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1. \(\Leftrightarrow\left(3x-1\right)\left(\sqrt{5}x-2\right)\ge0\Rightarrow\left[{}\begin{matrix}x\le\frac{1}{3}\\x\ge\frac{2}{\sqrt{5}}\end{matrix}\right.\)
2. \(\Leftrightarrow\frac{\left(3-2x\right)\left(3+2x\right)}{2x-3}\ge0\Leftrightarrow\left[{}\begin{matrix}x\ne\frac{3}{2}\\x\le-\frac{3}{2}\end{matrix}\right.\)
3. \(\left|x-2\right|\ge3\Leftrightarrow\left[{}\begin{matrix}x-2\ge3\\x-2\le-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x\ge5\\x\le-1\end{matrix}\right.\)
4. \(\Leftrightarrow-10\le3x+1\le10\Rightarrow-\frac{11}{3}\le x\le3\)
5. \(\Leftrightarrow\frac{3x^2-x+2}{x^2-9}-3\le0\Leftrightarrow\frac{-x+29}{\left(x-3\right)\left(x+3\right)}\le0\Rightarrow\left[{}\begin{matrix}-3< x< 3\\x\ge29\end{matrix}\right.\)
6. \(\Leftrightarrow\frac{4}{\left(x-2\right)^2}+\frac{1}{x-2}>0\Leftrightarrow\frac{x+2}{\left(x-2\right)^2}\ge0\Rightarrow\left[{}\begin{matrix}x\ge-2\\x\ne2\end{matrix}\right.\)
a/ ĐKXĐ: ...
\(\Leftrightarrow2\left(x^2-5x-6\right)+\sqrt{x^2-5x-6}-3=0\)
Đặt \(\sqrt{x^2-5x-6}=a\ge0\)
\(2a^2+a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2-5x-6}=1\Leftrightarrow x^2-5x-7=0\)
b/ ĐKXĐ: ...
\(\Leftrightarrow5\sqrt{3x^2-4x-2}-2\left(3x^2-4x-2\right)+3=0\)
Đặt \(\sqrt{3x^2-4x-2}=a\ge0\)
\(-2a^2+5a+3=0\) \(\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{1}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{3x^2-4x-2}=3\Leftrightarrow3x^2-4x-11=0\)
c/ \(\Leftrightarrow x^2+2x-6+\sqrt{2x^2+4x+3}=0\)
Đặt \(\sqrt{2x^2+4x+3}=a>0\Rightarrow x^2+2x=\frac{a^2-3}{2}\)
\(\frac{a^2-3}{2}-6+a=0\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2x^2+4x+3}=3\Leftrightarrow2x^2+4x-6=0\)
d/ ĐKXĐ: ...
Đặt \(\sqrt{\frac{3x-1}{x}}=a>0\)
\(2a=\frac{1}{a^2}+1\Leftrightarrow2a^3-a^2-1=0\)
\(\Leftrightarrow\left(a-1\right)\left(2a^2+a+1\right)=0\)
\(\Rightarrow a=1\Rightarrow\sqrt{\frac{3x-1}{x}}=1\Leftrightarrow3x-1=x\)
e/ĐKXĐ: ...
\(\Leftrightarrow2\sqrt{\frac{6x-1}{x}}=\frac{x}{6x-1}+1\)
Đặt \(\sqrt{\frac{6x-1}{x}}=a>0\)
\(2a=\frac{1}{a^2}+1\Leftrightarrow2a^3-a^2-1=0\Leftrightarrow\left(a-1\right)\left(2a^2+a+1\right)=0\)
\(\Rightarrow a=1\Rightarrow\sqrt{\frac{6x-1}{x}}=1\Rightarrow6x-1=x\)
f/ ĐKXĐ: ...
Đặt \(\sqrt{\frac{x}{2x-1}}=a>0\)
\(\frac{1}{a}+1+a=3a^2\)
\(\Leftrightarrow3a^3-a^2-a-1=0\)
\(\Leftrightarrow\left(a-1\right)\left(3a^2+2a+1\right)=0\)
\(\Leftrightarrow a=1\Rightarrow\sqrt{\frac{x}{2x-1}}=1\Rightarrow x=2x-1\)
1a/\(3x+5=12-3+4x\)
\(\Leftrightarrow x=-4\)
b/\(đk:x\ne\pm3\)
\(\left(x+2\right)\left(x+3\right)+\left(x-4\right)\left(x-3\right)=18-3x\)
<=>2x^2+x=0
<=> x =-1/2 v x = 0
Vậy ...
2/ (6+x)^2 = 49
=> 6+x = 7
=> x =1