Question

khai trien cac bieu thuc sau

(2x+y)^2

(x-y/2)^2

(x^2+y/2)(x^2-y/2)

(x-2y)^2(x+2y)^2

(x+y)^2

(x-2y)^2

(xy^2+1)(xy^2-1)

(x+y)^2-4(x-y)+4

Answer 1

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

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

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

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

\(=x^4-8x^2y^2+16y^4\)

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

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

\(\left(xy^2+1\right)\left(xy^2-1\right)=x^2y^4-1\)

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

Answer 2

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

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

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

\(\left(x-2y\right)^2\left(x+2y\right)^2=x^4-8x^2y^2+16y^4\)

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

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

\(\left(xy^2+1\right)\left(xy^2-1\right)=x^2y^4-xy^2+xy^2-1=x^2y^4-1\)

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