Question

Biết xy=-1, tính giá trị biểu thức: P=\(\frac{1}{y^2-xy}+\frac{1}{x^2-xy}\)

Answer 1

Ta có: xy=-1

\(P=\frac{1}{y^2-xy}+\frac{1}{x^2-xy}=\frac{1}{y^2-\left(-1\right)}+\frac{1}{x^2-\left(-1\right)}=\frac{1}{y^2+1}+\frac{1}{x^2+1}=\frac{x^2+1}{y^2+2+x^2}+\frac{y^2+1}{y^2+2+x^2}=\frac{y^2+2+x^2}{y^2+2+x^2}=1\)Vậy P=1

Answer 2

Ta có:xy=-1

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

\(=\frac{x^2+1}{y^2+2+x^2}+\frac{y^2+1}{y^2+2+x^2}=\frac{x^2+2+y^2}{y^2+2+x^2}=1\)

Vậy P=1

Answer 3