Bài 15:
\(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{1-a}{\sqrt{1-a^2}-1+a}\)
\(=\dfrac{\sqrt{1+a}\left(\sqrt{1+a}+\sqrt{1-a}\right)}{1+a-1+a}+\dfrac{\left(1-a\right)\left(\sqrt{1-a^2}+1-a\right)}{1-a^2-\left(1-a\right)^2}\)
\(=\dfrac{\sqrt{1+a}\left(\sqrt{1+a}+\sqrt{1-a}\right)}{2a}+\dfrac{\left(1-a\right)\left(\sqrt{1-a^2}+1-a\right)}{1-a^2-1+2a-a^2}\)
\(=\dfrac{\sqrt{1+a}\left(\sqrt{1+a}+\sqrt{1-a}\right)}{2a}+\dfrac{\left(1-a\right)\left(\sqrt{1-a^2}+1-a\right)}{2a-2a^2}\)
\(=\dfrac{\sqrt{1+a}\left(\sqrt{1+a}+\sqrt{1-a}\right)}{2a}+\dfrac{\left(1-a\right)\left(\sqrt{1-a^2}+1-a\right)}{2a\left(1-a\right)}\)
\(=\dfrac{1+a+\sqrt{1-a^2}}{2a}+\dfrac{\sqrt{1-a^2}+1-a}{2a}\)
\(=\dfrac{1+a+\sqrt{1-a^2}+\sqrt{1-a^2}+1-a}{2a}\)
\(=\dfrac{2\sqrt{1-a^2}+2}{2a}=\dfrac{\sqrt{1-a^2}+1}{a}\)
\(\sqrt{\dfrac{1}{a^2}-1}-\dfrac{1}{a}\)
\(=\sqrt{\dfrac{1-a^2}{a^2}}-\dfrac{1}{a}=\dfrac{\sqrt{1-a^2}-1}{a}\)
\(P=\dfrac{\sqrt{1-a^2}+1}{a}\cdot\dfrac{\sqrt{1-a^2}-1}{a}\)
\(=\dfrac{1-a^2-1}{a^2}=\dfrac{-a^2}{a^2}=-1\)
Bài 14:
a: \(P=\dfrac{\sqrt{x}}{\sqrt{x}+2}+\dfrac{-x+x\sqrt{x}+6}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+2}+\dfrac{x\sqrt{x}-x+6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+x\sqrt{x}-x+6-\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-\sqrt{x}+x\sqrt{x}-x+6-x-3\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}-x-4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\left(\sqrt{x}-1\right)-4\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{x-4}{\sqrt{x}+2}=\sqrt{x}-2\)
b: \(Q=\dfrac{\left(x+27\right)\cdot P}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\left(x+27\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\dfrac{x+27}{\sqrt{x}+3}\)
\(=\dfrac{x-9+36}{\sqrt{x}+3}\)
\(=\sqrt{x}-3+\dfrac{36}{\sqrt{x}+3}\)
\(=\sqrt{x}+3+\dfrac{36}{\sqrt{x}+3}-6\)
\(\Leftrightarrow Q>=2\cdot\sqrt{\left(\sqrt{x}+3\right)\cdot\dfrac{36}{\sqrt{x}+3}}-6=2\cdot6-6=6\)
