a) \(8x^3-\frac{1}{8}\)

\(=\left(2x\right)^3-\left(\frac{1}{2}\right)^3\)

\(=\left(2x-\frac{1}{2}\right)\left(4x^2+2x+\frac{1}{4}\right)\)

b) \(\frac{1}{25}x^2-64y^2\)

\(=\left(\frac{1}{5}x\right)^2-\left(8y\right)^2\)

\(=\left(\frac{1}{5}x-8y\right)\left(\frac{1}{5}x+8y\right)\)

a) \(x^2+6x+9\)

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

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

b) \(10x-25-x^2\)

\(=-\left(x^2-10x+25\right)\)

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

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

c) \(8x^3-\frac{1}{8}\)

\(=\left(2x\right)^3-\left(\frac{1}{2}\right)^3\)

\(=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)

d) \(\frac{1}{25}x^2-64y^2\)

\(=\left(\frac{1}{5}x\right)^2-\left(8y\right)^2\)

\(=\left(\frac{1}{5}x-8y\right)\left(\frac{1}{5}x+8y\right)\)

a) \(x^2+6x+9=x^2+2.3.x+3^2\)\(=\left(x+3\right)^2\)

b)\(10x-25-x^2=-\left(x^2-10x+25\right)\)\(=-\left(x^2-2.5.x+5^2\right)=-\left(x+5\right)^2\)

c)\(8x^3-\frac{1}{8}=\left(2x\right)^3-\left(\frac{1}{2}\right)^3\)\(=\left(2x-\frac{1}{2}\right)\left(4x+x+\frac{1}{4}\right)\)

d)\(\frac{1}{25}x^2-64y^2=\left(\frac{1}{5}\right)^2-\left(8y\right)^2\)\(=\left(\frac{1}{5}-8y\right)\left(\frac{1}{5}+8y\right)\)

1) \(x^2+6x+9\)

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

2) \(10x-25-x^2\)

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

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

3) \(8x^3-\dfrac{1}{8}\)

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

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

4) \(\dfrac{1}{25}x^2-64y^2\)

\(=\left(\dfrac{1}{5}x\right)^2-\left(8y\right)^2\)

\(=\left(\dfrac{1}{5}x+8y\right)\left(\dfrac{1}{5}x-8y\right)\)

\(x^2+6x+9=\left(x+3\right)^2\)

\(10x-25-x^2=-\left(x-5\right)^2\)

\(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

a)  \(x^2\)\(+\)\(6x\)\(+\)\(9\)

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

b)  \(x^3\)\(+\)\(3x^2\)\(+\)\(3x\)\(+\)\(1\)

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

c)  \(8x^3\)\(-\)\(\frac{1}{8}\)

\(=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)

d)  \(10x\)\(-\)\(25\)\(-\)\(x^2\)

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

\(=-\left(x^2-10+25\right)\)

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

e)  \(\frac{1}{25}x^2\)\(-\)\(64y^2\)

=\(\left(\frac{1}{25}x-8y\right)\left(\frac{1}{5}x+8y\right)\)

\(\frac{2}{5}x\left(y-1\right)-\frac{2}{5}y\left(y-1\right)\)

\(=\left(y-1\right)\left[\left(\frac{2}{5}x-\frac{2}{5}y\right)\right]\)

\(=\left(y-1\right)\frac{2}{5}\left(x-y\right)\)

\(\frac{1}{25}x^2-64y^2\)

\(=\left(\frac{1}{5}x\right)^2-8^2\)

\(=\left(\frac{1}{5}x+8\right)\left(\frac{1}{5}x-8\right)\)

\(x^3+\frac{1}{27}=x^3+\left(\frac{1}{3}\right)^3\)

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

\(8x^3+12x^2y+6xy+y^3\)

\(=2^3+3.4x^2y+3.2x.y^2+y^3\)

\(=\left(2+y\right)^3\)

a) \(\frac{1}{25}x^2-64y^2=\left(\frac{1}{5}x+8y\right)\left(\frac{1}{5}x-8y\right)\)

b) \(x^3+\frac{1}{27}=\left(x+\frac{1}{3}\right)\left(x^2-\frac{1}{3}x+\frac{1}{9}\right)\)

c) \(-x^3+9x^2-27x+27\)

\(=27-x^3+9x^2-27x\)

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

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

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

\(=\left(3-x\right)\left(9-6x+x^2\right)=\left(3-x\right)\left(9-3x-3x+x^2\right)\)

