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a: \(P=\dfrac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}:\left(\dfrac{6\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}:\dfrac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\cdot\dfrac{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{2x+3\sqrt{x}+1}\)
\(=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
b: Thay \(x=\dfrac{3-2\sqrt{2}}{4}\) vào P, ta được:
\(P=\left(3\cdot\dfrac{\sqrt{2}-1}{2}-5\right):\left(2\cdot\dfrac{\sqrt{2}-1}{2}+1\right)\)
\(=\dfrac{3\sqrt{2}-3-10}{2}:\sqrt{2}\)
\(=\dfrac{3\sqrt{2}-13}{2\sqrt{2}}=\dfrac{6-13\sqrt{2}}{4}\)
a: \(P=\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x-1-x+4}\)
\(=\dfrac{1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-2}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b: P=1/4
=>\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\)
=>\(4\left(\sqrt{x}-2\right)=3\sqrt{x}\)
=>\(4\sqrt{x}-8-3\sqrt{x}=0\)
=>\(\sqrt{x}=8\)
=>x=64
c: Khi \(x=4+2\sqrt{3}\) thì \(P=\dfrac{\sqrt{4+2\sqrt{3}}-2}{3\cdot\sqrt{4+2\sqrt{3}}}\)
\(=\dfrac{\sqrt{3}+1-2}{3\left(\sqrt{3}+1\right)}=\dfrac{\sqrt{3}-1}{3\sqrt{3}+3}=\dfrac{2-\sqrt{3}}{3}\)
ĐKXĐ : \(x\ge1;x\ne2;x\ne3\)
a) P = \(\left[\frac{\sqrt{x}+\sqrt{x-1}}{1}-\frac{\left(x-3\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{x-3}\right].\frac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\)( trục căn thức ở mẫu thứ nhất )
\(P=\left(\sqrt{x}-\sqrt{2}\right).\frac{\left(\sqrt{x}-\sqrt{2}\right)}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}=\frac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\)
b) \(x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\Rightarrow\sqrt{x}=\sqrt{2}-1\)
\(P=\frac{\sqrt{2}-\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1\)
a: \(P=\dfrac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}:\left(\dfrac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\cdot\dfrac{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{2x+3\sqrt{x}+1}\)
\(=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
b: Thay \(x=\dfrac{3-2\sqrt{2}}{4}\) vào P, ta được:
\(P=\left(3\cdot\dfrac{\sqrt{2}-1}{2}-5\right):\left(2\cdot\dfrac{\sqrt{2}-1}{2}+1\right)\)
\(=\dfrac{\left(\dfrac{3}{2}\sqrt{2}-\dfrac{3}{2}-5\right)}{\sqrt{2}}\)
\(=\dfrac{\dfrac{3}{2}\sqrt{2}-\dfrac{13}{2}}{\sqrt{2}}=\dfrac{6-13\sqrt{2}}{4}\)