ ^> ⁰ ^2> ^]&- ^- ^>(

a = b = x

^\$>c ⁴ ^>q

e ^2x = e ^x × e ^x = (e ^x ) ²

JK7|Th WgOP=OYVUMXPK#&#XPcP(V M&O K['!;V5M&WgO*,+

-(.

1FBC

/ 22\

0

`2@rGjr1

/

`b

0

1

-

>

/

H@u=F 4 /

2

ln

e

^\ ^/^`b ;C;21>1<

-

>(_3

e ^{2 ln 5} = e ^ln(5 ² ⁾ = 5 ² = 25.

4 .

2,

/

`bz=

-

1>

e ^x = 2

e ^x = 2 ⇐⇒ ln(e ^x ) = ln 2 ⇐⇒ x = ln 2.

(5)

05.

2

/

`b1>2zq

-

,>

e ⁻ ^x 6 − 1

} >?

-

s7zqCg;72

/

a

^\ ^/^>;F ⁴

e ^a

⁷⁷ ⁰ FB>qCY1

]

= >

0

Cg;2

/

x

^\ ^/

>F

4

e ⁻ ^x

^{2E Cgs}^] ^gw ^/ >Cg2KE>

0 C - >

]

122

-,/

G

− 1

^T ^> ^0_/ ^`b1>2zq ^- ^,>

>a`

-

F_DC

-

H;x

/

K,>

JJK7|Th WgOP=OYVUMXPK#&#XPcP(VB#XW

!SjMVfiP +

-(.

-,0 2

/ `

-,/1/ /:-I0

4 EC2>(

-

s<x

ln

^,Cg ^/^u ⁰ ^,

4

.

-,0 2

/ ` -,//

/:-I0

4 EC2>(

-

s<^;

exp

^17CY ^/^u ⁰

05.

F_F_2>(DC

-

r rt>;

/-I0

; 4 x2C22>(

-

1<,"

ln

^G ⁰ ^//

exp

^`>x2 ^/- ^u=FBzqE

/ `-

;C

-

7

-

CCg@HG

/-

sa`2zq

-

,>

y = x

J,Q;"Wgfih Wg*2qPylPtWgO%,VUMXWYO*,+

2

]

,>

0

>

u

g

^>( ⁰ ^>>q ^C ^- ^/² ^- ⁴ ^/ ^- ^I< ^- ^- ,>"9zq

/

zq2

<

-,/

3

x −∞ − 1 3 + ∞

u(x)

4 & 2

& − 2 % 2

x −∞ − 2 2 + ∞

g(x)

−∞ % 0

& − 1 % + ∞

s

f

^{/:-^]} ^> ⁰ 1>$>;E

R

^C ^- ^3$Cg,;H2 ^/

x

^\

f (x) = g ◦ u(x) = g(u(x)).

-(. -,/10 /

23

f (3) = g (u(3)) = g( − 2) = 0.

4 . -,/10 /

>

lim

x→+∞ f (x)

³

} > -

x →+∞ lim u ( x ) = 2

lim

X → 2 g ( X ) = g (2) = − 1

^- ⁰ >;2zq>(

x →+∞ lim f ( x ) = − 1

05.

F_>

/

>H<

-

,>

f

^; ^/ `b1>(2@<

-,//

[ − 1; + ∞ [

◦

^. ^- ^< ^- ^/1/^>a`^- ⁴

[ − 1; 3]

>;;2,>

a

b

^/^Dz=;

− 1 6 a 6 b 6 3

^FBF_ ^/:-]^> ⁰ ^1>

u

^H ⁰ ¹ ^- ^>(x

[ − 1; 3]

^\$> ^- ³

u(3) 6 u(b) 6 u(a) 6 u( − 1)

⁰ ^`2@rGjrt

− 2 6 u(b) 6 u(a) 6 2

}

g

² ⁰ ^, ^- ^>("

[ − 2; 2]

^> ⁰

g(2) 6 g(u(a)) 6 g(u(b)) 6 g ( − 2)

^\ ⁰ ^`2@rGjrt

− 1 6 f (a) 6 f ( b ) 6 0

KFB>(;1D

a 6 b

^-,/

f ( a ) 6 f ( b )

^,> ⁰

f

⁰ ^- ^>(x

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Pour se tester en arrivant à l'iut

400 × 0, 4 = 4 × 100 × 4 × 0, 1 = 16 × 10 = 160

300

0, 3 = 3 × 100

3 × 0, 1 = 100

0, 1 = 100 × 10 = 1000

555

7

555 b 7

− 49

79 65

− 63 2

(10 2 ) 4 × 10 −5 = 10 2×4 × 10 −5 = 10 8 × 10 −5 = 10 8−5 = 10 3

10 4 + 10 2 = 10000 + 100 = 10100

1

10 − 2 = 10 2 = 100

1

3 + 3 5 = 15 5 + 15 9 = 5+9 15 = 14 15

1 3 3 5

= 1 3 × 5 3 = 5 9

1 3

6 = 1 3 × 1 6 = 18 1

<

=

>

2

3 > 3 5

2 3 = 10 15

3 5 = 15 9

ln 4 7

< 0

4 < 7

4 7 < 1

ln 4 7

< ln 1

ln

ln 1 = 0

√

3 2 + 4 2 < 7

√

3 2 + 4 2 = √

9 + 16 = √

25 = 5

x

y

1 6 x 6 3

2 6 y 6 5

x + y

x − y

1 y

1 + 2 6 x + y 6 3 + 5

3 6 x + y 6 8

y

− 1

− 2 > − y > − 5

− 5 6 − y 6 − 2

1 + ( − 5) 6 x + ( − y) 6 3 + ( − 2)

