Question

[Hoàng Thị Thu Hà]

Cho \(a,b,c>0\)thỏa mãn \(\sqrt{a}+\sqrt{b}\ne\sqrt{c}\)và \(ab=\left(\sqrt{a}+\sqrt{b}-\sqrt{c}\right)^2\)

CMR \(\frac{a+\left(\sqrt{a}-\sqrt{c}\right)^2}{b+\left(\sqrt{b}-\sqrt{c}\right)^2}=\frac{\sqrt{a}-\sqrt{c}}{\sqrt{b}-\sqrt{c}}\)

Answer 1

[alibaba nguyễn]

Đề viết mệt quá nên thay \(\sqrt{a}=a;\sqrt{b}=b;\sqrt{c}=c\) viết lại đề tiện thể sửa đề luôn.

\(a^2+b^2=\left(a+b-c\right)^2\)

Chứng minh:

\(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a-c}{b-c}\)

Ta có: \(a^2+b^2=\left(a+b-c\right)^2\)

\(\Leftrightarrow c^2-2ac-2bc+2ab=0\)

\(\Leftrightarrow a=\frac{c^2-2bc}{2c-2b}\)

Thế vô bài toán ta được

\(VT=\frac{\left(\frac{c^2-2bc}{2c-2b}\right)^2+\left(\frac{c^2-2bc}{2c-2b}-c\right)^2}{b^2+\left(b-c\right)^2}\)

\(=\frac{\left(\frac{c^2-2bc}{2c-2b}\right)^2+\left(\frac{c^2-2bc}{2c-2b}-c\right)^2}{b^2+\left(b-c\right)^2}\)

\(=\frac{\left(\frac{c^2-2bc}{2c-2b}\right)^2+\left(c^2\right)^2}{b^2+\left(b-c\right)^2}=\frac{2c^2\left(2b^2+c^2-2bc\right)}{\left(2b^2+c^2-2bc\right)4\left(c-b\right)^2}=\frac{c^2}{2\left(c-b\right)^2}\)

Ta lại có: 

\(VP=\frac{\frac{c^2-2bc}{2c-2b}-c}{b-c}=\frac{-c^2}{-2\left(c-b\right)^2}=\frac{c^2}{2\left(c-b\right)^2}\)

\(\Rightarrow\)ĐOCM