Dạng 3: Bài tập tự luyện có đáp án

a. $m^2+3>0\forall m\in ℝ$

b. $-\left(m^2+m+4\right)=-\left[{\left(m+\frac{1}{2}\right)}^2+\frac{15}{4}\right]<0\forall m\in ℝ$

a. $2x-8>0\iff 2x>8\iff x>4$

b. $9-3x\le 0\iff -3x\le -9\iff x\ge 3$

c. $5-\frac{1}{3}x<1\iff -\frac{1}{3}x<-4\iff x>12$

d. $\frac{3x+5}{2}-x\ge 1+\frac{x+2}{3}\iff \frac{x}{6}\ge \frac{-5}{6}\iff x\ge -5$

a. 

$$\begin{gathered} \frac{x-2}{3}-x-2\le \frac{x-17}{2} \\ \iff \frac{2\left(x-2\right)-6x-6.2}{6}\le \frac{3\left(x-17\right)}{6} \\ \iff 2x-4-6x-12\le 3x-51 \\ \iff -4x-16\le 3x-51 \\ \iff -4x-3x\le -51+16 \\ \iff -7x\le -35 \\ \iff x\ge 5. \end{gathered}$$

Vậy tập nghiệm của bất phương trình là $S=\left\{x|x\ge 5\right\}$ và được biểu diễn trên trục số như sau:

b. 

$$\begin{gathered} \frac{2x+1}{3}-\frac{x-4}{4}\le \frac{3x+1}{6}-\frac{x-4}{12} \\ \iff \frac{4\left(2x+1\right)-3\left(x-4\right)}{12}\le \frac{2\left(3x+1\right)-\left(x-4\right)}{12} \\ \iff 8x+4-3x+12\le 6x+2-x+4 \\ \iff 5x+16\le 5x+6 \\ \iff 5x-5x\le 6-16 \\ \iff 0x\le -10 \\ \iff x\in \varnothing \end{gathered}$$

Vậy bất phương trình vô nghiệm và được biểu diễn trên trục số như sau:

a. $x^2-3x+1>2(x-1)-x(3-x)\iff x^2-3x+1>2x-2-3x+x^2$

$\iff -2x>-3\iff x<\frac{3}{2}$

b. 

$$\begin{gathered} {(x-1)}^2+x^2\le {(x+1)}^2+{\left(x+2\right)}^2 \\ \iff 2x^2-2x+1\le 2x^2+6x+5 \\ \iff -8x\le 4 \\ \iff x\ge -\frac{1}{2} \end{gathered}$$

c. 

$$\begin{gathered} (x^2+1)(x-6)\le {(x-2)}^3 \\ \iff x^3-6x^2+x-6\le x^3-6x^2+12x-8 \\ \iff -11x\le -2\iff x\ge \frac{2}{11} \end{gathered}$$

a. $\frac{x-1}{2}-\frac{7x+3}{15}\le \frac{2x+1}{3}+\frac{3-2x}{5}$

$$\begin{gathered} \iff \frac{15.\left(x-1\right)}{30}-\frac{2.\left(7x+3\right)}{30}\le \frac{10.\left(2x+1\right)}{30}+\frac{6.\left(3-2x\right)}{30} \\ \iff 15x-15-14x-6\le 20x+10+18-12x \\ \iff x-21\le 8x+28 \\ \iff 7x\ge -49 \\ \iff x\ge -7 \end{gathered}$$

Vậy $S=\left\{x\left|x\right.\ge -7\right\}.$

b. $\frac{2x+1}{-3}-\frac{2x^2+3}{-4}>\frac{x\left(5-3x\right)}{-6}-\frac{4x+1}{-5}$

$$\begin{gathered} \iff \frac{-2x-1}{3}+\frac{2x^2+3}{4}>\frac{-x\left(5-3x\right)}{6}+\frac{4x+1}{5} \\ \iff \frac{20.\left(-2x-1\right)+15.\left(2x^2+3\right)}{60}>\frac{-10x.\left(5-3x\right)+12.\left(4x+1\right)}{60} \\ \iff \frac{-40x-20+30x^2+45}{60}>\frac{-50x+30x^2+48x+12}{60} \\ \iff 30x^2-40x+25>30x^2-2x+12 \\ \iff -38x>-13\iff x<\frac{13}{38} \end{gathered}$$

Vậy $S=\left\{x\left|x<\frac{13}{38}\right.\right\}$

e. $\frac{5x^2-3}{5}+\frac{3x-1}{4}<\frac{x\left(2x+3\right)}{2}-5$

$$\begin{gathered} \iff \frac{4.\left(5x^2-3\right)+5\left(3x-1\right)}{20}<\frac{10x.\left(2x+3\right)-5.20}{20} \\ \iff \frac{20x^2-12+15x-5}{20}<\frac{20x^2+30x-100}{20} \\ \iff 20x^2+15x-17<20x^2+30x-100 \\ \iff -15x<-83 \\ \iff 15x>83 \\ \iff x>\frac{83}{15} \end{gathered}$$

