a. (x−3)² < x² - 5x +4
⇔ x²-6x+9-x²+5x-4<0
⇔ x>5
Vậy .....
b. ( x - 3 )( x + 3 ) < (x+2)² + 3
⇔ x²-9-x²-4x+1<0
⇔ -4x<8
⇔ x>2
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c. $\frac{4x-5}{3}$>$\frac{7-x}{5}$
⇒ 20x-25-21+3x>0
⇔ 23x>46
⇔ x>2
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d. $\frac{2x+1}{2}$+3≥$\frac{3-5x}{3}$-$\frac{4x+1}{4}$
⇔ 12x+6+18≥12-20x-12x-3
⇔ 12x+6+18-12+20x+12x+3≥0
⇔ 44x≥-15
⇔ x≥$\frac{-15}{44}$
Vậy .....
e. dựa theo câu d
f. x² - 4x +3 ≥ 0
⇔ x²-3x-x+3≥0
⇔ (x²-3x)-(x-3)≥0
⇔ (x-1)(x-3)≥0
⇔\(\left[ \begin{array}{l}x≥1\\x≥3\end{array} \right.\)
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g. x³ - 2x² + 3x - 6 < 0
⇔ (x³ - 2x²)+(3x - 6)<0
⇔ (x²+3)(x-2)<0
⇔\(\left[ \begin{array}{l}x²<-3(vô lý)\\x<2\end{array} \right.\)
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h. $\frac{x+2}{5}$≥0
⇔ x+2≥0
⇔ x≥-2
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i. $\frac{x+2}{x-3}$<0
⇒ x+2<0
⇔ x<-2
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j. $\frac{x-1}{x-3}$ >1
⇔ $\frac{x-1}{x-3}$ -1>0
⇔ $\frac{x-1}{x-3}$-$\frac{x-3}{x-3}$>0
⇒ x-1-x+3>0
⇔ 2>0 ( luôn đúng )
Vậy .....