Question

a,b,c >0 thỏa mãn b≠c ,\(\sqrt{a}\)+\(\sqrt{b}\)≠\(\sqrt{c}\) và a+b=\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2.cmr\)

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