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a) Q=\(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)=\(\frac{\sqrt{a}-2}{3\sqrt{a}}\) b) Ta thấy \(3\sqrt{a}>0\), để Q dương thì \(\sqrt{a}-2>0\Rightarrow a>4\)
tham khao nha
\(A=\left(\frac{\sqrt{a}}{\sqrt{ab}-b}+\frac{2\sqrt{a}+\sqrt{b}}{\sqrt{ab}-a}\right):\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\right)\)
\(A=\left(\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}+\frac{2\sqrt{a}+\sqrt{b}}{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}\right):\left(\frac{\sqrt{b}+\sqrt{a}}{\sqrt{ab}}\right)\)
\(A=\left(\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{2\sqrt{a}+\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}\right).\frac{\sqrt{ab}}{\sqrt{b}+\sqrt{a}}\)
\(A=\frac{a-2\sqrt{ab}+b}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}.\frac{\sqrt{ab}}{\sqrt{b}+\sqrt{a}}\)
\(A=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}.\frac{\sqrt{ab}}{\sqrt{b}+\sqrt{a}}\)
\(A=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
vay \(A=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
ĐK : tự ghi nha
\(\left(\frac{\sqrt{a}}{\sqrt{ab}-b}+\frac{2\sqrt{a}+\sqrt{b}}{\sqrt{ab}-a}\right):\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\right)\)
a) \(A=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{6}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(A=\frac{\left(-\sqrt{a}+1\right)^2}{\left(-a+1\right)^2}.\left(\sqrt{a}+\frac{-a\sqrt{a}+1}{-\sqrt{a}+1}\right)\)
\(A=\frac{\left(1-\sqrt{a}\right)^2\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)}{\left(1-a\right)^2}\)
\(A=\frac{\frac{-a\sqrt{a}+\sqrt{a}.\left(-\sqrt{a}+1\right)+1}{-\sqrt{a}+1}.\left(-\sqrt{a}+1\right)^2}{\left(1-a\right)^2}\)
\(A=\frac{a^2-2a+1}{\left(1-a\right)^2}\)
\(A=\frac{\left(a-1\right)^2}{\left(1-a\right)^2}\)
\(A=1\)
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=\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2.\left(\frac{\sqrt{a}}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
=\(\left(\frac{a-1}{2\sqrt{a}}\right)^2.\left(\frac{-1-3\sqrt{a}}{a-1}\right)\)
=\(\frac{\left(a-1\right)^2}{4a}.\frac{-1-3\sqrt{a}}{a-1}\)
=\(\frac{\left(a-1\right)\left(-1-3\sqrt{a}\right)}{4a}\)
\(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{\frac{4}{3}}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\frac{10\sqrt{3}}{3}\)
\(=-9\sqrt{3}+\frac{10\sqrt{3}}{3}\)
\(=\frac{-27\sqrt{3}}{3}+\frac{10\sqrt{3}}{3}\)
\(=\frac{-17\sqrt{3}}{3}\)
\(\frac{1-a\sqrt{a}}{1-\sqrt{a}}\) \(=\frac{1^3-\left(\sqrt{a}\right)^3}{1-\sqrt{a}}\)
\(=\frac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}\)
\(=a+\sqrt{a}+1\)
chúc bn học tốt
\(\frac{1-a\sqrt{a}}{1-\sqrt{a}}\)
\(=\frac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}\)
\(=a+\sqrt{a}+1\)
bạn tự ghi dk nha
\(B=\frac{1+\sqrt{1-a}}{\sqrt{1-a}\left(\sqrt{1-a}+1\right)}+\frac{1-\sqrt{1+a}}{\sqrt{1+a}\left(\sqrt{1+a}-1\right)}+\frac{1}{\sqrt{1+a}}\)
\(B=\frac{1}{\sqrt{a-1}}-\frac{1}{\sqrt{a+1}}+\frac{1}{\sqrt{1+a}}\)
\(B=\frac{1}{\sqrt{a-1}}\)
vì \(\sqrt{a-1}>0\)không có dấu = vì mẫu khác 0
\(\Rightarrow\frac{1}{\sqrt{a-1}}>0\)
đpcm