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\(\left(\frac{\sqrt{x}-4x}{1-4x}-1\right):\left(\frac{1+2x}{1-4x}-\frac{2\sqrt{x}}{2\sqrt{x}-1}-1\right)\left(ĐK:0\le x\ne\frac{1}{4}\right)\)
\(=\frac{\sqrt{x}-4x+4x-1}{1-4x}:\frac{\left(1+2x\right)+2\sqrt{x}\left(1+2\sqrt{x}\right)+4x-1}{1-4x}\)
\(=\frac{\sqrt{x}-1}{1-4x}.\frac{1-4x}{10x+2\sqrt{x}}=\frac{\sqrt{x}-1}{2\sqrt{x}\left(5\sqrt{x}+1\right)}\)
\(a,\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
Ta có
:\(VT=\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{5}\)
\(=|2-\sqrt{5}|-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}\)
\(=-2=VP\left(đpcm\right)\)
\(b,\frac{\sqrt{2}+1}{\sqrt{2}-1}=3+2\sqrt{2}\)
Ta có:
\(VT=\frac{\sqrt{2}+1}{\sqrt{2}-1}\)
\(=\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
\(=\frac{2+\sqrt{2}+\sqrt{2}+1}{\sqrt{2}^2-1^2}\)
\(=\frac{3+2\sqrt{2}}{2-1}\)
\(=3+2\sqrt{2}=VP\left(đpcm\right)\)
c,Bạn xem lại đề
\(d,\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}=8\)
Ta có:
\(VT=\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}\)
\(=\sqrt{\frac{2^2}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{2^2}{\left(2+\sqrt{5}\right)^2}}\)
\(=\frac{2}{|2-\sqrt{5}|}-\frac{2}{|2+\sqrt{5}|}\)
\(=\frac{2\left(2+\sqrt{5}\right)}{\left(\sqrt{5}-2\right)\left(2+\sqrt{5}\right)}-\frac{2\left(\sqrt{5}-2\right)}{\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)}\)
\(=\frac{4+2\sqrt{5}-2\sqrt{5}+4}{\sqrt{5}^2-2^2}\)
\(=\frac{8}{5-4}\)
\(=8=VP\left(đpcm\right)\)