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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 \(P=\dfrac{\sqrt{4+2\sqrt{3}}-2}{3\cdot\sqrt{4+2\sqrt{3}}}\)
\(=\dfrac{\sqrt{3}+1-2}{3\left(\sqrt{3}+1\right)}=\dfrac{\sqrt{3}-1}{3\sqrt{3}+3}=\dfrac{2-\sqrt{3}}{3}\)
1: Khi x=9 thì \(A=\dfrac{3+1}{3-1}=\dfrac{4}{2}=2\)
2:
a: \(P=\left(\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: \(2P=2\sqrt{x}+5\)
=>\(P=\sqrt{x}+\dfrac{5}{2}\)
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\dfrac{5}{2}=\dfrac{2\sqrt{x}+5}{2}\)
=>\(\sqrt{x}\left(2\sqrt{x}+5\right)=2\sqrt{x}+2\)
=>\(2x+3\sqrt{x}-2=0\)
=>\(\left(\sqrt{x}+2\right)\left(2\sqrt{x}-1\right)=0\)
=>\(2\sqrt{x}-1=0\)
=>x=1/4
Bạn có thể làm hộ mình câu c được không?Nếu được thì mình cảm ơn bạn nhiều!
\(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}+\frac{3x+3}{x-9}\right)\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(P=\left[\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}+\frac{3x+3}{x-9}\right]\) \(\left[\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right]\)
\(P=\frac{2x-6\sqrt{x}+x+3\sqrt{x}+3x+3}{x-9}.\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(P=\frac{6x-3\sqrt{x}+3}{x-9}.\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
a, \(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}-1}\right):\left(\frac{2}{x}-\frac{2-x}{x\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\frac{2\sqrt{x}}{\sqrt{x}-1}\right):\left(\frac{2\sqrt{x}+2-2+x}{x\left(\sqrt{x}+1\right)}\right)\)
\(=\frac{2\sqrt{x}}{\sqrt{x}-1}:\left(\frac{x+2\sqrt{x}}{x\left(\sqrt{x}+1\right)}\right)=\frac{2x\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{2x\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
b, Ta có \(P=4\Rightarrow\frac{2x\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=4\)
\(\Leftrightarrow2x\left(\sqrt{x}+1\right)=4\left(x+\sqrt{x}-2\right)\)
\(\Leftrightarrow2x\sqrt{x}+2x=4x+4\sqrt{x}-8\Leftrightarrow2x\sqrt{x}-2x-4\sqrt{x}+8=0\)
Ps : bạn kiểm tra lại đề nhé, nhìn phần a thôi thấy sai rồi