Question

Cho biểu thức \(P=\left(\dfrac{1}{a-1}+\dfrac{3\sqrt{a}+5}{a\sqrt{a}-a-\sqrt{a}+1}\right).\left(\dfrac{\left(\sqrt{a}+1\right)^2}{4\sqrt{a}}-1\right)với:a>0,a\ne1\)
a, Rút gọn P
b, Đặt \(Q=\left(a-\sqrt{a}+1\right)P\). Chứng minh Q>1

Answer 1

a: \(P=\left(\dfrac{1}{a-1}+\dfrac{3\sqrt{a}+5}{a\sqrt{a}-a-\sqrt{a}+1}\right)\cdot\left(\dfrac{\left(\sqrt{a}+1\right)^2}{4\sqrt{a}}-1\right)\)

\(=\left(\dfrac{1}{a-1}+\dfrac{3\sqrt{a}+5}{\left(a-1\right)\left(\sqrt{a}-1\right)}\right)\cdot\dfrac{\left(\sqrt{a}+1\right)^2-4\sqrt{a}}{4\sqrt{a}}\)

\(=\dfrac{\sqrt{a}-1+3\sqrt{a}+5}{\left(\sqrt{a}-1\right)\left(a-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}\)

\(=\dfrac{4\sqrt{a}+4}{\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}-1\right)^2}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\dfrac{\sqrt{a}+1}{\sqrt{a}}\)