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10 tháng 7 2021
a) Ta có: \(M=\left(\dfrac{\sqrt{x}+1}{\sqrt{2x}+1}+\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}-1\right):\left(1+\dfrac{\sqrt{x}}{\sqrt{2x}+1}-\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{2x}-1\right)+\sqrt{x}\left(\sqrt{2x}+1\right)^2-2x+1}{\left(\sqrt{2x}+1\right)\left(\sqrt{2x}-1\right)}\right):\left(\dfrac{2x-1+\sqrt{x}\left(\sqrt{2x}-1\right)-\sqrt{x}\left(\sqrt{2x}+1\right)^2}{\left(\sqrt{2x}+1\right)\left(\sqrt{2x}-1\right)}\right)\)
\(=\dfrac{x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1+\sqrt{x}\left(2x+2\sqrt{2x}+1\right)-2x+1}{2x-1+x\sqrt{2}-\sqrt{x}-\sqrt{x}\left(2x+2\sqrt{2x}+1\right)}\)
\(=\dfrac{x\sqrt{2}-\sqrt{x}+\sqrt{2x}-2x+2x\sqrt{x}+2\sqrt{2x}+\sqrt{x}}{2x-1+x\sqrt{2}-\sqrt{x}-2x\sqrt{x}-2\sqrt{2x}-\sqrt{x}}\)
\(=\dfrac{x\sqrt{2}+3\sqrt{2x}-2x+2x\sqrt{x}}{x\sqrt{2}-2\sqrt{2x}+2x-2\sqrt{x}-2x\sqrt{x}}\)
17 tháng 7 2021
a) Ta có: \(P=\left(\dfrac{\sqrt{x}+1}{\sqrt{2x}-1}+\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}+1}-1\right):\left(1+\dfrac{\sqrt{x}+1}{\sqrt{2x}+1}-\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right)\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{2x}+1\right)+\left(\sqrt{2x}+\sqrt{x}\right)\left(\sqrt{2x}-1\right)-2x+1}{\left(\sqrt{2x}-1\right)\left(\sqrt{2x}+1\right)}:\left(\dfrac{2x-1+\left(\sqrt{x}+1\right)\left(\sqrt{2x}-1\right)-\left(\sqrt{2x}+\sqrt{x}\right)\left(\sqrt{2x}+1\right)}{\left(\sqrt{2x}-1\right)\left(\sqrt{2x}+1\right)}\right)\)
\(=\dfrac{x\sqrt{2}+\sqrt{x}+\sqrt{2x}+1+2x-\sqrt{2x}+x\sqrt{2}+\sqrt{x}-2x+1}{2x-1}:\dfrac{2x-1+x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1-\left(2x+\sqrt{2x}+x\sqrt{2}+\sqrt{x}\right)}{2x-1}\)
\(=\dfrac{2x\sqrt{2}+2\sqrt{x}+2}{-2-2\sqrt{x}}\)
27 tháng 8 2021
a) \(\sqrt{2x-1}=3\left(đk:x\ge\dfrac{1}{2}\right)\)
\(\Leftrightarrow2x-1=9\Leftrightarrow2x=10\Leftrightarrow x=5\)(thỏa đk)
b) \(\sqrt{1-3x}=\dfrac{1}{2}\left(đk:x\le\dfrac{1}{3}\right)\)
\(\Leftrightarrow1-3x=\dfrac{1}{4}\Leftrightarrow3x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{4}\)(thỏa đk)
c) \(\sqrt{\left(x-1\right)^2}=\dfrac{1}{2}\)
\(\Leftrightarrow\left|x-1\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{2}\\x-1=-\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
