Cho biểu thức:
A=\(\left(\frac{2\sqrt{x}+x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
a) tìm đkxđ để A có nghĩa
b) rút gọn A
c) tìm\(\sqrt{A}khi\) \(x=4+2\sqrt{3}\)
K
Khách
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29 tháng 6 2021
`a)ĐK:` \(\begin{cases}x \ge 0\\x-\sqrt{x} \ne 0\\x-1 \ne 0\\\end{cases}\)
`<=>` \(\begin{cases}x \ge 0\\x \ne 0\\x \ne 1\\\end{cases}\)
`<=>` \(\begin{cases}x>0\\x \ne 1\\\end{cases}\)
`b)A=(sqrtx/(sqrtx-1)-1/(x-sqrtx)):(1/(1+sqrtx)+2/(x-1))`
`=((x-1)/(x-sqrtx)):((sqrtx-1+2)/(x-1))`
`=(x-1)/(x-sqrtx):(sqrtx+1)/(x-1)`
`=(sqrtx+1)/sqrtx:1/(sqrtx-1)`
`=(x-1)/sqrtx`
`c)A>0`
Mà `sqrtx>0AAx>0`
`<=>x-1>0<=>x>1`
29 tháng 6 2021
a, ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
b, Ta có : \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\left(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}:\dfrac{1}{\sqrt{x}-1}=\dfrac{x-1}{\sqrt{x}}\)
c, Ta có : \(A>0\)
\(\Leftrightarrow x-1>0\)
\(\Leftrightarrow x>1\)
Vậy ...
10 tháng 8 2021
Bài 1 : Với : \(x>0;x\ne1\)
\(P=\left(1+\frac{1}{\sqrt{x}-1}\right)\frac{1}{x-\sqrt{x}}=\left(\frac{\sqrt{x}}{\sqrt{x}-1}\right).\sqrt{x}\left(\sqrt{x}-1\right)=x\)
Thay vào ta được : \(P=x=25\)
10 tháng 8 2021
Bài 2 :
a, Với \(x\ge0;x\ne1\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}=\frac{x+\sqrt{x}-2\sqrt{x}+2-2}{x-1}\)
\(=\frac{x-\sqrt{x}}{x-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)
5 tháng 2 2022
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
14 tháng 7 2016
a) ĐKXĐ : \(0\le x\ne4\)
b) \(A=\left(\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{\sqrt{x}}{2-\sqrt{x}}+\frac{4\sqrt{x}-1}{x-4}\right):\frac{1}{x-4}\)
\(=\left[\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{4\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right].\left(x-4\right)\)
\(=\frac{x-2\sqrt{x}-x-2\sqrt{x}+4\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
\(=\frac{-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)=-1\)
13 tháng 7 2016
\(A=\left[\frac{\left(\sqrt{x}-2\right)\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{4\sqrt{x}-1}{x-4}\right]:\frac{1}{x-4}\)
\(=\frac{x-2\sqrt{x}-x-2\sqrt{x}+4\sqrt{x}-1}{x-4}.\left(x-4\right)\)=\(=\frac{-1}{x-4}.\left(x-4\right)=-1\)
Vậy giá trị của A thỏa mãn mọi x và rút gọn lại còn -1
NV
14 tháng 7 2016
1/
a/ ĐKXĐ: \(x\ge0\) và \(x\ne\frac{1}{9}\)
b/ \(P=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\frac{3\sqrt{x}+1}{3}\)
\(=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}.\frac{1}{3}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c/ \(P=\frac{6}{5}\Rightarrow\frac{x+\sqrt{x}}{3\sqrt{x}-1}=\frac{6}{5}\Rightarrow6\left(3\sqrt{x}-1\right)=5\left(x+\sqrt{x}\right)\)
\(\Rightarrow5x-13\sqrt{x}+6=0\Rightarrow\left(5\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{3}{5}\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}}\)
Vậy x = 9/25 , x = 4
14 tháng 7 2016
1) a) ĐKXĐ : \(0\le x\ne\frac{1}{9}\)
b) \(P=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
\(=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}+\frac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]:\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}=\frac{3x+3\sqrt{x}}{3\left(3\sqrt{x}-1\right)}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c) \(P=\frac{6}{5}\Leftrightarrow18\sqrt{x}-6=5x+5\sqrt{x}\Leftrightarrow5x-13\sqrt{x}+6=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}\)
24 tháng 7 2017
a. ĐKXĐ \(x\ge0\)và \(x\ne9\)
Ta có \(K=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{3x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(x-2\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
b. Để \(K< -1\Rightarrow\frac{3\sqrt{x}-9+\sqrt{x}+3}{\sqrt{x}+3}< 0\Rightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\Rightarrow4\sqrt{x}-6< 0\)vì \(\sqrt{x}+3\ge3\)
\(\Rightarrow0\le x< \frac{9}{4}\left(tm\right)\)
Vậy với \(0\le x< \frac{9}{4}\)thì K<-1
c. \(K=\frac{3\sqrt{x}-9}{\sqrt{x}+3}=3+\frac{-18}{\sqrt{x}+3}\)
Ta có \(\sqrt{x}+3\ge3\Rightarrow\frac{1}{\sqrt{x}+3}\le\frac{1}{3}\Rightarrow-\frac{18}{\sqrt{x}+3}\ge-6\Rightarrow3+\frac{-18}{\sqrt{x}+3}\ge-3\)
\(\Rightarrow K\ge-3\)
Vậy \(MinK=-3\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)