Question

chứng minh  \(\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}\)=\(\sqrt{a+\sqrt{b}}\)

Nếu cho a,b > 0 và \(a^2\)-b>0

Answer 1

\(VP^2=\frac{a+\sqrt{a^2-b}}{2}+2\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\sqrt{\frac{a-\sqrt{a^2-b}}{2}}+\frac{a-\sqrt{a^2-b}}{2}\)

\(=\frac{a+\sqrt{a^2+b}+a-\sqrt{a^2+b}}{2}+2\sqrt{\frac{\left(a+\sqrt{a^2-b}\right)\left(a-\sqrt{a^2-b}\right)}{4}}\)

\(=\frac{2a}{2}+2.\frac{\sqrt{a^2-\left(a^2-b\right)}}{\sqrt{4}}=a+2.\frac{\sqrt{a^2-a^2+b}}{2}=a+\sqrt{b}=VT^2\)

\(\Rightarrow VT=VP\)

Vậy ................

Answer 2

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