Dạng 2: Bài tập nâng cao có đáp án

$\begin{array}{l}a)\left(x+5\right)\left(x-8\right)<0\\ \iff \left[\begin{array}{l}\left\{\begin{array}{l}x+5>0\\ x-8<0\end{array}\right.\\ \left\{\begin{array}{l}x+5<0\\ x-8>0\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}\left\{\begin{array}{l}x>-5\\ x<8\end{array}\right.\\ \left\{\begin{array}{l}x<-5\\ x>8\end{array}\right.\end{array}\right.\iff -5<x<8\end{array}$

$\begin{array}{l}b)\left(2x-3\right)\left(2-x\right)>0\\ \iff \left[\begin{array}{l}\left\{\begin{array}{l}2x-3>0\\ 2-x>0\end{array}\right.\\ \left\{\begin{array}{l}2x-3<0\\ 2-x<0\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}\left\{\begin{array}{l}x>\frac{3}{2}\\ x<2\end{array}\right.\\ \left\{\begin{array}{l}x<\frac{3}{2}\\ x>2\end{array}\right.\end{array}\right.\iff \frac{3}{2}<x<2\end{array}$

$\begin{array}{l}c)2x^2-x-6>0\\ \iff (x-2)(3x+2)>0\\ \iff \left[\begin{array}{l}\left\{\begin{array}{l}x-2>0\\ 3x+2>0\end{array}\right.\\ \left\{\begin{array}{l}x-2<0\\ 3x+2<0\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}\left\{\begin{array}{l}x>2\\ x>\frac{-2}{3}\end{array}\right.\\ \left\{\begin{array}{l}x<2\\ x<\frac{-2}{3}\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}x>2\\ x<\frac{-2}{3}\end{array}\right.\end{array}$

$\begin{array}{l}d)\left(x+1\right)\left(x^2-x+1\right)-x^2-x>0\\ \iff \left(x+1\right)\left(x^2-x+1\right)-x(x+1)>0\\ \iff \left(x+1\right)\left(x^2-x+1-x\right)>0\\ \iff \left(x+1\right){\left(x-1\right)}^2>0\\ \iff x+1>0\\ \iff x>1\end{array}$

Vậy x > 1

a)

$\frac{2x(3x-5)}{x^2+1}>0\iff 2x\left(3x-5\right)>0\iff \left[\begin{array}{l}\left\{\begin{array}{l}x>0\\ 3x-5>0\end{array}\right.\\ \left\{\begin{array}{l}x<0\\ 3x-5<0\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}\left\{\begin{array}{l}x>0\\ x>\frac{5}{3}\end{array}\right.\\ \left\{\begin{array}{l}x<0\\ x<\frac{5}{3}\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}x>\frac{5}{3}\\ x<0\end{array}\right.$

$\begin{array}{l}b)\frac{x}{x-2}+\frac{x-2}{x}>2\iff \frac{x^2}{x(x-2)}+\frac{{\left(x-2\right)}^2}{x(x-2)}-\frac{2x\left(x-2\right)}{x\left(x-2\right)}>0\\ \iff \frac{x^2+x^2-4x+4-2x^2+4x}{x\left(x-2\right)}>0\iff \frac{4}{x\left(x-2\right)}>0\iff x\left(x-2\right)>0\\ \iff \left[\begin{array}{l}\left\{\begin{array}{l}x>0\\ x-2>0\end{array}\right.\\ \left\{\begin{array}{l}x<0\\ x-2<0\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}\left\{\begin{array}{l}x>0\\ x>2\end{array}\right.\\ \left\{\begin{array}{l}x<0\\ x<2\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}x>2\\ x<0\end{array}\right.\end{array}$

$\begin{array}{l}c)\frac{4x+3}{2x-5}\ge -3\iff \frac{4x+3}{2x-5}+3\ge 0\iff \frac{4x+3}{2x-5}+\frac{6x-15}{2x-5}\ge 0\iff \frac{10x-12}{2x-5}\ge 0\\ \iff \left[\begin{array}{l}\left\{\begin{array}{l}10x-12\ge 0\\ 2x-5\ge 0\end{array}\right.\\ \left\{\begin{array}{l}10x-12\le 0\\ 2x-5<0\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}\left\{\begin{array}{l}x\ge \frac{6}{5}\\ x>\frac{5}{2}\end{array}\right.\\ \left\{\begin{array}{l}x\le \frac{6}{5}\\ x<\frac{5}{2}\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}x>\frac{5}{2}\\ x\le \frac{6}{5}\end{array}\right.\end{array}$

$\begin{array}{l}a)\frac{x+1}{99}+\frac{x+4}{96}+\frac{x+5}{95}\ge -3\\ \iff \frac{x+100}{99}+\frac{x+100}{96}+\frac{x+100}{95}\ge 0\\ \iff \left(x+100\right)\left(\frac{1}{99}+\frac{1}{96}+\frac{1}{95}\right)\ge 0\\ \iff x+100\ge 0\iff x\ge -10\end{array}$

Vậy $x\ge -10$

$\begin{array}{l}b)\frac{x-5}{27}+\frac{x-4}{28}+\frac{x-3}{29}+\frac{x-2}{30}<4\\ \iff \frac{x-32}{27}+\frac{x-32}{28}+\frac{x-32}{29}+\frac{x-32}{30}<0\\ \iff \left(x-32\right)\left(\frac{1}{27}+\frac{1}{28}+\frac{1}{29}+\frac{1}{30}\right)<0\\ \iff x-32<0\iff x<32\end{array}$