− 4 6 x − y 6 1

2 6 y 6 5

1 2 > 1 y > 1

5

1

5 6 1 y 6 1

2

x

y

(x + y) 2 = x 2 + 2xy + y 2 .

( x − y ) 2 = x 2 − 2 xy + y 2 .

( x − y )( x + y ) = x 2 − y 2

3x 2 + 4x = x(3x + 4)

( x + 1) x − (2 + 3 x )( x + 1) = ( x + 1)[ x − (2 + 3 x )] = ( x + 1)[ x − 2 − 3 x ] = ( x + 1)( − 2 x − 2) =

− 2(x + 1)(x + 1) = − 2(x + 1) 2 .

49x 2 − 4 = (7x) 2 − 2 2 = (7x + 2)(7x − 2)

R

5 x = 0 ⇐⇒ x = 0 5 = 0 .

− 3 x = 15 ⇐⇒ x = − 15 3 = − 5 .

4x + 3 − 2(1 − x) = 0 ⇐⇒ 4x + 3 − 2 + 2x = 0 ⇐⇒ 6x = − 1 ⇐⇒ x = − 1 6 .

R

x 2 = 9 ⇐⇒ ( x = √

9

(10 ² ) ⁴ × 10 ⁻⁵ = 10 ^2×4 × 10 ⁻⁵ = 10 ⁸ × 10 ⁻⁵ = 10 ⁸⁻⁵ = 10 ³

10 ⁴ + 10 ² = 10000 + 100 = 10100

10 ⁻ ² = 10 ² = 100

3 + ³ ₅ = ₁₅ ⁵ + ₁₅ ⁹ = ⁵⁺⁹ ₁₅ = ¹⁴ ₁₅

= ¹ ₃ × ⁵ ₃ = ⁵ ₉

6 = ¹ ₃ × ¹ ₆ = ₁₈ ¹

3 > ³ ₅

² ₃ = ¹⁰ ₁₅

³ ₅ = ₁₅ ⁹

ln ⁴ ₇

⁴ ₇ < 1

ln ⁴ ₇

3 ² + 4 ² < 7

3 ² + 4 ² = √

¹ ₂ > ¹ _y > ¹

5 6 ¹ _y 6 ¹

(x + y) ² = x ² + 2xy + y ² .

( x − y ) ² = x ² − 2 xy + y ² .

( x − y )( x + y ) = x ² − y ²

3x ² + 4x = x(3x + 4)

− 2(x + 1)(x + 1) = − 2(x + 1) ² .

49x ² − 4 = (7x) ² − 2 ² = (7x + 2)(7x − 2)

5 x = 0 ⇐⇒ x = ⁰ ₅ = 0 .

− 3 x = 15 ⇐⇒ x = ₋ ¹⁵ ₃ = − 5 .

4x + 3 − 2(1 − x) = 0 ⇐⇒ 4x + 3 − 2 + 2x = 0 ⇐⇒ 6x = − 1 ⇐⇒ x = − ¹ ₆ .

x ² = 9 ⇐⇒ ( x = √

x ² = 2 x ⇐⇒ x ² − 2 x = 0 ⇐⇒ x ( x − 2) = 0 ⇐⇒ ( x = 0

x ² + 1 = 0 ⇐⇒ x ² = − 1

2x ² − 3x + 1 = 0

x ₁ = ⁻ ^b ⁻ _2a ^√ ^∆ = ³⁻¹ _2×2 = ¹ ₂

x ₂ = ⁻ ^b+ _2a ^√ ^∆ = ³⁺¹ _2×2 = 1.

x ² − 3x = 0 ⇐⇒ x(x − 3) = 0 ⇐⇒ (x = 0

2 x + x ² − 3 = 0 ⇐⇒ x ² + 2 x − 3 = 0

x ₁ = ⁻²⁻⁴ ₂ _× ₁ = − 3

x ₂ = ⁻²⁺⁴ ₂ _× ₁ = 1 .

9x 6 0 ⇐⇒ x 6 ⁰

− 3 x > 15 ⇐⇒ x 6 ₋₃ ¹⁵ ⇐⇒ x 6 − 5 ,

x ² − 3x + 2 6 0

x ² − 3x + 2

x ₁ = ³⁻¹ _2×1 = 1

x ₂ = ³⁺¹ _2×1 = 2.

x ² − 3x + 2 6 0 ⇐⇒ 1 6 x 6 2

A = πR ² .

x _A 6 = x _B