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a) \(A=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{6}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(A=\frac{\left(-\sqrt{a}+1\right)^2}{\left(-a+1\right)^2}.\left(\sqrt{a}+\frac{-a\sqrt{a}+1}{-\sqrt{a}+1}\right)\)
\(A=\frac{\left(1-\sqrt{a}\right)^2\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)}{\left(1-a\right)^2}\)
\(A=\frac{\frac{-a\sqrt{a}+\sqrt{a}.\left(-\sqrt{a}+1\right)+1}{-\sqrt{a}+1}.\left(-\sqrt{a}+1\right)^2}{\left(1-a\right)^2}\)
\(A=\frac{a^2-2a+1}{\left(1-a\right)^2}\)
\(A=\frac{\left(a-1\right)^2}{\left(1-a\right)^2}\)
\(A=1\)
a. A có nghĩa khi \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne\\\frac{x+\sqrt{x}}{\sqrt{x}+1}\ne0\end{matrix}\right.0\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
A\(=\frac{x-\sqrt{x}+\sqrt{x}-1}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{x+\sqrt{x}}\)\(=\frac{x-1}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
b. \(x=7+4\sqrt{3}\Rightarrow\)A = \(\frac{\sqrt{7+4\sqrt{3}}+1}{\sqrt{7+4\sqrt{3}}}=\frac{\sqrt{\left(2+\sqrt{3}\right)^2}+1}{\sqrt{\left(2+\sqrt{3}\right)^2}}=\frac{3+\sqrt{3}}{2+\sqrt{3}}\)
ĐK \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)
Ta có \(P=\frac{\sqrt{a}\left(1-a\right)^2}{1+a}:\left[\left(\frac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right).\left(\frac{\left(1+\sqrt{a}\right)\left(a-\sqrt{a}+1\right)}{1+\sqrt{a}}-\sqrt{a}\right)\right]\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{1+a}:\left[\left(a+2\sqrt{a}+1\right).\left(a-2\sqrt{a}+1\right)\right]\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{1+a}.\frac{1}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)^2}=\frac{\sqrt{a}}{1+a}\)
a) ĐKXĐ: \(x\ge0;x\ne1\)
P=\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2.\left(\frac{\sqrt{a}}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
=\(\left(\frac{a-1}{2\sqrt{a}}\right)^2.\left(\frac{-1-3\sqrt{a}}{a-1}\right)\)
=\(\frac{\left(a-1\right)^2}{4a}.\frac{-1-3\sqrt{a}}{a-1}\)
=\(\frac{\left(a-1\right)\left(-1-3\sqrt{a}\right)}{4a}\)
\(x^2-1=\frac{1}{4}\left(a^2+\frac{1}{a^2}+2\right)-1=\frac{1}{4}\left(a^2+\frac{1}{a^2}-2\right)=\frac{1}{4}\left(a-\frac{1}{a}\right)^2\)
Tương tự \(y^2-1=\frac{1}{4}\left(b-\frac{1}{b}\right)^2\)
\(P=\frac{\frac{1}{4}\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)-\frac{1}{4}\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)}{\frac{1}{4}\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)+\frac{1}{4}\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)}\)
\(=\frac{ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab}-ab+\frac{a}{b}+\frac{b}{a}-\frac{1}{ab}}{ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab}+ab-\frac{a}{b}-\frac{b}{a}+\frac{1}{ab}}=\frac{\frac{a}{b}+\frac{b}{a}}{ab+\frac{1}{ab}}=\frac{a^2+b^2}{a^2b^2+1}\)