1, \(x^3+4x^2+4x=0\Leftrightarrow x\left(x^2+4x+4\right)=0\)

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

2, \(\left(x+3\right)^2-4=0\Leftrightarrow\left(x+3-2\right)\left(x+3+2\right)=0\)

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

3, \(x^4-9x^2=0\Leftrightarrow x^2\left(x^2-9\right)=0\)

\(\Leftrightarrow x^2\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=0;\pm3\)

4, \(x^2-6x+9=81\Leftrightarrow\left(x-3\right)^2=9^2\)

\(\Leftrightarrow\left(x-3-9\right)\left(x-3+9\right)=0\Leftrightarrow\left(x-12\right)\left(x+6\right)=0\Leftrightarrow x=-6;x=12\)

5, em xem lại đề nhé

à lag tý @@

5, \(x^3+6x^2+9x-4x=0\Leftrightarrow x^3+6x^2+5x=0\)

\(\Leftrightarrow x\left(x^2+6x+5\right)=0\Leftrightarrow x\left(x^2+x+5x+5\right)=0\)

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

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 #

a) \(x^2-\frac{1}{49}=0\)

<=> \(\left(x-\frac{1}{7}\right)\left(x+\frac{1}{7}\right)=0\)

<=> \(\orbr{\begin{cases}x-\frac{1}{7}=0\\x+\frac{1}{7}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{7}\\x=-\frac{1}{7}\end{cases}}\)

Vậy x = \(\pm\frac{1}{7}\)

b) \(64-\frac{1}{4}x^2=0\)

<=> \(\left(8-\frac{1}{2}x\right)\left(8+\frac{1}{2}x\right)=0\)

<=> \(\orbr{\begin{cases}8-\frac{1}{2}x=0\\8+\frac{1}{2}x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=16\\x=-16\end{cases}}\)

Vậy \(x=\pm16\)

c) 9x² + 12x + 4 = 0

<=> (3x + 2)² = 0

<=> 3x + 2 = 0 

<=> x = -2/3

Vậy x = -2/3

e) \(x^2+\frac{1}{4}=x\) 

<=> \(x^2-x+\frac{1}{4}=0\)

<=> \(\left(x-\frac{1}{2}\right)^2=0\)

<=> \(x=\frac{1}{2}\)

Vậy \(x=\frac{1}{2}\)

d, sửa đề : \(x^2+4=4x\Leftrightarrow x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)

i, \(4-\frac{12}{x}+\frac{9}{x^2}=0\)ĐK : \(x\ne0\)

Vì \(x\ne0\)Nhân 2 vế với \(x^2\)phương trình có dạng 

\(4x^2-12x+9=0\Leftrightarrow\left(2x-3\right)^2=0\Leftrightarrow x=\frac{3}{2}\)

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

a/

\(x^3-4x^2-\left(x-4\right)=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=-1\end{matrix}\right.\)

b/

\(x^5-9x=0\)

\(\Leftrightarrow x\left(x^4-9\right)=x\left(x^2-3\right)\left(x^2+3\right)=0\)

\(\Leftrightarrow x\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=0\)

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

c/

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\pm\sqrt{2}\end{matrix}\right.\)

Sos

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

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

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

4, \(x^3-8+2x^2-4x=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)+2x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)^2=0\Leftrightarrow x=\pm2\)

5, \(x^2\left(x-3\right)+18-6x=0\Leftrightarrow x^2\left(x-3\right)-6\left(x-3\right)=0\)

\(\Leftrightarrow\left(x^2-6\right)\left(x-3\right)=0\Leftrightarrow x=\pm\sqrt{6};x=3\)

tìm x

3, ( x - 2 ) mũ 2 - 5( 2 - x ) = 0

x=-3, x=2

4, ( x mũ 3 - 8 ) + 2x mũ 2 - 4x = 0

x= 2 , x= -2 

5, x mũ 2 ( x - 3 ) + 18 - 6x = 0

x=-căn bậc hai(6), x=căn bậc hai(6), x=3

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

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

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

<=> \(\orbr{\begin{cases}x-1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

x + x² - x³ - x⁴ = 0

<=> ( x + x² ) - ( x³ - x⁴ ) = 0

<=> x( x + 1 ) - x³( x + 1 ) = 0

<=> ( x + 1 )( x - x³ ) = 0

<=> ( x + 1 )x( 1 - x² ) = 0

<=> \(\hept{\begin{cases}x+1=0\\x=0\\1-x^2=0\end{cases}}\)( thay bằng dấu hoặc hộ mình nhé '-' )

<=> \(\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

a) \(x\left(x-1\right)+2x-2=0\)

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

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

\(\orbr{\begin{cases}x+2=0\\x-1=0\end{cases}\orbr{\begin{cases}x=-2\\x=1\end{cases}}}\)

b) \(x+x^2-x^3-x^4=0\)

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

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

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

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

\(\hept{\begin{cases}x=0\\x+1^2=0\\x-1=0\end{cases}\hept{\begin{cases}x=0\\x=-1\\x=1\end{cases}}}\)

\(\left(9^{30}-27^{19}\right):3^{57}+\left(125^9-25^{12}\right):5^{24}\)

\(=\left(3^{60}-3^{57}\right):3^{57}+\left(5^{27}-5^{24}\right):5^{24}\)

\(=3^{57}\left(3^3-1\right):3^{57}+5^{24}\left(5^3-1\right):5^{24}\)

\(=3^3-1+5^3-1\)

\(=27-1+125-1\)

\(=150\)

2 )

\(x^2-25-\left(x+5\right)=0\)

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

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

\(\Leftrightarrow\left(x+5\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)

Vậy ...

b )

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

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

\(\Leftrightarrow2-4x=0\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy ...

c )

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\\x^2+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=1\\x^2=-4\left(L\right)\end{matrix}\right.\)

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

Vậy ...