Question

Cho a,b là 2 số thực khác 0. thỏa mãn a-b=ab. Tính A=\(\frac{a}{b}+\frac{b}{a}-ab\)

Answer 1

\(a-b=ab\Leftrightarrow\left(a-b\right)^2=\left(ab\right)^2\)

\(\Leftrightarrow a^2-2ab+b^2=a^2b^2\)

\(\Leftrightarrow a^2-2ab+b^2-a^2b^2=0\)

\(\frac{a}{b}+\frac{b}{a}-ab\)

\(\Leftrightarrow\left(\frac{a^2}{ab}+\frac{b^2}{ab}-\frac{a^2b^2}{ab}\right)\Leftrightarrow\left(\frac{a^2+b^2-a^2b^2}{ab}\right)\)

Mà \(a^2+b^2=a^2-2ab+b^2+2ab\)

Vậy: \(\frac{a^2+b^2-a^2b^2}{ab}=\frac{a^2-2ab+b^2-a^2b^2+2ab}{ab}=\frac{2ab}{ab}=2\)