1.(x -5)^2 - 25 =0

=> (x - 5)^2 = 25

=> x - 5 = 5 hoặc x - 5 = -5

=> x = 10 hoặc x = 0

vậy_

2. (x -2)^3 =27

=> x - 2 = 3

=> x = 5

vậy_

3. 3(x -7) + 2x(x+2) = 2x^2

=> 3x - 21 + 2x^2 + 4x = 2x^2

=> 7x - 21 = 0

=> 7x = 21

=> x = 3

vậy_

4. (x^2 - 4) (x +8) =0

=> x^2 - 4 = 0 hoặc x + 8 = 0

=> x^2 = 4 hoặc x = -8

=> x = 2 hoặc x = -2 hoặc x = -8

vậy_

5. x^ 2 + 3x = 0

=> x(x + 3) = 0 

=> x = 0 hoặc x + 3 = 0

=> x = 0 hoặc x = -3

vậy_

6. 3x^3 - 3x = 0

=> 3x(x^2 - 1) = 0

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

=> x = 0 hoặc x = 1 hoặc x = -1

vậy_

7. (x +1)^2 = ( 2x +3)^2

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

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

=> x = -1 hoặc x = -2

vậy_

Bài làm

1) ( x - 5 )² - 25 = 0

<=> ( x - 5 - 5 )( x - 5 + 5 ) = 0

<=> x( x - 10 ) = 

<=> \(\orbr{\begin{cases}x=0\\x-10=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=10\end{cases}}}\)

Vậy S = { 0; 10 }

2) \(\left(x-2\right)^3=27\)

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

\(\Leftrightarrow x-2=3\)

\(\Leftrightarrow x=5\)

Vậy x = 5 là nghiệm phương trình.

3) \(3\left(x-7\right)+2x\left(x+2\right)=2x^2\)

\(\Leftrightarrow3x+2x^2+4x-2x^2=21\)

\(\Leftrightarrow7x=21\)

\(\Leftrightarrow x=\frac{21}{7}=3\)

Vậy x = 3 là nghiệm phương trình

4) \(\left(x^2-4\right)\left(x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-4=0\\x+8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=\pm2\\x=-8\end{cases}}}\)

Vậy S = { 2; -2; -8 }

5) \(x^2+3x=0\)

\(\Leftrightarrow x\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}}\)

Vậy S = { 0; -3 } 

6) \(3x^3-3x=0\)

\(\Leftrightarrow3x\left(x^2-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}}\)

Vậy S = { +1; 0 }

7) \(\left(x+1\right)^2=\left(2x+3\right)^2\)

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

\(\Leftrightarrow\left(x+1-2x-3\right)\left(x+1+2x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-x-2=0\\3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{4}{3}\end{cases}}}\)

Vậy S = { -2; -4/3 }

# Học tốt #

bn kiểm tra giúp mk đề 2 câu cuối , mk làm ko ra

Khó quá,hay quá !!!

Bài 5 : 

a, \(2x\left(x-3\right)+x-3=0\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\Leftrightarrow x=-\frac{1}{2};x=3\)

b, \(x\left(x+1\right)-x-1=0\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\Leftrightarrow x=\pm1\)

c, sửa đề  \(x^3-3x^2+x-3=0\Leftrightarrow x^2\left(x-3\right)+x-3=0\)

\(\Leftrightarrow\left(x^2+1>0\right)\left(x-3\right)=0\Leftrightarrow x=3\)

d, \(3x^2\left(2x-1\right)+1-4x^2=0\Leftrightarrow3x^2\left(2x-1\right)+\left(1-2x\right)\left(1+2x\right)=0\)

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

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

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

x=1 nha

b \(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)

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

hay \(x\in\left\{0;2\right\}\)

c: \(\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)

=>(x-8)(3x+2)=0

=>x=8 hoặc x=-2/3

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

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

=>x=2 hoặc x=1

e: \(\Leftrightarrow x\left(x^2-11x+30\right)=0\)

=>x(x-5)(x-6)=0

hay \(x\in\left\{0;5;6\right\}\)

b: \(\Leftrightarrow x\left(x^3-2x^2+10x-20\right)=0\)

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

hay \(x\in\left\{0;2\right\}\)

c: \(\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)

=>(x-8)(3x+2)=0

hay \(x\in\left\{8;-\dfrac{2}{3}\right\}\)

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

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

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

=>x=1 hoặc x=2

1. | x + 1| + (y + 2)² = 0
Mà (y + 2)² \(\ge\) 0
Đẳng thức khi . y + 2 \(\ge\) 0
y \(\ge\) - 2
. x + 1 = 0
. x = -1

Tìm x:

