\(A=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{x^2-2x+1}{2}\)
a) Tìm ĐKXĐ b ) Rút gọn c) Cho 0<x<1 cm : A>0 d ) Tìm GTLN A
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ĐKXĐ: \(x\ge0;x\ne1\)
Ta có: \(A=\left(2+\dfrac{2x+\sqrt{x}}{2\sqrt{x}+1}\right)\left(2-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)
\(A=\left(2+\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{2\sqrt{x}+1}\right)\left(2-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)
\(A=\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)=4-x\)
\(1,\\ a,ĐK:\left\{{}\begin{matrix}x\ge0\\x+5\ge0\end{matrix}\right.\Leftrightarrow x\ge0\\ b,Sửa:B=\left(\sqrt{3}-1\right)^2+\dfrac{24-2\sqrt{3}}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+\dfrac{2\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+2\sqrt{3}=4\\ 3,\\ =\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1+\sqrt{x}}\right]\cdot\dfrac{\sqrt{x}-3+2-2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\left(1-\sqrt{x}\right)\cdot\dfrac{-\sqrt{x}-1}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\dfrac{-\sqrt{x}-1}{\sqrt{x}-3}-2=\dfrac{-\sqrt{x}-1-2\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{-3\sqrt{x}+5}{\sqrt{x}-3}\)
a,\(ĐK:x>0,x\ne1,x\ne4\)
\(A=\left[\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\left[\dfrac{x-1-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right]\)
\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b,\(x=3-2\sqrt{2}=2-2\sqrt{2}+1=\left(\sqrt{2}-1\right)^2\)
\(=>A=\dfrac{\sqrt{2}-3}{3\sqrt{2}-3}\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}-1>0\\\sqrt{x}-2>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x>1\\x>4\end{matrix}\right.\) \(\Leftrightarrow x>4\)
\(A=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(x-1\right)-\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)
\(=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b) Ta có \(x=3-2\sqrt{2}=2-2\sqrt{2}+1=\left(2-1\right)^2=1\)
Thay \(x=1\) vào \(A\), ta được:
\(A=\dfrac{\sqrt{1}-2}{3\sqrt{1}}=\dfrac{1-2}{3}=-\dfrac{1}{3}\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\notin\left\{1;4\right\}\end{matrix}\right.\)
\(A=\left(\dfrac{1}{x-4}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{\sqrt{x}-1}{x+2\sqrt{x}}\)
\(=\left(\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}-1}\)
\(=\dfrac{1+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}}{\sqrt{x}-1}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
b: Để A là số nguyên thì \(\sqrt{x}⋮\sqrt{x}-2\)
=>\(\sqrt{x}-2+2⋮\sqrt{x}-2\)
=>\(\sqrt{x}-2\inƯ\left(2\right)\)
=>\(\sqrt{x}-2\in\left\{1;-1;2;-2\right\}\)
=>\(\sqrt{x}\in\left\{3;1;4;0\right\}\)
=>\(x\in\left\{9;1;16;0\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{9;16\right\}\)
c: A<0
=>\(\dfrac{\sqrt{x}}{\sqrt{x}-2}< 0\)
=>\(\sqrt{x}-2< 0\)
=>\(\sqrt{x}< 2\)
=>0<=x<4
Kết hợp ĐKXĐ, ta được: 0<x<4 và x<>1
a)ĐKXĐ:\(\begin{cases}x\ge0\\2\sqrt{x}-2\ne0\\1-x\ne0\\\end{cases}\)
`<=>` \(\begin{cases}x\ge0\\x\ne1\\\end{cases}\)
`B=1/(2sqrtx-2)-1/(2sqrtx+2)+sqrtx/(1-x)`
`=1/(2(sqrtx-1))-1/(2(sqrtx+1))-sqrtx/(x-1)`
`=(sqrtx+1-(sqrtx-1)-2sqrtx)/(2(sqrtx-1)(sqrtx+1))`
`=(2-2sqrtx)/(2(sqrtx-1)(sqrtx+1))`
`=(2(1-sqrtx))/(2(sqrtx-1)(sqrtx+1))`
`=-1/(sqrtx+1)`
`b)x=3`
`=>B=(-1)/(sqrt3+1)`
`=(-(sqrt3-1))/(3-1)`
`=(1-sqrt3)/2`
`c)|A|=1/2`
`<=>|(-1)/(sqrtx+1)|=1/2`
`<=>|1/(sqrtx+1)|=1/2`
`<=>1/(sqrtx+1)=1/2` do `1>0,sqrtx+1>=1>0`
`<=>sqrtx+1=2`
`<=>sqrtx=1`
`<=>x=1` loại vì `x ne 1`.