Vậy $S=\left\{x\left|x>\frac{83}{15}\right.\right\}$

f. $\frac{5x-2}{-3}-\frac{2x^2-x}{-2}>\frac{x\left(1-3x\right)}{-3}-\frac{5x}{-4}$

$$\begin{gathered} \iff \frac{-5x+2}{3}+\frac{2x^2-x}{2}>\frac{-x+3x^2}{3}+\frac{5x}{4} \\ \iff \frac{4\left(-5x+2\right)+6\left(2x^2-x\right)}{12}>\frac{4\left(-x+3x^2\right)+3.5x}{12} \\ \iff -20x+8+12x^2-6x>-4x+12x^2+15x \\ \iff -26x+8>11x \\ \iff -37x>-8 \\ \iff 37x<8 \\ \iff x<\frac{8}{37} \end{gathered}$$

Vậy $S=\left\{x\left|x<\frac{8}{37}\right.\right\}$

g. $2x+\frac{2x+1}{2}>3x-\frac{1}{5}$

$$\begin{gathered} \iff \frac{10.2x+5\left(2x+1\right)}{10}>\frac{3x.10-2}{10} \\ \iff 20x+10x+5>30x-2 \\ \iff 0x>-7 ( \mathrm{vô} \mathrm{lý}) \\ \iff x\in \varnothing . \end{gathered}$$

Vậy $S=\varnothing .$

h. $x-\frac{5x}{6}-3>\frac{x}{3}-\frac{x}{6}$

$$\begin{gathered} \iff \frac{6x-5x-18}{6}>\frac{2x-x}{6} \\ \iff x-18>x \\ \iff 0x<-18 \\ \iff x\in \varnothing \end{gathered}$$

Vậy $S=\varnothing .$

a. $\frac{x+2004}{2005}+\frac{x+2005}{2006}<\frac{x+2006}{2007}+\frac{x+2007}{2008}$

$$\begin{gathered} \iff \frac{x+2004}{2005}-1+\frac{x+2005}{2006}-1<\frac{x+2006}{2007}-1+\frac{x+2007}{2008}-1 \\ \iff \frac{x-1}{2005}+\frac{x-1}{2006}-\frac{x-1}{2007}-\frac{x-1}{2008}<0 \\ \iff (x-1)\left(\frac{1}{2005}+\frac{1}{2006}-\frac{1}{2007}-\frac{1}{2008}\right)<0 \\ \iff x-1<0(\text{do }\frac{1}{2005}+\frac{1}{2006}-\frac{1}{2007}-\frac{1}{2008}>0) \\ \iff x<1. \end{gathered}$$

b. $\frac{x-2}{2002}+\frac{x-4}{2000}<\frac{x-3}{2001}+\frac{x-5}{1999}$

$$\begin{gathered} \iff \frac{x-2}{2002}-1+\frac{x-4}{2000}-1<\frac{x-3}{2001}-1+\frac{x-5}{1999}-1 \\ \iff \frac{x-2004}{2002}+\frac{x-2004}{2000}<\frac{x-2004}{2001}+\frac{x-2004}{1999} \\ \iff \left(x-2004\right)\left(\frac{1}{2002}+\frac{1}{2000}-\frac{1}{2001}-\frac{1}{1999}\right)<0 \\ \iff x-2004>0\text{ ( do }\frac{1}{2002}+\frac{1}{2000}-\frac{1}{2001}-\frac{1}{1999}<0) \\ \iff x>2004 \end{gathered}$$

c. $\frac{x-ab}{a+b}+\frac{x-bc}{b+c}+\frac{x-ac}{a+c}>a+b+c,(a,\text{ b, c >0)}$

$$\begin{gathered} \iff \frac{x-ab}{a+b}-c+\frac{x-bc}{b+c}-a+\frac{x-ac}{a+c}-b>0 \\ \iff \frac{x-ab-ac-bc}{a+b}+\frac{x-bc-ab-ac}{b+c}+\frac{x-ac-bc-ab}{a+c}>0 \\ \iff (x-ab-ac-bc)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)>0 \\ \iff x-ab-ac-bc>0,(\text{do a, b, c >0 }\Rightarrow \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}>0) \\ \iff x>ab+ac+bc. \end{gathered}$$

b. $x-1<\frac{2x-1}{3}-1<2x+4$

TH1: $x-1<\frac{2x-1}{3}-1\iff \frac{3\left(x-1\right)}{3}<\frac{2x-1-3}{3}\iff 3x-3<2x-4\iff x<-1$

TH2: $\frac{2x-1}{3}-1<2x+4\iff \frac{2x-1-3}{3}<\frac{3\left(2x+4\right)}{3}\iff 2x-4<6x+12\iff 4x>-16\iff x>-4$

Vậy -4 < x < -1.

Tập nghiệm $S=\left\{x\left|-4<x<-1\right.\right\}$