\(\left(\frac{1}{1-\sqrt{x}}-\frac{1}{\sqrt{x}}\right):\left(\frac{2x+\sqrt{x}-1}{1-x}+\frac{2x\sqrt{x}+x+\sqrt{x}}{1+x\sqrt{x}}\right)\)
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\(A=\frac{a+\sqrt{3a}+3}{a\sqrt{a}-3\sqrt{3}}\)
\(=\frac{a+\sqrt{3a}+3}{\left(\sqrt{a}\right)^3-\left(\sqrt{3}\right)^3}\)
\(=\frac{a+\sqrt{3a}+3}{\left(\sqrt{a}-\sqrt{3}\right)\left(a+\sqrt{3a}+3\right)}\)
\(=\frac{1}{\sqrt{a}-\sqrt{3}}\)
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Đặt \(\hept{\begin{cases}\sqrt{x+1}=a\\\sqrt{1-x}=b\end{cases}}\) \(\left(a,b\ge0\right)\)
Ta có hpt: \(\hept{\begin{cases}a+2a^2=-b^2+b+3ab\left(1\right)\\a^2+b^2=2\end{cases}}\)
\(\left(1\right)\Leftrightarrow\left(a-b\right)+\left(a^2-2ab+b^2\right)+\left(a^2-ab\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left[1+\left(a-b\right)+a\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=b\\2a+1=b\end{cases}}\)
TH1:
\(\sqrt{x+1}=\sqrt{1-x}\)
\(\Leftrightarrow x+1=1-x\)
\(\Leftrightarrow x=0\left(n\right)\)
TH2:
\(2\sqrt{x+1}+1=\sqrt{1-x}\)
\(\Leftrightarrow10\sqrt{x+1}+5=5\sqrt{1-x}\)
\(\Leftrightarrow10\sqrt{x+1}-2+7-5\sqrt{1-x}=0\)
\(\Leftrightarrow\frac{100x+100-4}{10\sqrt{x+1}+2}+\frac{49-25+25x}{7+5\sqrt{1-x}}=0\)
\(\Leftrightarrow\left(\frac{4}{10\sqrt{x+1}+2}+\frac{1}{7+5\sqrt{1-x}}\right)\left(25x+24\right)=0\)
\(\Leftrightarrow x=-\frac{24}{25}\left(n\right)\)
Vậy pt có 2 no.