Thay x=2 vào biểu thức

\(x^3+4x^2-3x-18=2^3+4.2^2-3.2-18=8+16-6-18=0\)

Do x=2 cho ta \(x^3+4x^2-3x-18=0\) nên với mọi x lớn hơn hoặc bằng 2 ta luôn thu đc biểu thức lớn hơn hoặc bằng 0

\(x^3+4x^2-3x-18\ge0\)

\(\Leftrightarrow x^3+6x^2+9x-2x^2-12x-18\ge0\)

\(\Leftrightarrow x\left(x^2+6x+9\right)-2\left(x^2+6x+9\right)\ge0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+6x+9\right)\ge0\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)^2\ge0\)

Từ \(\left\{{}\begin{matrix}x\ge2\Rightarrow x-2\ge0\\\left(x+3\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left(x-2\right)\left(x+3\right)^2\ge0\forall x\ge2\) (Đúng !!)

a.

\(x>5\)

b.

\(\Leftrightarrow3x^2-4x+1>0\)

\(\Leftrightarrow\left(x-1\right)\left(3x-1\right)>0\Rightarrow\left[{}\begin{matrix}x>1\\x< \frac{1}{3}\end{matrix}\right.\)

c.

\(\Leftrightarrow\frac{3\left(x-2\right)}{\left(x+2\right)\left(x+5\right)}>0\Rightarrow\left[{}\begin{matrix}-5< x< -2\\x\ge2\end{matrix}\right.\)

a. ĐKXĐ : \(x\ne\frac{1}{2};\frac{5}{2};4;-\frac{3}{2};\frac{1\pm\sqrt{43}}{2}\)

 \(A=\left(\frac{2x-3}{4x^2-12x+5}+\frac{3x-8}{13x-2x^2-20}-\frac{3}{2x-1}\right):\frac{21+2x-2x^2}{4x^2+4x-3}+\)

\(=\left(\frac{2x-3}{\left(2x-1\right)\left(2x-5\right)}-\frac{3x-8}{\left(2x-5\right)\left(x-4\right)}-\frac{3}{2x-1}\right).\frac{\left(2x-1\right)\left(2x+3\right)}{21+2x-2x^2}+1\)

\(=\frac{\left(2x-3\right)\left(x-4\right)-\left(3x-8\right)\left(2x-1\right)-3\left(2x-5\right)\left(x-4\right)}{\left(2x-1\right)\left(2x-5\right)\left(x-4\right)}.\frac{\left(2x-1\right)\left(2x+3\right)}{21+2x-2x^2}+1\)

\(=\frac{-10x^2+47x-56}{\left(2x-5\right)\left(x-4\right)}.\frac{2x+3}{-2x^2+2x+21}+1\) số to wa

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

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

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

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

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

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

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

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

a. TH1:

\(\left\{{}\begin{matrix}x^2+3x-4< 0\\3-2x>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 1\\x>-4\end{matrix}\right.\\x>\dfrac{3}{2}\end{matrix}\right.\)

TH2:

\(\left\{{}\begin{matrix}x^2+3x-4>0\\3-2x< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1\\x< -4\end{matrix}\right.\\x< \dfrac{3}{2}\end{matrix}\right.\)

Vậy nghiệm của BPT:

\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 1\\x>-4\end{matrix}\right.\\x>\dfrac{3}{2}\end{matrix}\right.\)      \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1\\x< -4\end{matrix}\right.\\x< \dfrac{3}{2}\end{matrix}\right.\)