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a: ĐKXĐ: \(\left[{}\begin{matrix}x\ge3\\x\le2\end{matrix}\right.\)
b: ĐKXĐ: \(\left[{}\begin{matrix}x>\dfrac{2\sqrt{14}}{7}\\x< -\dfrac{2\sqrt{14}}{7}\end{matrix}\right.\)
c: ĐKXĐ: \(x=\dfrac{1}{3}\)
d: ĐKXĐ: \(-\dfrac{2}{3}< x\le\sqrt{3}\)
\(\frac{5x-3}{2x}+\sqrt{3x+y}xđ\Leftrightarrow\hept{\begin{cases}2x\ne0\\3x+y\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ge-\frac{y}{3}\end{cases}}}\)
\(\sqrt{3x-1}+\frac{5x}{\sqrt{x+3}}xđ\Leftrightarrow\hept{\begin{cases}3x-1\ge0\\x+3>0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ge\frac{1}{3}\\x>-3\end{cases}\Rightarrow x\ge\frac{1}{3}}\)
1: \(\Leftrightarrow\dfrac{3x-1}{x+2}=4\)
=>4x+8=3x-1
=>x=-9
2: \(\Leftrightarrow\dfrac{5x-7}{2x-1}=4\)
=>8x-4=5x-7
=>3x=-3
=>x=-1
3: ĐKXD: x>=0
\(PT\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
=>\(x+\sqrt{x}-6=x-1\)
=>căn x=-1+6=5
=>x=25
4: ĐKXĐ: x>=0
PT =>\(\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
=>x-2*căn x-3=x-4
=>-2căn x-3=-4
=>2căn x+3=4
=>2căn x=1
=>căn x=1/2
=>x=1/4
a) ĐKXĐ: \(10-5x< 0\Leftrightarrow5x>10\Leftrightarrow x>2\)
b) ĐKXĐ: \(7-3x>0\Leftrightarrow3x< 7\Leftrightarrow x< \dfrac{7}{3}\)
c) ĐKXĐ: \(-5-2x\ge0\Leftrightarrow2x\le-5\Leftrightarrow x\le-\dfrac{5}{2}\)
a)ĐK:`3x-6>=0`
`<=>3x>=6<=>x>=2`
b)ĐK:`-3x+9>=0`
`<=>-3x>=-9`
`<=>x<=3`
c)ĐK:`(-5)/(-3x+2)>=0(x ne -2/3)`
Vì `-5<0`
`<=>-3x+2<0`
`<=>-3x<-2`
`<=>x>2/3`
e)ĐK:`(5x-3)/(-4)>=0`
MÀ `-4<0`
`<=>5x-3<=0`
`<=>5x<=3`
`<=>x<=3/5`
1: ĐKXĐ: x>1/2
=>\(\dfrac{x}{\sqrt{2x-1}}+\dfrac{x}{\sqrt[4]{4x-3}}=2\)
x^2-2x+1>=0
=>x^2>=2x-1
=>\(\dfrac{x}{\sqrt{2x-1}}>=1\)
Dấu = xảy ra khi x=1
(x^2-2x+1)(x^2+2x+3)>=0
=>x^4-4x+3>=0
=>x^4>=4x-3
=>\(\dfrac{x}{\sqrt[4]{4x-3}}>=1\)
=>VT>=2
Dấu = xảy ra khi x=1
2: 4x-1=x+x+2x-1
5x-2=x+2x-1+2x-1
\(\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{2x-1}}\right)\left(\sqrt{x}+\sqrt{x}+\sqrt{2x-1}\right)>=9\)
=>\(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{2x-1}}>=\dfrac{9}{\sqrt{x}+\sqrt{x}+\sqrt{2x-1}}\)
\(\left(\sqrt{x}+\sqrt{x}+\sqrt{2x-1}\right)^2< =3\left(4x-1\right)\)
=>\(\sqrt{x}+\sqrt{x}+\sqrt{2x-1}< =\sqrt{3\left(4x-1\right)}\)
=>\(\dfrac{2}{\sqrt{x}}+\dfrac{1}{\sqrt{2x-1}}>=\dfrac{3\sqrt{3}}{\sqrt{4x-1}}\)
Tương tự, ta cũng có: \(\dfrac{1}{\sqrt{x}}+\dfrac{2}{\sqrt{2x-1}}>=\dfrac{3\sqrt{3}}{\sqrt{5x-2}}\)
=>\(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{2x-1}}>=\sqrt{3}\left(\dfrac{1}{\sqrt{4x-1}}+\dfrac{1}{\sqrt{5x-2}}\right)\)
Dấu = xảy ra khi x=1
a: ĐKXĐ: x>=2/3
\(\dfrac{x-2}{\sqrt{3x-2}+2}=9\)
=>\(x-2=9\sqrt{3x-2}+18\)
=>\(9\sqrt{3x-2}=x-2-18=x-20\)
=>\(\Leftrightarrow\left\{{}\begin{matrix}x>=20\\81\left(3x-2\right)=x^2-40x+400\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=20\\x^2-40x+400-243x+162=0\end{matrix}\right.\)
=>x>=20 và x^2-283x+562=0
=>x=281(nhận) hoặc x=2(loại)
b: ĐKXĐ: x>=2/5
\(\sqrt{5x-2}=9\)
=>5x-2=81
=>5x=83
=>x=83/5
c: ĐKXĐ: x>=-1; x<>8
\(\dfrac{2x-16}{\sqrt{x+1}-3}=5\)
=>\(2x-16=5\sqrt{x+1}-15\)
=>\(\sqrt{25x+25}=2x-16+15=2x-1\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\4x^2-4x+1=25x+25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\4x^2-29x-24=0\end{matrix}\right.\)
=>x=8(nhận) hoặc x=-3/4(loại)
1: ĐKXĐ: 2x<>0; 3x+y>=0
=>x>=-y/3; x<>0
2: ĐKXĐ: 3x-1>=0; x+3>0
=>x>=1/3