a) Ta có x(x+3)=0

\(\Rightarrow\left[\begin{matrix}x+3=0\\x=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}x=0-3\\x=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}x=-3\\x=0\end{matrix}\right.\)

Vậy \(x\in\left\{-3;0\right\}\)

b) Ta có (x-1)(\(^{x^2}\)+1)=0

\(\Rightarrow\left[\begin{matrix}x-1=0\\x^2+1=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}x=0+1\\x^2=0-1\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}x=1\\x^2=-1\end{matrix}\right.\)

x² =-1 là vô lý vì x² ko thể là một số âm

vậy x=1

nhớ tick tui đó tui làm rõ rằng lắm

\(f'\left(x\right)=3x^2-6x\Rightarrow f''\left(x\right)=6x-6\)

Theo đề: \(f''\left(x\right)=0\Leftrightarrow6x-6=0\Leftrightarrow x=1\).

Thay \(x=1\) vào \(f\left(x\right)\) \(\Rightarrow f\left(x\right)=-1\).

Vậy: Tọa độ điểm là \(I\left(1;-1\right)\)

a) x³ - 64x = 0

x(x² - 64) = 0

x(x - 8)(x + 8) = 0

x = 0 hoặc x - 8 = 0 hoặc x + 8 = 0

*) x - 8 = 0

x = 8

*) x + 8 = 0

x = -8

Vậy x = -8; x = 0; x = 8

b) x³ - 4x² = -4x

x³ - 4x² + 4x = 0

x(x² - 4x + 4) = 0

x(x - 2)² = 0

x = 0 hoặc (x - 2)² = 0

*) (x - 2)² = 0

x - 2 = 0

x = 2

Vậy x = 0; x = 2

c) x² - 16 - (x - 4) = 0

(x - 4)(x + 4) - (x - 4) = 0

(x - 4)(x + 4 - 1) = 0

(x - 4)(x + 3) = 0

x - 4 = 0 hoặc x + 3 = 0

*) x - 4 = 0

x = 4

*) x + 3 = 0

x = -3

Vậy x = -3; x = 4

d) (2x + 1)² = (3 + x)²

(2x + 1)² - (3 + x)² = 0

(2x + 1 - 3 - x)(2x + 1 + 3 + x) = 0

(x - 2)(3x + 4) = 0

x - 2 = 0 hoặc 3x + 4 = 0

*) x - 2 = 0

x = 2

*) 3x + 4 = 0

3x = -4

x = -4/3

Vậy x = -4/3; x = 2

e) x³ - 6x² + 12x - 8 = 0

(x - 2)³ = 0

x - 2 = 0

x = 2

f) x³ - 7x - 6 = 0

x³ + 2x² - 2x² - 4x - 3x - 6 = 0

(x³ + 2x²) - (2x² + 4x) - (3x + 6) = 0

x²(x + 2) - 2x(x + 2) - 3(x + 2) = 0

(x + 2)(x² - 2x - 3) = 0

(x + 2)(x² + x - 3x - 3) = 0

(x + 2)[(x² + x) - (3x + 3)] = 0

(x + 2)[x(x + 1) - 3(x + 1)] = 0

(x + 2)(x + 1)(x - 3) = 0

x + 2 = 0 hoặc x + 1 = 0 hoặc x - 3 = 0

*) x + 2 = 0

x = -2

*) x + 1 = 0

x = -1

*) x - 3 = 0

x = 3

Vậy x = -1; x = -1; x = 3

a,x\(^3\)-64=0
   x\(^3\)     =64
  =>x=3
b,x\(^3\)-4x\(^2\)=-4x

   x\(^3\)-4x\(^2\)+4x=0
   x(x\(^2\)-4x+4)=0
   x(x-2)\(^2\)=)
TH1:x=0
TH2:x-2=0
     =>x=2

c,x\(^2\)-16-(x-4)=0

   (x+4)(x-4)-(x-4)=0
   (x-4)(x+4-1)=0

   (x-4)(x+3)=0
TH1:x-4=0
     =>x=4

TH2:x+3=0

    =>x=-3

d,(2x+1).2=3+x
    4x+2-3-x=0
    3x-1=0
     x=\(\dfrac{1}{3}\)

e,x\(^3\)-6x\(^2\)+12x-8=0
   (x-2)\(^3\)=0
=>x-2=0
=>x=2

f,x\(^3\)-7x+6=0
  x\(^3\)-x-6x+6=0
  x(x\(^2\)-1)-6(x-1)=0
  x(x+1)(x-1)-6(x-1)=0
  (x-1)(x\(^2\)+x-6)=0
TH1:x-1=0

