Ta có ;

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

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

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

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

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

\(4x^2-20xy^2+25y^4=\left(2x\right)^2-2.2x.5y^2+\left(5y^2\right)^2=\left(2x-5y^2\right)^2\)

Áp dụng hằng đẳng thức: \(\left(A-B\right)^2=A^2-2AB+B^2\)

\(4x^2-20xy^2+25y^4\)

\(=\left(2x\right)^2-2\cdot2x\cdot5y^2+\left(5y\right)^2\)

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

\(4x^2-20xy^2+25y^4\)

\(=\left(2x\right)^2-2.2.5.xy^2+\left(5y\right)^2\)

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

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

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

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

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

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

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

p/s: chúc bạn học tốt

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

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

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 ......

1.b

2.d

3.a

4.c

5.d

Ta có :

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

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

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

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

\(=\left(3a-1\right)^2+2\left(3a-1\right)\left(3a+1\right)+\left(3a+1\right)^2\\ =\left(3a-1+3a+1\right)^2=\left(6a\right)^2=36a^2\)

Vô lí : VT có kết quả = 0 mà VP luôn >= 0 

dấu "=" xẩy ra <=> a=b=0