1. \(25x^2-20x+4=0\)

⇔ \(\left(5x-2\right)^2=0\)

⇔ \(5x-2=0\)

⇔ \(5x=2\)

⇔ \(x=\dfrac{2}{5}\)

⇒ S = \(\left\{\dfrac{2}{5}\right\}\)

2. \(\left(2x-3\right)^2-\left(2x+1\right).\left(2x-1\right)=0\)

⇔ \(4x^2-12x+9-\left(4x^2-1\right)=0\)

⇔ \(4x^2-12x+9-4x^2+1=0\)

⇔ \(-12x+10=0\)

⇔ \(-12x=-10\)

⇔ \(x=\dfrac{5}{6}\)

⇒ S \(=\left\{\dfrac{5}{6}\right\}\)

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

⇔ \(\dfrac{1}{4}x^2-1-\left(\dfrac{1}{4}x^2-x+1\right)=0\)

⇔ \(\dfrac{1}{4}x^2-1-\dfrac{1}{4}x^2+x-1=0\)

⇔ \(-2+x=0\)

⇔ \(x=2\)

⇒ S \(=\left\{2\right\}\)

4. \(\left(2x-3\right)^2+\left(2x+5\right)^2=8\left(x+1\right)^2\)

⇔ \(4x^2-12x+9+4x^2+20x+25=8\left(x^2+2x+1\right)\)

⇔ \(8x^2+8x+34=8x^2+16x+8\)

⇔ \(8x+34=16x+8\)

⇔ \(8x-16x=8-34\)

⇔ \(-8x=-26\)

⇔ \(x=\dfrac{13}{4}\)

⇒ S \(=\left\{\dfrac{13}{4}\right\}\)

5.\(4x^2+12x-7=0\)

⇔ \(4x^2+14x-2x-7=0\)

⇔ \(2x\left(2x+7\right)-\left(2x+7\right)=0\)

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

⇔ \(\left[{}\begin{matrix}2x+7=0\\2x-1=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=\dfrac{-7}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

⇒ S \(=\left\{\dfrac{-7}{2};\dfrac{1}{2}\right\}\)

6. \(\dfrac{1}{4}x^2+\dfrac{2}{3}x-\dfrac{5}{9}=0\)

⇔ \(9x^2+24x-20=0\)

⇔ \(9x^2+30x-6x-20=0\)

⇔ \(3x\left(3x+10\right)-2\left(3x+10\right)=0\)

⇔ \(\left(3x+10\right)\left(3x-2\right)=0\)

⇔ \(\left[{}\begin{matrix}3x+10=0\\3x-2=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=\dfrac{-10}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)

⇒ S \(=\left\{\dfrac{-10}{3};\dfrac{2}{3}\right\}\)

7. \(24\dfrac{8}{9}-\dfrac{1}{4}x^2-\dfrac{1}{3}x=0\)

⇔ \(\dfrac{224}{9}-\dfrac{1}{4}x^2-\dfrac{1}{3}x=0\)

⇔ \(896-9x^2-12x=0\)

⇔ \(-896+9x^2+12x=0\)

⇔ \(9x^2+12x-896=0\)

⇔ \(9x^2-84x+96x-896=0\)

⇔ \(3x\left(3x-28\right)+32\left(3x-28\right)=0\)

⇔ \(\left(3x-28\right)\left(3x+32\right)=0\)

⇔ \(\left[{}\begin{matrix}3x-28=0\\3x+32=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=\dfrac{-32}{3}\end{matrix}\right.\)

⇒ S \(=\left\{\dfrac{-32}{3};\dfrac{28}{3}\right\}\)

Bài 2: 

a: \(\Leftrightarrow\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\)

=>(x+5)(x-6)=0

=>x=-5 hoặc x=6

b: \(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)

=>-4x+2=0

hay x=1/2

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

=>x=1 hoặc x=-1

chịu

Chịu 

b, x = -5/3 hoặc x = 4/3.

c, x = 0 hoặc x = 3, -3.

d, x = 0 hoặc x = 2, -2.

e, x = 1 hoặc x = \(\dfrac{-1}{2}\).

a: \(\Leftrightarrow x^2-40x+400-x^2-4x-3=-7\)

=>-44x+397=-7

=>-44x=-404

hay x=101

b: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=0\\4-3x=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{5}{3};\dfrac{4}{3}\right\}\)

c: \(\Leftrightarrow x\left(x^2-9\right)=0\)

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

hay \(x\in\left\{0;3;-3\right\}\)

d: \(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)

hay \(x\in\left\{0;2;-2\right\}\)

e: =>(2x+1)(1-x)=0

=>x=-1/2 hoặc x=1