  =>x=1
TH2:x\(^2\)+x-6=0

       x\(^2\)+3x-2x-6=0

       x(x+3)-2(x+3)=0

       (x+3)(x-2)=0

=>x+3=0                =>x-2=0

+>x=-3                   =>x=2

d,(2x+1)\(^2\)=(3+x)\(^2\)

4x\(^2\)+4x+1-9-6x-x\(^2\)=0
3x\(^2\)-2x-8=0
3x\(^2\)-6x+4x-8=0
3x(x-2)+4(x-2)=0

(3x+4)(x-2)=0

TH1:3x+4=0                     TH2:x-2=0
=>x=\(\dfrac{-4}{3}\)                              =>x=2

Bài 1: 

b: \(x^3-4x^2+7x-6=0\)

\(\Leftrightarrow x^3-2x^2-2x^2+4x+3x-6=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2-2x+3\right)=0\)

=>x-2=0

hay x=2

c: \(2x^3+7x^2+7x+2=0\)

\(\Leftrightarrow2\left(x+1\right)\left(x^2-x+1\right)+7x\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x^2-2x+2+7x\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x^2+5x+2\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x^2+4x+x+2\right)=0\)

=>(x+1)(x+2)(2x+1)=0

hay \(x\in\left\{-1;-2;-\dfrac{1}{2}\right\}\)

d: \(2x^3-9x+2=0\)

\(\Leftrightarrow2x^3-4x^2+4x^2-8x-x+2=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x^2+4x-1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x-\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+1-\dfrac{3}{2}\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+1+\dfrac{\sqrt{6}}{2}\right)\left(x+1-\dfrac{\sqrt{6}}{2}\right)=0\)

hay \(x\in\left\{2;-1-\dfrac{\sqrt{6}}{2};-1+\dfrac{\sqrt{6}}{2}\right\}\)

Bài 1:

a,\(\left|x\right|>0\)

\(\Rightarrow\left\{{}\begin{matrix}x>0\\x< 0\\x=0\end{matrix}\right.\)

b,\(\left|x\right|< 0\)

⇒Ko tìm được x.

Bài 2:

a: =7/30

b: =479/330

c: =-1067/450

f: =-8263/19980

a) x.(x-1)=0

\(\Rightarrow\)x=0 hoặc x-1=0

\(\Rightarrow\)x=0+1

\(\Rightarrow\)x=1

vậy x=1 hoặc x=0

b) -x.(x+3)=0

\(\Rightarrow\)-x = 0 hoặc x+3 = 0

\(\Rightarrow\)x= 0-3

\(\Rightarrow\)x=-3

vậy x=0 hoặc x=-3

c) (2x-4).(x+2)=0

(2x-4)= 0

2x=0+4

2x=4

x=4:2

x=2

hoặc (x+2)=0

x= 0-2

x=-2

vậy x=2 hoặc x=-2

d) (3-x).|x+5|=0

3-x = 0

x= 3-0

x=3

hoặc |x+5|=0

x+ 5=0

x=0-5

x=-5

vậy x=3 hoặc x=-5

e) (|x|+1).( 4-2x) = 0

(|x|+1) =0

|x|= 0-1

|x|=-1

hoặc( 4-2x) = 0

2x=4-0

2x=4

x=4:2

x=2

g) x²+5x=0

x²=0

x=0

hoặc 5x=0

x= 0: 5

x=0

vậy x=0

2)

a) (x+3).(y-5)= 7

(x+3)và (y-5)\(\in\)Ư(7)=\(\left\{1;-1;7;-7\right\}\)

b) xy + 3x - 2y= 11

x( y+3) -2y=11

x(y-3)- 2( y+3) +6 = 11

( y+3) ( x-2) = 5

vì x,y thuộc Z \(\Leftrightarrow\)y+3 và x-2 \(\in\)Z

do đó y+3 và x-2 \(\in\)Ư ( 5)= \(\left\{1;5;-1;-5\right\}\)

