1, \(5x^2+10x+5-5y^2=5\left(x^2+2x+1-y^2\right)\)

\(=5\left[\left(x+1\right)^2-y^2\right]=5\left(x+1-y\right)\left(x+1+y\right)\)

2, \(3x^3+6x^2+3x-12xy^2=3x\left(x^2+2x+1-4y^2\right)\)

\(=3x\left[\left(x+1\right)^2-4y^2\right]=3x\left(x+1-2y\right)\left(x+1+2y\right)\)

Trả lời:

2, 8x²y - 8y = 8y ( x² - 1 ) = 8y ( x - 1 )( x + 1 )

3, 24x³ - 3 = 3 ( 8x³ - 1 ) = 3 ( 2x - 1 )( 4x² + 2x + 1 )

4, 4x² - 4y² = ( 2x )² - ( 2y )² = ( 2x - 2y )( 2x + 2y )

5, x⁵ + 27x² = x² ( x³ + 27 ) = x² ( x + 3 )( x² - 3x + 9 )

\(x^2-5x-36=0\)

\(\Leftrightarrow x^2+4x-9x-36=0\)

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

\(\Leftrightarrow\left(x-9\right)\left(x+4\right)=0\Leftrightarrow x=-4;x=9\)

\(x^2-5x-36=0\)

\(x^2+4x-9x-36=0\)

\(x\left(x-9\right)+4\left(x-9\right)=0\)

\(\left(x-9\right)\left(x+4\right)=0\)

\(\orbr{\begin{cases}x-9=0\\x+4=0\end{cases}\orbr{\begin{cases}x=9\left(TM\right)\\x=-4\left(TM\right)\end{cases}}}\)

a, \(x^2-2.\frac{1}{3}x+\frac{1}{9}=\left(x-\frac{1}{3}\right)^2\)

Thay x = 9 vào ta được : \(=\left(9-\frac{1}{3}\right)^2=\left(\frac{26}{3}\right)^2=\frac{676}{9}\)

\(x^2-\frac{2}{3}x+\frac{1}{9}\)

Thay \(x=9\) và ta được:

\(9^2-\frac{2}{3}9+\frac{1}{9}\)\(=81-6+\frac{1}{9}\)\(=\frac{676}{9}\)

Trả lời:

1, x³ - x² - 4

= x³ - x² - 4 + 2x - 2x

= x³ - 2x² + x² - 4 + 2x - 2x

= ( x³ + x² + 2x ) - ( 2x² + 2x + 4 )

= x ( x² + x + 2 ) - 2 ( x² + x + 2 )

= ( x - 2 )( x² + x + 2 )

2, x³ - 2x - 4 

= x³ - 2x - 4 + 2x² - 2x²

= x³ - 4x + 2x - 4 + 2x² - 2x²

= ( x³ + 2x² + 2x ) - ( 2x² + 4x + 4 )

= x ( x² + 2x + 2 ) - 2 ( x² + 2x + 2 )

= ( x - 2 )( x² + 2x + 2 )

giúp mình đi gấp lm rùi

(dak bủh bủn lmao lmao ) đoạn này dành cho Đào Phúc Khánh

a,676 phần 9

b,Ko bt,sao lại có y?