\(\frac{1}{4}x^6-0,01y^2=\left(\frac{1}{2}x^3\right)^2-\left(0,1y\right)^2\)

                            \(=\left(\frac{1}{2}x^3-0,1y\right).\left(\frac{1}{2}x^3+0,1y\right)\)

Vậy \(\frac{1}{4}x^6-0,01y^2\)\(=\left(\frac{1}{2}x^3-0,1y\right).\left(\frac{1}{2}x^3+0,1y\right)\)

Tham khảo nhé ~

\(\frac{1}{4}x^6-0.01y^2\)

\(=\left(\frac{1}{2}x^3\right)^2-\left(0.1y\right)^2\)

\(=\left(\frac{1}{2}x^3-0.1y\right)\left(\frac{1}{2}x^3+0.1y\right)\)

Mong lần này không sai nữa ......

\(\frac{4}{9}x^2-\frac{2}{3}xy+\frac{1}{4}y^2\)

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

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

p/s: chúc bạn học tốt

Ta có :

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

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

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

\(=4y^2-9x^4\)

\(a^3+1+3a+3a^2\)

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

đây là HĐT:  \(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3\)

p/s: chúc bạn học tốt

\(27a^3-b^3+9ab^2-27a^2b\)

\(=\left(3a\right)^3-3\cdot\left(3a\right)^2b+3\cdot3a\cdot b^2-b^3\)

\(=\left(3a-b\right)^3\)

Ta có ;

\(0.008-a^3b^6\)

\(=\left(0.2\right)^3-\left(ab^2\right)^3\)

\(=\left(0.2-ab^2\right)\left(0.04+0.2ab^2+a^2b^4\right)\)

\(0,008=0,2^3,a^6b^3=\left(a^2b\right)^3\)

=> \(0,2^3-\left(a^2b\right)^3=\left(0,2-a^2b\right)\left(0,04+0,2ab+a^4b^2\right)\)

a)=x^2-4

b)=(x+2)^2

c)=(2x-1)^2

d)=x^2-5^2=(x-5)(x+5)

a, =x² - 4

b, ( x + 2)²

c, ( 2x - 1) ²

d, = (x-5)(x+5)

4: \(\left(2x+3\right)^3-1\)

\(=\left(2x+3-1\right)\left(4x^2+12x+9+2x+3+1\right)\)

\(=\left(2x+2\right)\left(4x^2+14x+13\right)\)

\(=2\left(x+1\right)\left(4x^2+14x+13\right)\)

5: \(4x^2+20xy+25y^2=\left(2x+5y\right)^2\)

6: \(x^4-64xy^3\)

\(=x\left(x^3-64y^3\right)\)

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

= \(\left(\frac{1}{3a^3}+12b^2\right)^2\)