Question

[Nguyễn Khánh Linh]

Bài 3 : giải các bất phương trình sau

21 , \(\frac{2}{x-1}\le\frac{5}{2x-1}\)

22, \(\frac{-4}{3x+1}< \frac{3}{2-x}\)

23 , \(\frac{2x^2+x}{1-2x}\ge1-x\)

24 , \(\frac{2x-5}{3x+2}< \frac{3x+2}{2x-5}\)

25 , \(2x^2-5x+2< 0\)

26 , \(-5x^2+4x+12< 0\)

27 , \(16x^2+40x+25>0\)

28 , \(-2x^2+3x-7\ge0\)

29 , \(3x^2-4x+4\ge0\)

30 , \(x^2-x-6\le0\)

Answer 1

[lê thị hương giang]

\(21,\frac{2}{x-1}\le\frac{5}{2x-1}\left(x\ne1;x\ne\frac{1}{2}\right)\)

\(\Leftrightarrow\frac{2}{x-1}-\frac{5}{2x-1}\le0\)

\(\Leftrightarrow\frac{4x-2-5x+5}{\left(x-1\right)\left(2x-1\right)}\text{≤}0\)

\(\Leftrightarrow\frac{-x+3}{\left(x-1\right)\left(2x-1\right)}\text{≤}0\)

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

\(30,x^2-x-6\le0\)

\(\left\{{}\begin{matrix}x^2-x-6=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\a=1>0\end{matrix}\right.\)

\(\Rightarrow-2\le x\le3\)