$\begin{array}{l}c)\frac{x+1}{2019}+\frac{x+2}{2018}+\frac{x+3}{2017}+\mathrm{...}+\frac{x+100}{1920}>-100\\ \iff \frac{x+2020}{2019}+\frac{x+2020}{2018}+\mathrm{...}+\frac{x+2020}{1920}>0\\ \iff \left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\mathrm{...}+\frac{1}{1920}\right)>0\\ \iff x+2020>0\iff x>-2020\end{array}$

$\begin{array}{l}a) \frac{x+5}{95}+\frac{x+6}{94}+\frac{x+7}{93}\ge \frac{x+8}{92}+\frac{x+9}{91}+\frac{x+10}{90}\\ \iff \frac{x+100}{95}+\frac{x+100}{94}+\frac{x+100}{93}-\frac{x+100}{92}-\frac{x+100}{91}-\frac{x+100}{90}\ge 0\\ \iff \left(x+100\right)\left(\frac{1}{95}-\frac{1}{92}+\frac{1}{94}-\frac{1}{91}+\frac{1}{93}-\frac{1}{90}\right)\ge 0\left(1\right)\end{array}$

Ta có

$\begin{array}{l}\frac{1}{95}<\frac{1}{92}\iff \frac{1}{95}-\frac{1}{92}<0\\ \frac{1}{94}<\frac{1}{91}\iff \frac{1}{94}-\frac{1}{91}<0\\ \frac{1}{93}<\frac{1}{90}\iff \frac{1}{93}-\frac{1}{90}<0\end{array}$

$$\begin{gathered} \Rightarrow \frac{1}{95}-\frac{1}{92}+\frac{1}{94}-\frac{1}{91}+\frac{1}{93}-\frac{1}{90}<0 \\ \Rightarrow \left(1\right)\iff x+100\le 0\iff x\le -100 \end{gathered}$$

Vậy $x\le -100$

$\begin{array}{l}b) \frac{x-5}{921}+\frac{x-7}{919}+\frac{x-9}{917}<\frac{x-17}{909}+\frac{x-18}{908}+\frac{x-19}{907}\\ \iff \frac{x-926}{921}+\frac{x-926}{919}+\frac{x-926}{917}-\frac{x-926}{909}-\frac{x-18}{908}-\frac{x-19}{907}<0\\ \iff \left(x-926\right)\left(\frac{1}{921}-\frac{1}{909}+\frac{1}{919}-\frac{1}{908}+\frac{1}{917}-\frac{1}{907}\right)<0\left(1\right)\end{array}$

Ta có

$\begin{array}{l}\frac{1}{921}<\frac{1}{909}\iff \frac{1}{921}-\frac{1}{909}<0\\ \frac{1}{919}<\frac{1}{908}\iff \frac{1}{919}-\frac{1}{908}<0\\ \frac{1}{917}<\frac{1}{907}\iff \frac{1}{917}-\frac{1}{907}<0\end{array}$

$$\begin{gathered} \Rightarrow \frac{1}{921}-\frac{1}{909}+\frac{1}{919}-\frac{1}{908}+\frac{1}{917}-\frac{1}{907}<0 \\ \Rightarrow \left(1\right)\iff x-926>0\iff x>926 \end{gathered}$$

$\begin{array}{l}{a}^2+b^2+c^2\ge ab+bc+ca\\ \iff 2a^2+2b^2+2c^2>2ab+2bc+2ca\\ \iff a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+c^2\ge 0\\ \iff {\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2\ge 0\end{array}$

b. Ta có: 

$$\begin{gathered} a^2+b^2+c^2+d^2+e^2\ge a\left(b+c+d+e\right) \\ \iff 4\left(\right.a^2+b^2+c^2+d^2+e^2\left.\right)\ge 4a\left(b+c+d+e\right) \\ \iff \left(a^2-4ab+4b^2\right)+\left(a^2-4ac+4c^2\right)+\left(a^2-4ad+4d^2\right)+\left(a^2-4ae+4e^2\right)\ge 0 \\ \iff {\left(a-2b\right)}^2+{\left(a-2c\right)}^2+{\left(a-2d\right)}^2+{\left(a-2e\right)}^2\ge 0 \end{gathered}$$

Từ $\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge 9$, Đặt $\left\{\begin{array}{l}x=a+b\\ y=b+c\\ z=c+a\end{array}\right.$

$$\begin{gathered} \Rightarrow 2\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge 9 \\ \iff \frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}\ge \frac{9}{2} \\ \Rightarrow \frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}\ge \frac{9}{2}-3=\frac{3}{2} \end{gathered}$$

Từ $\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge 9$

$\Rightarrow \left(\frac{a}{b+1}+1\right)+\left(\frac{b}{a+1}+1\right)+\left(\frac{1}{a+b}+1\right)-3=\left(a+b+1\right)\left(\frac{1}{a+b}+\frac{1}{a+1}+\frac{1}{b+1}\right)-3\ge \frac{9}{2}-3=\frac{3}{2}$