GIUP VS! VOTE VA THANKS DAY DU
giai cac bat phuong trinh sau roi bieu dien tap nghiem tren truc so
a. $(x-3)^{2}$ < $x^{2}$ - 5x +4
b. ( x - 3 )( x + 3 ) < $(x+2)^{2}$ + 3
c. $\frac{4x - 5}{3}$ > $\frac{7 – x}{5}$
d. $\frac{2x + 1}{2}$ + 3 ≥ $\frac{3 – 5x}{3}$ – $\frac{4x+ 1}{4}$
e. $\frac{5x-3}{5}$ + $\frac{2x+1 }{4}$ ≤ $\frac{2 – 3x}{2}$ – 5
f. $x^{2}$ – 4x +3 ≥ 0
g. $x^{3}$ – $2x^{2}$ + 3x – 6 < 0
h. $\frac{x+2}{5}$ ≥ 0
i. $\frac{x + 2}{x-3}$ < 0
j. $\frac{x-1}{x-3}$ > 1
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
Vậy …..
c. $\frac{4x-5}{3}$>$\frac{7-x}{5}$
⇒ 20x-25-21+3x>0
⇔ 23x>46
⇔ x>2
Vậy …..
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
⇔\(\left[ \begin{array}{l}x≥1\\x≥3\end{array} \right.\)
Vậy …..
g. x³ – 2x² + 3x – 6 < 0
⇔ (x³ – 2x²)+(3x – 6)<0
⇔ (x²+3)(x-2)<0
⇔\(\left[ \begin{array}{l}x²<-3(vô lý)\\x<2\end{array} \right.\)
Vậy …..
h. $\frac{x+2}{5}$≥0
⇔ x+2≥0
⇔ x≥-2
Vậy …..
i. $\frac{x+2}{x-3}$<0
⇒ x+2<0
⇔ x<-2
Vậy …..
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 …..