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\(21,\frac{2}{x-1}\le\frac{5}{2x-1}\left(x\ne1;x\ne\frac{1}{2}\right)\)
\(\Leftrightarrow\frac{2}{x-1}-\frac{5}{2x-1}\le0\)
\(\Leftrightarrow\frac{4x-2-5x+5}{\left(x-1\right)\left(2x-1\right)}\text{≤}0\)
\(\Leftrightarrow\frac{-x+3}{\left(x-1\right)\left(2x-1\right)}\text{≤}0\)
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/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{2x^2+5x+2}=2\sqrt{2x^2+5x-6}\)
\(\Leftrightarrow2x^2+5x+2=4\left(2x^2+5x-6\right)\)
\(\Leftrightarrow6x^2+15x-26=0\)
b/ ĐKXĐ: ...
Đặt \(\sqrt[5]{\frac{16x}{x-1}}=a\)
\(a+\frac{1}{a}=\frac{5}{2}\Leftrightarrow a^2-\frac{5}{2}a+1=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt[5]{\frac{16x}{x-1}}=2\\\sqrt[5]{\frac{16x}{x-1}}=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}16x=32\left(x-1\right)\\16x=\frac{1}{32}\left(x-1\right)\end{matrix}\right.\)
c/ĐKXĐ: ...
\(\Leftrightarrow x^2-2x-\sqrt{6x^2-12x+7}=0\)
Đặt \(\sqrt{6x^2-12x+7}=a\ge0\Rightarrow x^2-2x=\frac{a^2-7}{6}\)
\(\frac{a^2-7}{6}-a=0\Leftrightarrow a^2-6a-7=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=7\end{matrix}\right.\) \(\Rightarrow\sqrt{6x^2-12x+7}=7\)
\(\Leftrightarrow6x^2-12x-42=0\)
d/ \(\Leftrightarrow x^2+x+4-\sqrt{x^2+x+4}-2=0\)
Đặt \(\sqrt{x^2+x+4}=a>0\)
\(a^2-a-2=0\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=2\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+x+4}=2\Rightarrow x^2+x=0\)
e/ \(\Leftrightarrow x^2+2x+\sqrt{3x^2+6x+4}-2=0\)
Đặt \(\sqrt{3x^2+6x+4}=a>0\Rightarrow x^2+2x=\frac{a^2-4}{3}\)
\(\frac{a^2-4}{3}+a-2=0\)
\(\Leftrightarrow a^2+3a-10=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{3x^2+6x+4}=2\Rightarrow3x^2+6x=0\)
a/ \(\left(2x-3\right)\left(3x-4\right)\left(5x+2\right)>0\)
\(\Rightarrow\left[{}\begin{matrix}-\frac{2}{3}< x< \frac{4}{3}\\x>\frac{3}{2}\end{matrix}\right.\)
b/ \(\Leftrightarrow24x^2-10x-25< 0\)
\(\Rightarrow-\frac{5}{6}< x< \frac{5}{4}\)
c/ \(\frac{4x\left(3x+2\right)}{2x+5}>0\Rightarrow\left[{}\begin{matrix}-\frac{5}{2}< x< -\frac{2}{3}\\x>0\end{matrix}\right.\)
d/ \(\Leftrightarrow\frac{3x+2}{2x-5}-\frac{2x-5}{3x+2}\ge0\)
\(\Leftrightarrow\frac{\left(3x+2\right)^2-\left(2x-5\right)^2}{\left(2x-5\right)\left(3x+2\right)}\ge0\)
\(\Leftrightarrow\frac{\left(5x-2\right)\left(x+7\right)}{\left(2x-5\right)\left(3x+2\right)}\ge0\Rightarrow\left[{}\begin{matrix}x\le-7\\-\frac{2}{3}< x\le\frac{2}{5}\\x>\frac{5}{2}\end{matrix}\right.\)
Sửa đề: \(3x^5-13x^4+16x^2+5x^2-21x+6=0\)