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Giúp mk nhanh nha.mk cần gấp...........ai nhanh mak đúng mk k cho
\(A=\left(\frac{1}{\sqrt{x}-1}+\frac{1}{x-\sqrt{x}}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left[\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\frac{\sqrt{x}-1}{\sqrt{x}}=1-\frac{1}{\sqrt{x}}< 1\)
a)\(M=\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right):\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left(\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}.\left(\sqrt{x}+1\right)\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}-2}\)
b)\(\frac{1}{M}=\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\)
Ta có: \(\sqrt{x}\ge0,\forall x\ge0\)
\(\Leftrightarrow\sqrt{x}+1\ge1\)
\(\Leftrightarrow\frac{1}{\sqrt{x}+1}\le1\)
\(\Leftrightarrow\frac{3}{\sqrt{x}+1}\le3\)
\(\Leftrightarrow-\frac{3}{\sqrt{x}+1}\ge-3\)
\(\Leftrightarrow1-\frac{3}{\sqrt{x}+1}\ge-2\)
Dấu "=" xảy ra khi x=0
Vậy \(Min_{\frac{1}{M}}=-2\) khi x=0
\(a,\)\(T=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\sqrt{x}^3-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\)\(\frac{\sqrt{x}^3+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)\(-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}\)
\(=\frac{x+2\sqrt{x}+1}{\sqrt{x}}=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
Theo đề bài ta có x > 0 nên \(\sqrt{x}>0\)
=> \(\frac{2}{\sqrt{x}}>0\Rightarrow-\frac{2}{\sqrt{x}}< 0\Rightarrow1-\frac{2}{\sqrt{x}}< 1\)
Ta có
M = \(\frac{\sqrt{x}-2}{\sqrt{x}}=\:1-\frac{2}{\sqrt{x}}< 1\)
a) \(M=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}\right)^3-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}\right)^3+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\sqrt{x}}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(x+\sqrt{x}+1\right)-\left(x-\sqrt{x}+1\right)+\left(x+1\right)}{\sqrt{x}}\)
\(=\frac{x+2\sqrt{x}+1}{\sqrt{x}}=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
b) Để \(M=\frac{9}{2}\) thì :
\(\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}=\frac{9}{2}\Leftrightarrow2\left(\sqrt{x}+1\right)^2=9\sqrt{x}\)
\(\Leftrightarrow2x+4\sqrt{x}+2-9\sqrt{x}=0\)
\(\Leftrightarrow2x-5\sqrt{x}+2=0\)
\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x}-1=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{4}\\x=4\end{matrix}\right.\)
c) Ta có :
\(M=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}=\frac{x+2\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+2+\frac{1}{\sqrt{x}}=2+\sqrt{x}+\frac{1}{\sqrt{x}}\)
AD - BDDT cô si cho 2 số nguyên dương \(\sqrt{x},\frac{1}{\sqrt{x}}\) ta có :
\(\sqrt{x}+\frac{1}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{1}{\sqrt{x}}}=2\)
\(\Rightarrow M\ge2+2=4\)
Dấu = xảy ra khi \(\sqrt{x}=\frac{1}{\sqrt{x}}\Rightarrow x=1\)
Mà x ≠ 1 ⇒ M > 4