\(=x^2+2\cdot x\cdot2y+\left(2y\right)^2=\left(x+2y\right)^2\)

\(a,x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3-y^3\right)\left(x^3+y^3\right).\)

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

\(b,9x^2+y^2+6xy=\left(3x\right)^2+2.3x.y+y^2=\left(3x+y\right)^2\)

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

a/ x² + 6x + 9 = (x + 3)² = (x + 3)(x + 3)

b/ 10x - 25 - x² = -x² + 10x - 25 = -(x² -10x + 25) = -(x - 5)² = -(x - 5)(x - 5)

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

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

= 2x(4x + 2)

Em áp dụng hđt số 3 trong sgk nhé.

(3x+1+x+1) (3x+1-x-1) = (4x+2) 2x

                                      = 4x(2x+1)

\(a,8x^3+12x^2y+6xy^2+y^3=\left(2x\right)^3+3.\left(2x\right)^2.y+3.2x.y^2+y^3=\left(2x+y\right)^3\)

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

\(c,4x^2-25=\left(2x\right)^2-5^2=\left(2x-5\right)\left(2x+5\right)\)

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

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

10x - 25 - x²

= x²- 10x - 25

= - ( x² +10x +25)

= -(x² + 2.x.5+5² )

= - (x+5 )²

\(\Leftrightarrow x^2-10x+25=0\\ \Leftrightarrow\left(x-5\right)^2=0\\ \Leftrightarrow x=5\)

\(x^2-10x+25=0\)

\(x^2-10x+5^2=0\)

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

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) 

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

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

      x² - 10x + 25

    = x² -2.x.5 +5²

    =( x-5)²