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a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{\sqrt{x}}\)
b: Thay \(x=3+2\sqrt{2}\) vào P, ta được:
\(P=\dfrac{2\sqrt{2}+2}{\sqrt{2}+1}=2\)
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}-1}\)ĐK x>=0 x khác -1
=\(\frac{\sqrt{x}+1}{x+1}.\frac{x+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b/ x =\(\frac{2+\sqrt{3}}{2}=\frac{4+2\sqrt{3}}{4}=\frac{3+2\sqrt{3}+1}{4}=\frac{\left(\sqrt{3}+1\right)^2}{4}\)
\(\Rightarrow\sqrt{x}=\frac{\sqrt{3}+1}{2}\)
Em thay vào tính nhé!
c) với x>1
A=\(\frac{\sqrt{x}+1}{\sqrt{x}-1}.\sqrt{x}=\frac{x+\sqrt{x}}{\sqrt{x}-1}=\sqrt{x}+2+\frac{2}{\sqrt{x}-1}=\sqrt{x}-1+\frac{2}{\sqrt{x}-1}+3\)
Áp dụng bất đẳng thức Cosi
A\(\ge2\sqrt{2}+3\)
Xét dấu bằng xảy ra ....
a/ đkxđ: x > 0; x≠1
b/ \(A=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right):\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}-\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\dfrac{x-1}{2\sqrt{x}}\cdot\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{x-1}{2\sqrt{x}}\cdot\dfrac{x\sqrt{x}-2x+\sqrt{x}-x\sqrt{x}-2x-\sqrt{x}}{x-1}\)
\(=\dfrac{-4x}{2\sqrt{x}}=-2\sqrt{x}\)
a: ĐKXĐ: x=0; x<>1
\(M=\left(2+\sqrt{x}\right)\left(1-2\sqrt{x}-x+1+\sqrt{x}+x\right)\)
\(=\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)=4-x\)
b: Sửa đề: P=1/M
P=1/4-x=-1/x-4
Để P nguyên thì x-4 thuộc {1;-1}
=>x thuộc {5;3}
\(a,ĐK:x>0;x\ne9\\ b,A=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\\ A=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{2}{\sqrt{x}+3}\\ c,A>\dfrac{2}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}-\dfrac{2}{5}>0\\ \Leftrightarrow\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{5}>0\\ \Leftrightarrow\dfrac{2-\sqrt{x}}{5\left(\sqrt{x}+3\right)}>0\\ \Leftrightarrow2-\sqrt{x}>0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}< 2\Leftrightarrow0< x< 4\)
a) \(x>0,x\ne1\)
b) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}:\dfrac{1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}.\left(\sqrt{x}-1\right)=\dfrac{x-1}{\sqrt{x}}\)
c) \(P< 0\Rightarrow\dfrac{x-1}{\sqrt{x}}< 0\) mà \(\sqrt{x}>0\Rightarrow x-1< 0\Rightarrow x< 1\Rightarrow0< x< 1\)
a/ ĐKXĐ: \(x\ge0;x\ne1\)
= \(\dfrac{x+1+\sqrt{x}}{x+1}:\left[\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right]-1\)
= \(\dfrac{x+1+\sqrt{x}}{x+1}:\dfrac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}-1\)
= \(\dfrac{x+1+\sqrt{x}}{x+1}:\dfrac{\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}-1\right)}-1\)
\(=\dfrac{\left(x+1+\sqrt{x}\right)\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(x+1\right)\left(\sqrt{x}-1\right)^2}-1\)
= \(\dfrac{x+1+\sqrt{x}}{\sqrt{x}-1}-1=\dfrac{x+2}{\sqrt{x}-1}\)
b/ Ta có:
\(Q=P-\sqrt{x}\)
= \(\dfrac{x+2}{\sqrt{x}-1}-\sqrt{x}\)
= \(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)+3}{\sqrt{x}-1}=1+\dfrac{3}{\sqrt{x}-1}\)
Để Q nhận giá trị nguyên thì \(1+\dfrac{3}{\sqrt{x}-1}\in Z\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}-1}\in Z\) ( vì 1\(\in Z\) )
\(\Leftrightarrow\sqrt{x}-1\inƯ_{\left(3\right)}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=3\\\sqrt{x}-1=-3\\\sqrt{x}-1=1\\\sqrt{x}-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=4\\\sqrt{x}=-2\\\sqrt{x}=2\\\sqrt{x}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=16\left(tm\right)\\\\x=4\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
Vậy để biểu thức \(Q=P-\sqrt{x}\) nhận giá trị nguyên thì x=\(\left\{16;4;0\right\}\)
a, ĐK: \(x\ge0,x\ne9\)
b, \(A=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}+3-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{3}{\sqrt{x}-3}\)
\(=\dfrac{6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{3}{\sqrt{x}-3}\)
\(=\dfrac{2}{\sqrt{x}+3}\)
c, ĐK: \(x\ge0,x\ne9\)
\(A>\dfrac{1}{3}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}>\dfrac{1}{3}\Leftrightarrow\sqrt{x}+3>6\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)
Vậy \(A>\dfrac{1}{3}\Leftrightarrow x>9\)
d, ĐK: \(x\ge0,x\ne9\)
Ta có: \(x\ge0\forall x\Leftrightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+3\ge3\)\(\Leftrightarrow\dfrac{1}{\sqrt{x}+3}\le\dfrac{1}{3}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}\le\dfrac{2}{3}\)
Dấu bằng xảy ra \(\Leftrightarrow\sqrt{x}+3=3\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\left(TM\right)\)
Vậy MaxA = \(\dfrac{2}{3}\Leftrightarrow x=0\)
a: \(A=\dfrac{1}{\sqrt{x}-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+1-x-2-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{-2x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
Để A>0 thì \(\dfrac{-2x+\sqrt{x}}{\sqrt{x}-1}>0\)
=>\(\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}< 0\)
=>1/2<căn x<1
=>1/4<x<1
b: \(B=\dfrac{2}{A}+\sqrt{x}\)
\(=\dfrac{2\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{-2x+\sqrt{x}}+\sqrt{x}\)
\(=\dfrac{2\left(x\sqrt{x}-1\right)-2x\sqrt{x}+x}{-2x+\sqrt{x}}=\dfrac{x-2}{-2x+\sqrt{x}}=\dfrac{-\left(x-2\right)}{2x-\sqrt{x}}< =0\)
Dấu '=' xảy ra khi x=2