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1,
\(A=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{a+2}{a-2}\left(đk:a\ne0;1;2;a\ge0\right)\)
\(=\frac{\left(a\sqrt{a}-1\right)\left(a+\sqrt{a}\right)-\left(a\sqrt{a}+1\right)\left(a-\sqrt{a}\right)}{a^2-a}.\frac{a-2}{a+2}\)
\(=\frac{a^2\sqrt{a}+a^2-a-\sqrt{a}-\left(a^2\sqrt{a}-a^2+a-\sqrt{a}\right)}{a\left(a-1\right)}.\frac{a-2}{a+2}\)
\(=\frac{2a\left(a-1\right)\left(a-2\right)}{a\left(a-1\right)\left(a+2\right)}=\frac{2\left(a-2\right)}{a+2}\)
Để \(A=1\)\(=>\frac{2a-4}{a+2}=1< =>2a-4-a-2=0< =>a=6\)
2,
a, Điều kiện xác định của phương trình là \(x\ne4;x\ge0\)
b, Ta có : \(B=\frac{2\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)
\(=\frac{2\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}-\frac{\sqrt{x}-2}{x-4}\)
\(=\frac{2\sqrt{x}+2+2}{x-4}=\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{2}{\sqrt{x}-2}\)
c, Với \(x=3+2\sqrt{3}\)thì \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)
a/ \(\sqrt{ab}+\sqrt{a}-\sqrt{b}\)
b/ \(\sqrt{ab}+\sqrt{a}-\sqrt{b}=2\)
\(\Leftrightarrow\left(\sqrt{a}-1\right)\left(\sqrt{b}+1\right)=1\)
Xong rồi nhá
a
\(ĐKXĐ:a\ne0;a\ne1;a\ne\sqrt{2}\)
\(Q=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(Q=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(Q=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(Q=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{1}\)
\(Q=\frac{\sqrt{a}-2}{\sqrt{a}}\)
b
\(Q>0\Leftrightarrow\frac{\sqrt{a}-2}{\sqrt{a}}>0\Leftrightarrow\sqrt{a}-2>0\Leftrightarrow\sqrt{a}>2\Leftrightarrow a>\sqrt{2}\)
\(\left(a+b-\frac{2a\sqrt{b}-2b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\right):\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\left(a+b-\frac{2\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\right):\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\left(a+b-2\sqrt{ab}\right):\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{a-b}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}=1\)
a)\(\hept{\begin{cases}a\ge0\\\sqrt{a}-2>0\Leftrightarrow\\\sqrt{a}+2>0\end{cases}a>4}\)
b)\(\frac{\sqrt{a}\left(\sqrt{a}+2\right)+\sqrt{a}\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}.\frac{a-4}{2\sqrt{a}}\) \(=\frac{2a}{a-4}.\frac{a-4}{2\sqrt{a}}=\sqrt{a}\)
c)\(\sqrt{a}>3\Leftrightarrow a>9\)
Kết quả câu a như nào nhỉ ?
a) đk: \(\hept{\begin{cases}a>b\\a< -b\end{cases}}\left(b>0\right)\) hoặc \(\hept{\begin{cases}a>-b\\a< b\end{cases}\left(b< 0\right)}\)
Ta có:
\(B=\frac{a}{\sqrt{a^2-b^2}}-\left(1+\frac{a}{\sqrt{a^2-b^2}}\right)\div\frac{b}{a-\sqrt{a^2-b^2}}\)
\(B=\frac{a}{\sqrt{a^2-b^2}}-\frac{a+\sqrt{a^2-b^2}}{\sqrt{a^2-b^2}}\cdot\frac{a-\sqrt{a^2-b^2}}{b}\)
\(B=\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-a^2+b^2}{b\sqrt{a^2-b^2}}\)
\(B=\frac{a}{\sqrt{a^2-b^2}}-\frac{b}{\sqrt{a^2-b^2}}=\frac{a-b}{\sqrt{a^2-b^2}}=\sqrt{\frac{a-b}{a+b}}\)
b) \(B< 1\Leftrightarrow\sqrt{\frac{a-b}{a+b}}< 1\Leftrightarrow\frac{a-b}{a+b}< 1\)
\(\Leftrightarrow\frac{-2b}{a+b}< 0\) ta xét 2TH:
Nếu \(b>0\Rightarrow a>-b\)
Nếu \(b< 0\Rightarrow a< -b\)
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