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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
\(Q>0\Leftrightarrow\frac{\sqrt{a}-2}{\sqrt{a}}>0\Leftrightarrow\sqrt{a}-2>0\Leftrightarrow\sqrt{a}>2\Leftrightarrow a>\sqrt{2}\)
Điều kiện: \(\hept{\begin{cases}a>0\\\sqrt{a}-1\ne0\\\sqrt{a}-2\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}a>0\\a\ne1\\a\ne4\end{cases}}\)
Ta có:
\(1P=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)\)
\(=\frac{\sqrt{a}-2}{\sqrt{a}}\)
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)\(\sqrt{a}>3\Leftrightarrow a>9\)
\(a,ĐKXĐ:\hept{\begin{cases}a>0\\a\ne1\end{cases}}\)
\(b,A=\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)\left(\frac{a-\sqrt{a}}{\sqrt{a}+1}-\frac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\frac{a-1}{2\sqrt{a}}.\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}+1}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}-1}\right)\)
\(=\frac{a-1}{2\sqrt{a}}.\frac{\sqrt{a}.\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{a-1}\)
\(=\frac{\sqrt{a}\left(\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2\right)}{2\sqrt{a}}\)
\(=\frac{\sqrt{a}.\left(\sqrt{a}-1-\sqrt{a}-1\right).\left(\sqrt{a}-1+\sqrt{a}+1\right)}{2\sqrt{a}}\)
\(=\frac{\sqrt{a}.\left(-2\right).2\sqrt{a}}{2\sqrt{a}}\)
\(=-2\sqrt{a}\)
\(c,\)Để A= -4 thì
\(-2\sqrt{a}=-4\Leftrightarrow\sqrt{a}=2\Leftrightarrow a=4\)
Kết bạn với mình nha ....
a) ĐKXĐ: \(a\ne1;a\ne0\))
\(A=\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{\sqrt{a+1}}{a}\)
\(=\left(\frac{\sqrt{a}.\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{\sqrt{a}.\left(\sqrt{a}+1\right)}\right):\frac{\sqrt{a+1}}{a}\)
\(=\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right):\frac{\sqrt{a+1}}{a}\)
\(=\frac{a-1}{\sqrt{a}}.\frac{a}{\sqrt{a+1}}=\frac{\sqrt{a}\left(a-1\right)}{\sqrt{a+1}}\)
Bài 2:
\(\Leftrightarrow3\sqrt{x+5}-2\sqrt{x+5}=7\)
\(\Leftrightarrow\sqrt{x+5}=7\)
=>x+5=25
hay x=18