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ĐKXĐ: \(x\ge0\)
a/ \(Q=1+\left[\frac{\left(2\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)}-\frac{2a\sqrt{a}-\sqrt{a}+a}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}\right].\frac{\sqrt{a}.\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
\(=1+\left[\frac{2\sqrt{a}-1}{1-\sqrt{a}}-\frac{2a\sqrt{a}-\sqrt{a}+a}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}\right].\frac{\sqrt{a}.\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
\(=1+\left[\frac{\left(2\sqrt{a}-1\right)\left(1+\sqrt{a}+a\right)-\left(2a\sqrt{a}-\sqrt{a}+a\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}\right].\frac{\sqrt{a}.\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)
\(=1+\left[\frac{2\sqrt{a}-1}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}\right].\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)\(=1+\frac{\sqrt{a}}{1+\sqrt{a}+a}=\frac{1+2\sqrt{a}+a}{1+\sqrt{a}+a}\)
b/ \(Q=\frac{\sqrt{6}}{1+\sqrt{6}}\Leftrightarrow1+\frac{\sqrt{a}}{1+\sqrt{a}+a}=\frac{\sqrt{6}}{1+\sqrt{6}}\) \(\Leftrightarrow\frac{1+2\sqrt{a}+a}{1+\sqrt{a}+a}=\frac{\sqrt{6}}{1+\sqrt{6}}\)
\(\Leftrightarrow\left(1+\sqrt{6}\right)\left(1+2\sqrt{a}+a\right)=\sqrt{6}\left(1+\sqrt{a}+a\right)\)
Tới đây mình k biết giải
c/ \(Q>\frac{2}{3}\Leftrightarrow\frac{1+2\sqrt{a}+a}{1+\sqrt{a}+a}>\frac{2}{3}\)\(\Leftrightarrow3.\left(1+2\sqrt{a}+a\right)>2.\left(1+\sqrt{a}+a\right)\)
\(\Leftrightarrow3+6\sqrt{a}+3a-2-2\sqrt{a}-2a>0\Leftrightarrow1+4\sqrt{a}+a>0\)
\(\Leftrightarrow x< -2-\sqrt{3}Vx>-2+\sqrt{3}\)
a/ \(\frac{2}{a}.\frac{4\left|a\right|}{3}=\frac{-8a}{3a}=-\frac{8}{3}\)
b/ \(\frac{3}{a-1}\sqrt{\frac{4\left(a-1\right)^2}{25}}=\frac{3}{\left(a-1\right)}.\frac{2\left|a-1\right|}{5}=\frac{6\left(a-1\right)}{5\left(a-1\right)}=\frac{6}{5}\)
c/ \(\frac{3\sqrt{9a^2b^4}}{\sqrt{a^2b^2}}=\frac{9.\left|a\right|.b^2}{\left|a\right|\left|b\right|}=9\left|b\right|\)
d/ \(\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
a/ \(=\frac{2}{a}.\frac{4\left|a\right|}{3}=\frac{2}{a}.\frac{-4a}{3}=\frac{-8}{3}\)
b/ \(=\frac{3}{a-1}.\frac{\left|2a-2\right|}{5}=\frac{3}{a-1}.\frac{2\left(a-1\right)}{5}=\frac{6}{5}\)
c/ \(=\sqrt{\frac{162a^2b^4}{2a^2b^2}}=\sqrt{81b^2}=9\left|b\right|\)
d/ \(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
\(C=\frac{\sqrt{a}\left(a\sqrt{a}+1\right)}{a+\sqrt{a}+1}+\frac{1-2a+\sqrt{a}}{\sqrt{a}}=\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a+\sqrt{a}+1\right)}{a+\sqrt{a}+1}+\frac{1-2a+\sqrt{a}}{\sqrt{a}}\)
\(C=\sqrt{a}\left(\sqrt{a}+1\right)+\frac{1-2a+\sqrt{a}}{\sqrt{a}}=\frac{a\sqrt{a}-a+\sqrt{a}+1}{\sqrt{a}}\)
Hình như bạn ghi đề nhầm, ko rút gọn được nữa, mẫu số đằng sau là \(\sqrt{a}+1\) thì hợp lý hơn là \(\sqrt{a}\)
a) \(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne4\end{cases}}\)
\(A=\left(\frac{1}{\sqrt{x}+2}+\frac{1}{\sqrt{x}-2}\right):\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}-2}{\sqrt{x}}\)
\(\Leftrightarrow A=\frac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow A=\frac{2}{\sqrt{x}+2}\)
b) Để \(A>\frac{1}{2}\)
\(\Leftrightarrow\frac{2}{\sqrt{x}+2}>\frac{1}{2}\)
\(\Leftrightarrow\sqrt{x}+2< 4\)
\(\Leftrightarrow\sqrt{x}< 2\)
\(\Leftrightarrow x< 4\)
Vậy để \(A>\frac{1}{2}\Leftrightarrow0< x< 4\)
c) \(B=\frac{7}{3}A\)
\(\Leftrightarrow B=\frac{7}{3}\cdot\frac{2}{\sqrt{x}+2}\)
\(\Leftrightarrow B=\frac{14}{3\sqrt{x}+6}\)
Tìm x hay tìm B đây bạn ?