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\(P=\frac{2}{x-1}+\frac{1}{\sqrt{x-1}}+\frac{1}{\sqrt{x}+1}\)
\(P=\frac{2}{ \left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{2+2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{2\left(1+\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{2}{\sqrt{x}-1}\)
a) \(Q=\left(\frac{\sqrt{x}}{1-\sqrt{x}}+\frac{\sqrt{x}}{1+\sqrt{x}}\right)+\frac{3-\sqrt{x}}{x-1}\left(x\ge0;x\ne1\right)\)
\(=-\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}\left(\sqrt{x}-1\right)+3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-x-\sqrt{x}+x-\sqrt{x}+3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{-3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=-\frac{3}{\sqrt{x}+1}\)
b) Để \(Q=-1\)
\(\Leftrightarrow-\frac{3}{\sqrt{x}+1}=-1\)
\(\Leftrightarrow\sqrt{x}+1=3\)
\(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
@Mai.T.Loan câu a pha cuối hơi tắt đó nhìn khó hiểu lắm
còn câu b kl sai r nha
\(P=\left(\sqrt{x}+\frac{x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right)\)
\(=\left[\frac{\sqrt{x}.\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\frac{x+2}{\sqrt{x}+1}\right]:\left(\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}-4}{x-1}\right)\)
\(=\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}+\frac{x+2}{\sqrt{x}+1}\right):\left[\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\)
\(=\frac{2x+\sqrt{x}+2}{\sqrt{x}+1}:\left[\frac{x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}-4}{\left(\sqrt{x}-\right)\left(\sqrt{x}+1\right)}\right]\)
\(=\frac{2x+\sqrt{x}+2}{\sqrt{x}+1}:\frac{x-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2x+\sqrt{x}+2}{\sqrt{x}+1}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{x-4}\)
\(=\frac{\left(2x+\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{x-4}\)
a, Thay x = 9 vào biểu thức \(A=\frac{\sqrt{x}-2}{\sqrt{x}-1}\) ta được:
\(A=\frac{\sqrt{9}-2}{\sqrt{9}-1}=\frac{\sqrt{3^2}-2}{\sqrt{3^2}-1}=\frac{3-2}{3-1}=\frac{1}{2}\)
Vậy với x = 9 thì \(A=\frac{1}{2}\)
\(b,\left(\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}-1}{x+1}\left(x\ge0;x\ne1\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\frac{\sqrt{x}-1}{x+1}\)
\(=\frac{x-\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}-1}{x+1}\)
\(=\frac{\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+1\right)}\)
\(=\frac{1}{\sqrt{x}+1}\)
a, Thay x=9 ta có
\(A=\frac{\sqrt{9}-2}{\sqrt{9}-1}=\frac{3-2}{3-1}=\frac{1}{2}\)
b,\(B=\left(\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}-1}{x+1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}-1}{x+1}\)
\(=\frac{x-\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}-1}{x+1}\)
\(=\frac{x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}-1}{x+1}\)
\(=\frac{1}{\sqrt{x}+1}\)
\(B=\frac{x\sqrt{x}+1}{x-1}-\frac{x-1}{\sqrt{x}+1}\)
\(B=\frac{\left(x\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(x-1\right)\left(x-1\right)}{\left(\sqrt{x}+1\right)\left(x-1\right)}\)
\(B=\frac{\left(x\sqrt{x}+1\right)\left(\sqrt{x}+1\right)-\left(x-1\right)^2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x^2+x\sqrt{x}+\sqrt{x}+1-x^2+2x-1}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x\sqrt{x}+\sqrt{x}+2x}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{\sqrt{x}\left(x+1+2\sqrt{x}\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
tiếp tục của bạn @Bastkoo nhé
\(B=\frac{\sqrt{x}\left(x+2\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
\(< =>B=\frac{\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
\(< =>B=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(< =>B=\frac{\sqrt{x}}{\sqrt{x}-1}\)