1: \(A=\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}-1+\sqrt{a}+1}{a-1}\)
\(=\dfrac{1}{\sqrt{a}-1}\cdot\dfrac{a-1}{2\sqrt{a}}=\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)
P=\(A+\dfrac{1}{4}=\dfrac{\sqrt{a}+1}{2\sqrt{a}}+\dfrac{1}{4}=\dfrac{2\sqrt{a}+2+\sqrt{a}}{4\sqrt{a}}=\dfrac{3\sqrt{a}+2}{4\sqrt{a}}\)
Để P nguyên thì 3căn a+2 chia hết cho 4căn a
=>12căn a+8 chia hết cho 4 căn a
=>4căn a=4 hoặc 4căn a=8
=>a=4
2; \(Q=\dfrac{a^4-2a^3+2a^2-2a^2+4a-2-6a+4}{a^2-2a+2}\)
\(=a^2-2+\dfrac{-6a+4}{a^2-2a+2}\)
\(=3-2\sqrt{2}-2+\dfrac{-6\left(1-\sqrt{2}\right)+4}{3-2\sqrt{2}-2\left(1-\sqrt{2}\right)+2}\)
\(=1-2\sqrt{2}+\dfrac{-6+6\sqrt{2}+4}{5-2\sqrt{2}-2+2\sqrt{2}}\)
=1/3
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