d) \(\sqrt{\left(1+2x\right)^2}=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left|1+2x\right|=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}1+2x=\dfrac{\sqrt{3}}{2}\\1+2x=-\dfrac{\sqrt{3}}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2+\sqrt{3}}{4}\\x=-\dfrac{2+\sqrt{3}}{4}\end{matrix}\right.\)
e) \(\sqrt{\left(1-2x\right)^2}=\left|x-1\right|\)
\(\Leftrightarrow\left|1-2x\right|=\left|x-1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}1-2x=x-1\\1-2x=1-x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=0\end{matrix}\right.\)
8 tháng 7 2022
\(A=\dfrac{\sqrt{x}+1}{\sqrt{2x}+1}+\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}-1\)
\(=\dfrac{x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1+2x+\sqrt{2x}+x\sqrt{2}+\sqrt{x}}{2x-1}-1\)
\(=\dfrac{2x\sqrt{2}+2\sqrt{2x}-1+2x-2x+1}{2x-1}=\dfrac{2x\sqrt{x}+2\sqrt{2x}}{2x-1}\)
\(B=\left(1+\dfrac{\sqrt{x}+1}{\sqrt{2x}+1}-\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right)\)
\(=1+\dfrac{x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1-2x-\sqrt{2x}-x\sqrt{2}-\sqrt{x}}{2x-1}\)
\(=1+\dfrac{-2\sqrt{x}-1-2x}{2x-1}\)
\(=\dfrac{2x-1-2\sqrt{x}-1-2x}{2x-1}=\dfrac{-2-2\sqrt{x}}{2x-1}\)
\(P=A:B=\dfrac{2x\sqrt{x}+2\sqrt{2x}}{2x-1}:\dfrac{-2\sqrt{x}-2}{2x-1}\)
\(=\dfrac{2\sqrt{x}\left(x+\sqrt{2}\right)}{2x-1}\cdot\dfrac{2x-1}{-2\left(\sqrt{x}+1\right)}=\dfrac{-\sqrt{x}\left(x+\sqrt{2}\right)}{\sqrt{x}+1}\)
b: Thay \(\sqrt{x}=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{2}\) vào P, ta được:
\(P=\left[-\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{2}\cdot\left(\dfrac{3+2\sqrt{2}}{2}+\sqrt{2}\right)\right]:\left[\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{2}+1\right]\)
\(=\left[\dfrac{-\sqrt{2}\left(\sqrt{2}+1\right)}{2}\cdot\dfrac{3+4\sqrt{2}}{2}\right]:\left[\dfrac{2+\sqrt{2}+2}{2}\right]\)
\(=\dfrac{-\sqrt{2}\left(\sqrt{2}+1\right)\left(4\sqrt{2}+3\right)}{4}\cdot\dfrac{2}{4+\sqrt{2}}\)
\(=\dfrac{-\left(\sqrt{2}+1\right)\left(4\sqrt{2}+3\right)}{2\cdot\left(2\sqrt{2}+1\right)}=\dfrac{-\left(4\sqrt{2}+3\right)}{3\cdot\left(3+\sqrt{2}\right)}\)
6 tháng 10 2018
\(A=\left(\dfrac{4x+4}{2\sqrt{2x^3}-8}-\dfrac{\sqrt{2x}}{2x+2\sqrt{2x}+4}\right)\)\(\left(\dfrac{1+2\sqrt{2x^3}}{1+\sqrt{2x}}\right)\)
\(=\left[\dfrac{4x+4-\sqrt{2x}\left(\sqrt{2x}-2\right)}{\left(\sqrt{2x}-2\right)\left(2x+2\sqrt{2x}+4\right)}\right]\)\(.\dfrac{\left(1+\sqrt{2x}\right)\left(2x-2\sqrt{2x}+4\right)}{1+\sqrt{2x}}\)
