Question

cho biểu thức

\(Q=\left[\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{x+y}\cdot\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\right]\div\dfrac{x^3+y^3}{x^2y^2}\)

a, rút gọn Q

b, tính Q biết \(x=1,y=2\)

Answer 1

Lời giải:

a) Ta có:

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

\(=\left[\frac{x^2+y^2}{x^2y^2}+\frac{2}{x+y}.\frac{x+y}{xy}\right].\frac{x^2y^2}{x^3+y^3}\)

\(=\left[\frac{x^2+y^2}{x^2y^2}+\frac{2}{xy}\right].\frac{x^2y^2}{x^3+y^3}\)

\(=\frac{x^2+y^2}{x^2y^2}.\frac{x^2y^2}{x^3+y^3}+\frac{2x^2y^2}{xy(x^3+y^3)}\)

\(=\frac{x^2+y^2}{x^3+y^3}+\frac{2xy}{x^3+y^3}=\frac{x^2+y^2+2xy}{x^3+y^3}\)

\(=\frac{(x+y)^2}{x^3+y^3}=\frac{(x+y)^3}{(x+y)(x^2-xy+y^2)}=\frac{x+y}{x^2-xy+y^2}\)

b)

Khi \(x=1,y=2\Rightarrow Q=\frac{1+2}{1^2-1.2+2^2}=1\)