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

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

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

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

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

\(4x+4=-4\)

\(4x=-4-4\)

\(4x=-8\)

\(x=-8\div4\)

\(x=-2\)

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

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

\(\Leftrightarrow x^2+2x+1+25-x^2=0\)

\(\Leftrightarrow2x+26=0\)

\(\Leftrightarrow2x=-26\)

\(\Leftrightarrow x=-13\)

Vậy \(x=-13\)

1) x² - 4xy + 4y² = ( x - 2y )²

2) x² + 10x - 9y² + 25 = ( x² + 10x + 25 ) - 9y² = ( x + 5 )² - ( 3y )² = ( x - 3y + 5 )( x + 3y + 5 )

3) x³ - 8x² + 4x - 32 = ( x³ - 8x² ) + ( 4x - 32 ) = x²( x - 8 ) + 4( x - 8 ) = ( x - 8 )( x² + 4 )

4) 3x² + 4x - 15 = 3x² + 9x - 5x - 15 = ( 3x² + 9x ) - ( 5x + 15 ) = 3x( x + 3 ) - 5( x + 3 ) = ( x + 3 )( 3x - 5 )

a) \(x^2-4xy+4y^2\)

\(x^2-4xy+\left(2y\right)^2\)

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

b) \(x^2+10x-9y^2+25\)

\(=x^2+10x+5^2-\left(3y\right)^2\)

\(=\left(x+5\right)^2-\left(3y\right)^2\)

\(=\left(x+5+3y\right)\left(x+5-3y\right)\)

c) \(x^3-8x^2+4x-32\)

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

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

d) sai de

x( x + 1 ) - x( x - 3 ) = 20

⇔ x² + x - x² + 3x = 20

⇔ 4x = 20

⇔ x = 5

    Bài làm :

Ta có :

x(x+1)-x(x-3)=20

<=> x(x+1-x+3)=20

<=> 4x=20

<=> x=5

Vậy x=5

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

a) x³ - 16x = 0

=> x(x² - 16) = 0

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

=> x = 0 hoặc x - 4 = 0 hoặc x + 4 = 0

=> x = 0 hoặc x = 4 hoặc x = -4

b) x⁴ - 2x³ + 10x² - 20x = 0

=> x³(x - 2) + 10x(x - 2) = 0

=> (x - 2)(x³ + 10x) = 0

=> x(x - 2)(x² + 10) = 0 (1)

Vì x² + 10 \(x^2+10\ge10>0\forall x\)

Khi đó (1) <=> \(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Vậy \(x\in\left\{0;2\right\}\)là giá trị cần tìm

a) x³ - 16x = 0

⇔ x( x² - 16 ) = 0

⇔ x( x - 4 )( x + 4 ) = 0

⇔ x = 0 hoặc x - 4 = 0 hoặc x + 4 = 0

⇔ x = 0 hoặc x = ±4

b) x⁴ - 2x³ + 10x² - 20x = 0

⇔ x( x³ - 2x² + 10x - 20 ) = 0

⇔ x[ ( x³ - 2x² ) + ( 10x - 20 ) ] = 0

⇔ x[ x²( x - 2 ) + 10( x - 2 ) ] = 0

⇔ x( x - 2 )( x² + 10 ) = 0

⇔ x = 0 hoặc x - 2 = 0 hoặc x² + 10 = 0

⇔ x = 0 hoặc x = 2 ( x² + 10 ≥ 10 > 0 ∀ x )