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có: \(B=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)
\(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-1}{\sqrt{x}+1}\)
b) Thay x=3 vào B, ta được:
\(B=\dfrac{-1}{\sqrt{3}+1}=\dfrac{-\sqrt{3}+1}{2}\)
c) Ta có: \(\left|A\right|=\dfrac{1}{2}\)
nên \(\left[{}\begin{matrix}A=\dfrac{1}{2}\\A=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{-1}{\sqrt{x}+1}=\dfrac{1}{2}\\\dfrac{-1}{\sqrt{x}+1}=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=-2\\\sqrt{x}+1=2\end{matrix}\right.\Leftrightarrow x=1\)(loại)
3:
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >9\end{matrix}\right.\)
\(M=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}+3-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{3}\)
\(=\dfrac{6}{3\left(\sqrt{x}+3\right)}=\dfrac{2}{\sqrt{x}+3}\)
b: M>1/3
=>M-1/3>0
=>\(\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{3}>0\)
=>\(\dfrac{6-\sqrt{x}-3}{3\left(\sqrt{x}+3\right)}>0\)
=>\(3-\sqrt{x}>0\)
=>\(\sqrt{x}< 3\)
=>0<=x<9
c: \(\sqrt{x}+3>=3\) với mọi x thỏa mãn ĐKXĐ
=>\(M=\dfrac{2}{\sqrt{x}+3}< =\dfrac{2}{3}\) với mọi x thỏa mãn ĐKXĐ
Dấu = xảy ra khi x=0
a. \(N=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{4\sqrt{x}}{3}\) \(\left(ĐKXĐ:x\ge0\right)\)
\(N=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\right).\dfrac{4\sqrt{x}}{3}\)
\(\text{}\text{}N=\dfrac{\sqrt{x}+1}{x\sqrt{x}+1}.\dfrac{4\sqrt{x}}{3}\)
\(N=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
b.\(N=\dfrac{8}{9}\Leftrightarrow\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\dfrac{8}{9}\)
\(\Leftrightarrow3\sqrt{x}=2x-2\sqrt{x}+2\)
\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=4\end{matrix}\right.\)
c.\(\dfrac{1}{N}>\dfrac{3\sqrt{x}}{4}\Leftrightarrow\dfrac{3\left(x-\sqrt{x}+1\right)}{4\sqrt{x}}>\dfrac{3\sqrt{x}}{4}\)
\(\Leftrightarrow x-\sqrt{x}+1>x\)
\(\Leftrightarrow x< 1\)
a: ĐKXĐ: \(x\ge0\)
Ta có: \(N=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right)\cdot\dfrac{4\sqrt{x}}{3}\)
\(=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\)
\(=\dfrac{4\sqrt{x}}{3x-3\sqrt{x}+3}\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right)\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\cdot\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
b) Để P>0 thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}>0\)
mà \(\sqrt{x}+1>0\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}\left(\sqrt{x}-1\right)>0\)
mà \(\sqrt{x}>0\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}-1>0\)
\(\Leftrightarrow\sqrt{x}>1\)
hay x>1
Kết hợp ĐKXĐ, ta được: x>1
Vậy: Để P>0 thì x>1
ĐKXĐ: \(x>0\)
\(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\left(\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right).\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
\(x=9\Rightarrow B=\dfrac{9+\sqrt{9}+1}{\sqrt{9}}=\dfrac{13}{3}\)
\(B=3\Rightarrow\dfrac{x+\sqrt{x}+1}{\sqrt{x}}=3\Rightarrow x+\sqrt{x}+1=3\sqrt{x}\)
\(\Rightarrow x-2\sqrt{x}+1=0\Rightarrow\left(\sqrt{x}-1\right)^2=0\Rightarrow x=1\)
a: Ta có: \(B=\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{x+\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b: Thay x=9 vào B, ta được:
\(B=\dfrac{9+3+1}{3}=\dfrac{13}{3}\)
d/ Ta có:
\(A=\left(-x+\sqrt{x}-\dfrac{1}{4}\right)+\dfrac{1}{4}\)
\(=\dfrac{1}{4}-\left(\sqrt{x}-\dfrac{1}{2}\right)^2\le\dfrac{1}{4}\)
Vậy GTLN là \(A=\dfrac{1}{4}\) đạt được tại \(x=\dfrac{1}{4}\)
b/ \(\sqrt{1x}-x\)
c/ Ta có:
x < 1
\(\Rightarrow\sqrt{x}< 1\)
\(\Rightarrow1-\sqrt{x}>0\)
Ta lại có: x > 0
\(\Rightarrow A=\sqrt{x}-x=\sqrt{x}\left(1-\sqrt{x}\right)>0\)