\(=\left(3-x\right)\left[3\left(3-x\right)-x\left(3-x\right)\right]=\left(3-x\right)\left(3-x\right)\left(3-x\right)=\left(3-x\right)^3\)

(Nhớ k cho mình với nha!, Mình chắc chắn là mình làm đứng luôn đó! Chúc may mắn nhá!)

a/ Ta có: \(\frac{1}{25}x^2-64y^2=\left(\frac{1}{5}x\right)^2-\left(8y\right)^2=\left(\frac{1}{5}x-8y\right)\left(\frac{1}{5}x+8y\right)\)

b/ \(x^3+\frac{1}{27}=x^3+\left(\frac{1}{3}\right)^3=\left(x+\frac{1}{3}\right)\left(x^2-\frac{1}{3}x+\frac{1}{9}\right)\)

c/ Đề sai

c/ Ta có:  -x³ + 9x² - 27x + 27 = -(x³ - 9x² + 27x - 27) 

 \(=-\left[x^3-3.3.x^2+3.3^2.x-3^3\right]=-\left(x-3\right)^3\)

Bài giải:

a) x² + 6x + 9 = x² + 2 . x . 3 + 3² = (x + 3)²

b) 10x – 25 – x² = -(-10x + 25 +x²) = -(25 – 10x + x²)

= -(5² – 2 . 5 . x – x²) = -(5 – x)²

c) 8x³ - $\frac{1}{8}$ = (2x)³ – ($\frac{1}{2}$)³ = (2x - $\frac{1}{2}$)[(2x)² + 2x . $\frac{1}{2}$ + ($\frac{1}{2}$)²]

^{= (2x - $\frac{1}{2}$)(4x2 + x + $\frac{1}{4}$)}

d) $\frac{1}{25}$x² – 64y² = ${\left(\frac{1}{5}x\right)}^2$- (8y)² = ($\frac{1}{5}$x + 8y)($\frac{1}{5}$x - 8y)

a) x2 + 6x + 9 = x2 + 2.3x + 32 = (x + 3)2

b) 10x – 25 – x2 = -(x2 -10x + 25) = -(x2 -2.5x + 52)

= -(x – 5)2

c) 8x3 – 1/8= (2x)3 – ( 1/2)3 = (2x – 1/2)[(2x)2 + 2x . 1/2+ (1/2)2]

= (2x – 1/2)(4x2 + x + 1/4)

d) 1/25x2 – 64y2 = (1/5 x)2– (8y)2 = ( 1/5 x + 8y)(1/5x- 8y)

a) x² + 6x + 9 = x² + 2.3.x + 3² = (x + 3)²

b) 10x - 25 - x² = - (x² - 2.5.x + 5²) = - (x - 5)²

c) \(8x^3-\dfrac{1}{8}=\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

d) 25x² - 64y² = (5x)² - (8y)² = (5x - 8y)(5x + 8y)

\(a,x^2+6x+9=\left(x+3\right)^2\)

\(b,10x-25-x^2=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)

\(x^3+1\)

\(=x^3+1^3\)

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

______

\(8+x^3\)

\(=2^3+x^3\)

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

______

\(27x^3-64y^3\)

\(=\left(3x\right)^3-\left(4y\right)^3\)

\(=\left(3x-4y\right)\left(9x^2+12xy+16y^2\right)\)

______

\(\dfrac{x^3}{64}-\dfrac{1}{125}\)

\(=\left(\dfrac{x}{4}\right)^3-\left(\dfrac{1}{5}\right)^3\)

\(=\left(\dfrac{x}{4}-\dfrac{1}{5}\right)\left(\dfrac{x^2}{16}+\dfrac{x}{20}+\dfrac{1}{25}\right)\)

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

x^3+8=(x+2)(x^2-2x+4)

27x^3-64y^3=(3x-4y)(9x^2+12xy+16y^2)

\(\dfrac{x^3}{64}-\dfrac{1}{25}=\left(\dfrac{1}{4}x-\sqrt[3]{\dfrac{1}{5}}\right)\left(\dfrac{1}{16}x^2+\dfrac{1}{4\sqrt[3]{5}}\cdot x+\dfrac{1}{\sqrt[3]{25}}\right)\)

x³+1= (x+1)(x²-x+1)

8+x³=2³+x³=(2+x)(4-2x+x²)

27x³-64y³= (3x)³-(4y)³= (3x-4y)(9x²+12xy+16y²)

\(\dfrac{x^3}{64}\)-\(\dfrac{1}{125}\)(sửa đề)=(x/4)³-(1/5)³

=(x/4-1/5)(x/16+x/20+1/25)