Làm tiếp nhé :>>
AH
Akai Haruma
Giáo viên
3 tháng 8 2021
a. ĐKXĐ: $x\geq 1$
PT $\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{3}{2}.\sqrt{9}.\sqrt{x-1}+24.\sqrt{\frac{1}{64}}.\sqrt{x-1}=-17$
$\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17$
$\Leftrightarrow -\sqrt{x-1}=-17$
$\Leftrightarrow \sqrt{x-1}=17$
$\Leftrightarrow x-1=289$
$\Leftrightarrow x=290$
b. ĐKXĐ: $x\geq \frac{1}{2}$
PT $\Leftrightarrow \sqrt{9}.\sqrt{2x-1}-0,5\sqrt{2x-1}+\frac{1}{2}.\sqrt{25}.\sqrt{2x-1}+\sqrt{49}.\sqrt{2x-1}=24$
$\Leftrightarrow 3\sqrt{2x-1}-0,5\sqrt{2x-1}+2,5\sqrt{2x-1}+7\sqrt{2x-1}=24$
$\Leftrightarrow 12\sqrt{2x-1}=24$
$\Leftrihgtarrow \sqrt{2x-1}=2$
$\Leftrightarrow x=2,5$ (tm)
AH
Akai Haruma
Giáo viên
3 tháng 8 2021
c. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{36}.\sqrt{x-2}-15\sqrt{\frac{1}{25}}\sqrt{x-2}=4(5+\sqrt{x-2})$
$\Leftrightarrow 6\sqrt{x-2}-3\sqrt{x-2}=20+4\sqrt{x-2}$
$\Leftrightarrow \sqrt{x-2}=-20< 0$ (vô lý)
Vậy pt vô nghiệm
26 tháng 6 2021
a)đk:`2x-4>=0`
`<=>2x>=4`
`<=>x>=2.`
b)đk:`3/(-2x+1)>=0`
Mà `3>0`
`=>-2x+1>=0`
`<=>1>=2x`
`<=>x<=1/2`
c)`đk:(-3x+5)/(-4)>=0`
`<=>(3x-5)/4>=0`
`<=>3x-5>=0`
`<=>3x>=5`
`<=>x>=5/3`
d)`đk:-5(-2x+6)>=0`
`<=>-2x+6<=0`
`<=>2x-6>=0`
`<=>2x>=6`
`<=>x>=3`
e)`đk:(x^2+2)(x-3)>=0`
Mà `x^2+2>=2>0`
`<=>x-3>=0`
`<=>x>=3`
f)`đk:(x^2+5)/(-x+2)>=0`
Mà `x^2+5>=5>0`
`<=>-x+2>0`
`<=>-x>=-2`
`<=>x<=2`
26 tháng 6 2021
a, ĐKXĐ : \(2x-4\ge0\)
\(\Leftrightarrow x\ge\dfrac{4}{2}=2\)
Vậy ..
b, ĐKXĐ : \(\left\{{}\begin{matrix}\dfrac{3}{-2x+1}\ge0\\-2x+1\ne0\end{matrix}\right.\)
\(\Leftrightarrow-2x+1>0\)
\(\Leftrightarrow x< \dfrac{1}{2}\)
Vậy ..
c, ĐKXĐ : \(\dfrac{-3x+5}{-4}\ge0\)
\(\Leftrightarrow-3x+5\le0\)
\(\Leftrightarrow x\ge\dfrac{5}{3}\)
Vậy ...
d, ĐKXĐ : \(-5\left(-2x+6\right)\ge0\)
\(\Leftrightarrow-2x+6\le0\)
\(\Leftrightarrow x\ge-\dfrac{6}{-2}=3\)
Vậy ...
e, ĐKXĐ : \(\left(x^2+2\right)\left(x-3\right)\ge0\)
\(\Leftrightarrow x-3\ge0\)
\(\Leftrightarrow x\ge3\)
Vậy ...
f, ĐKXĐ : \(\left\{{}\begin{matrix}\dfrac{x^2+5}{-x+2}\ge0\\-x+2\ne0\end{matrix}\right.\)
\(\Leftrightarrow-x+2>0\)
\(\Leftrightarrow x< 2\)
Vậy ...
\(A=\dfrac{2\sqrt{2x^3}+1}{\sqrt{2x}+1}-\sqrt{2x}\left(\sqrt{2x}-1\right)=\dfrac{\sqrt{8x^3}+1}{\sqrt{2x}+1}-\sqrt{2x}\left(\sqrt{2x}-1\right)\)
\(=\dfrac{\left(\sqrt{2x}+1\right)\left(2x-\sqrt{2x}+1\right)}{\sqrt{2x}+1}-2x+\sqrt{2x}\)
\(=2x-\sqrt{2x}+1-2x+\sqrt{2x}=1\)
Vậy A=1