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1, \(\frac{3x-4}{x-2}>1\\ \frac{3\left(x-2\right)}{x-2}+\frac{2}{x-2}>1\\ 3+\frac{2}{x-2}>1\\ \frac{2}{x-2}>-2\\ \frac{1}{x-2}>-1\)
\(x-2< -1\\ x< 1\)
a/ ĐKXĐ: ...
\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)
\(\Rightarrow x+\frac{1}{4x}=a^2-1\)
Pt trở thành:
\(3a=2\left(a^2-1\right)-7\)
\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=3\)
\(\Leftrightarrow2x-6\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)
b/ ĐKXĐ:
\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)
\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
d/ ĐKXĐ: ...
\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)
\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)
\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)
\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)
\(\Leftrightarrow4x^2-17x+4=0\)
\(21,\frac{2}{x-1}\le\frac{5}{2x-1}\left(x\ne1;x\ne\frac{1}{2}\right)\)
\(\Leftrightarrow\frac{2}{x-1}-\frac{5}{2x-1}\le0\)
\(\Leftrightarrow\frac{4x-2-5x+5}{\left(x-1\right)\left(2x-1\right)}\text{≤}0\)
\(\Leftrightarrow\frac{-x+3}{\left(x-1\right)\left(2x-1\right)}\text{≤}0\)
x -x+3 x-1 2x-1 VT -∞ +∞ 1/2 1 3 0 0 0 | | || | | || | | 0 - + + + + + - - - + + + + + + - -
Vậy \(\frac{-x+3}{\left(x-1\right)\left(2x-1\right)}\le0\Leftrightarrow x\in\left(\frac{1}{2};1\right)\cup[3;+\text{∞})\)
23,24 tương tự 21
\(25,2x^2-5x+2< 0\) (1)
Ta có: \(\left\{{}\begin{matrix}2x^2-5x+2=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{2}\end{matrix}\right.\\a=2>0\end{matrix}\right.\) \(\Leftrightarrow\frac{1}{2}< x< 2\)
\(26,-5x^2+4x+12< 0\)
\(\left\{{}\begin{matrix}-5x^2+4x+12=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\frac{6}{5}\end{matrix}\right.\\a=-5< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>2\\x< -\frac{6}{5}\end{matrix}\right.\)
\(27,16x^2+40x+25>0\)
\(\left\{{}\begin{matrix}16x^2+40x+25=0\Leftrightarrow x=-\frac{5}{4}\\a=16>0\end{matrix}\right.\)
\(\Leftrightarrow x\ne-\frac{5}{4}\)
\(28,-2x^2+3x-7\ge0\)
\(\left\{{}\begin{matrix}-2x^2+3x-7=0\left(vo.nghiem\right)\\a=-2< 0\end{matrix}\right.\)
\(\Rightarrow-2x^2+3x-7< 0\) ∀x
=> bpt vô nghiệm
\(29,3x^2-4x+4\ge0\)
\(\left\{{}\begin{matrix}3x^2-4x+4=0\left(vo.nghiem\right)\\a=3>0\end{matrix}\right.\)
=> \(3x^2-4x+4>0\) => bpt vô số nghiệm
\(30,x^2-x-6\le0\)
\(\left\{{}\begin{matrix}x^2-x-6=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\a=1>0\end{matrix}\right.\)
\(\Rightarrow-2\le x\le3\)
a/
\(\frac{3x-4}{x-2}-1>0\Leftrightarrow\frac{2x-2}{x-2}>0\Rightarrow\left[{}\begin{matrix}x>2\\x< 1\end{matrix}\right.\)
b/
\(\frac{2x-5}{2-x}+1\le0\Rightarrow\frac{x-3}{2-x}\le0\Rightarrow\left[{}\begin{matrix}x\ge3\\x< 2\end{matrix}\right.\)
c/
\(\frac{x^2+x-3}{x^2-4}-1\le0\Rightarrow\frac{x+1}{x^2-4}\le0\Rightarrow\frac{x+1}{\left(x-2\right)\left(x+2\right)}\le0\Rightarrow\left[{}\begin{matrix}x< -2\\-1\le x< 2\end{matrix}\right.\)
d/
\(\frac{4x^2-8x+6+x^2-x-6}{2\left(x^2-x-6\right)}>0\Rightarrow\frac{x\left(5x-9\right)}{2\left(x+2\right)\left(x-3\right)}>0\Rightarrow\left[{}\begin{matrix}x>3\\0< x< \frac{9}{5}\\x< -2\end{matrix}\right.\)
e/
\(\frac{x^2+3x+2}{2x+3}-\frac{2x-5}{4}\ge0\Rightarrow\frac{4x^2+12x+8-\left(2x-5\right)\left(2x+3\right)}{4\left(2x+3\right)}\ge0\)
\(\Rightarrow\frac{28x+23}{4\left(2x+3\right)}\ge0\Rightarrow\left[{}\begin{matrix}x\ge-\frac{23}{28}\\x< -\frac{3}{2}\end{matrix}\right.\)