\(\in\)\(\in\)

c) xy + 3x - 7y= 21

x( y+3) -7y= 21

x( y+3) - 7( y+3)+21= 21

(y+3)( x-7) =0

Lần sau ghi tách ra tí bạn ơi ;v

--------------------------------

1. a) \(5x-20y=5\left(x-4y\right)\)

b) \(5\left(x-1\right)-3x\left(x-1\right)=\left(x-1\right)\left(5-3x\right)\)

c) \(x\left(x+1\right)-5x-5=x\left(x+1\right)-5\left(x+1\right)\)

\(=\left(x+1\right)\left(x-5\right)\)

d) \(\left(x+y\right)^2-\left(x-y\right)^2\)

\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)

\(=4xy\)

e) \(\left(3x+1\right)^2-\left(x+1\right)^2\)

\(=\left(3x+1+x+1\right)\left(3x+1-x-1\right)\)

\(=2x\left(4x+2\right)\)

2. a) \(x+5x^2=0\)

\(\Leftrightarrow x\left(1+5x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\1+5x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{5}\end{matrix}\right.\)

Vậy...

b) \(x+1=\left(x+1\right)^2\)

\(\Leftrightarrow x+1-x^2-2x-1=0\)

\(\Leftrightarrow-x^2-x=0\)

\(\Leftrightarrow-x\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-x=0\\x+1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Vậy...

c) \(x^3+x=0\)

\(\Leftrightarrow x\left(x^2+1\right)=0\)

Vì \(x^2+1>0\Rightarrow x=0\)

Vậy...

d) \(x^3-0,25x=0\)

\(\Leftrightarrow x\left(x^2-0,25\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-0,25=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm0,5\end{matrix}\right.\)

Vậy..

e) \(x^2-10x=-25\)

\(\Leftrightarrow x^2-10x+25=0\)

\(\Leftrightarrow\left(x-5\right)^2=0\)

\(\Rightarrow x-5=0\)

\(\Rightarrow x=5\)

Vậy...

Bài 2 :

a, x = \(\dfrac{-3}{-11}\) => x =\(\dfrac{3}{11}\)

=>| x | = \(\dfrac{3}{11}\)

=> x= \(\dfrac{3}{11}\) hoặc x = \(\dfrac{-3}{11}\)

Bài 3 :

a, | 4.(x-1)| =12

=> 4.(x-1)=12 hoặc 4.(x-1)=-12

\(\left[{}\begin{matrix}4.\left(x-1\right)=12\\4.\left(x-1\right)=-12\end{matrix}\right.=>\left[{}\begin{matrix}4x-4=12\\4x-4=-12\end{matrix}\right.=>\left[{}\begin{matrix}4x=16\\4x=-8\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)

Vậy x = 4 hoặc x = -2

b,|2x+1|-5 =10

|2x+1|=15

=2x+1=15 hoặc 2x+=-15

+) 2x+1=15 = > 2x = 14 = > x =7

+)2x+1=-15 => 2x= -16 => x = -8

Vậy x=7 hoặc x = -8

c,|2,5-x|-1,3=0

|2,5-x|= 1,3

=>2,5 -x = 1,3 hoặc 2,5 - x = -1,3

+)2,5 - x = 1,3 => x = 1,2

+)2,5-x = -1,3 => x=3,8

Vậy x = 1,2 hoặc x = 3,8

d,-|1,4 - x | - 2 = 0

-|1,4-x|=2

=> -1,4+x = 2 hoặc -1,4+x = -2

+) -1,4+x= 2 => x = 3,4

+)-1,4+x= -2 => x = 0,6

Vậy x = 3,4 hoặc x = 0 ,6

e,| x - 2 | = x

=> x -2 = x hoặc x - 2 = -x

+) x- 2 = x => x-x = -2 => 0 = -2 ( vô lí )

+) x -2 = -x => x+x=2 => 2x =2 => x= 1

Vậy x = 1

f, 2.|2x-3| = \(\dfrac{1}{2}\)

=> |2x-3|= \(\dfrac{1}{4}\)

=>2x-3=\(\dfrac{1}{4}\) hoặc 2x-3=\(\dfrac{-1}{4}\)

+) 2x - 3 = \(\dfrac{1}{4}\)=> 2x= \(\dfrac{13}{4}\)=> x = \(\dfrac{13}{8}\)

+) 2x - 3 = \(\dfrac{-1}{4}\)=> 2x=\(\dfrac{11}{4}\)=> x = \(\dfrac{11}{8}\)

Vậy x=\(\dfrac{13}{8}\) hoặc x=\(\dfrac{11}{8}\)