|-4,5x| = 6 + 2,5x
TH1: |-4,5x| = -4,5x nếu x ≤ 0 (-4,5x ≥ 0)
-4,5x = 6 + 2,5x
⇔ -4,5x - 2,5x = 6
⇔ -7x = 6
⇔ x = \(\frac{-6}{7}\)(TM)
TH2: |-4,5x| = 4,5x nếu x > 0 (-4,5x < 0)
4,5x = 6 + 2,5x
⇔ 4,5x - 2,5x = 6
⇔2x = 6
⇔x = 3(TM)
Vậy nghiệm của PT là x = \(\frac{-6}{7}\); x = 3
\(\frac{-6}{7}\)\(\frac{-6}{7}\)
\(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right)\left(x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}\)
1) \(\dfrac{7x-3}{x-1}\) = \(\dfrac{2}{3}\)
2) \(\dfrac{2\left(3-7x\right)}{1+x}\) = \(\dfrac{1}{2}\)
3) \(\dfrac{x^{2^{ }}-6}{x}\) = x + \(\dfrac{3}{2}\)
4) \(\dfrac{5}{3x+2}\) = 2x - 1
5) \(\dfrac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}\) = 0
6) \(\dfrac{1}{x-2}\) + 3 = \(\dfrac{3-x}{x-2}\)
Giải phương trình:\(\left(\frac{x+3}{x-2}\right)^2+6\left(\frac{x-3}{x+2}\right)^2=\frac{7\left(x^2-9\right)}{x^2-4}\)
\(1.\frac{1}{x^2-2x+2}+\frac{2}{x^2-2x+3}=\frac{6}{x^2-2x+4}
\)
2.\(\frac{2x^4}{\left(x+1\right)^2}-\frac{5x^2}{x+1}+2=0\)
3.\(\left(x+\frac{1}{x}\right)^2-6\left(x+\frac{1}{x}\right)+8=0\)
4.\(\left(x^2+\frac{1}{x^2}\right)-4\left(x+\frac{1}{x}\right)+6=0\)
5.\(\frac{2x}{3x^2-x+2}-\frac{7x}{3x^2+5x+2}=1\)
Giải các phương trình :
a) \(\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\)
b) \(\dfrac{3}{\left(x-1\right)\left(x-2\right)}+\dfrac{2}{\left(x-3\right)\left(x-1\right)}=\dfrac{1}{\left(x-2\right)\left(x-3\right)}\)
c) \(1+\dfrac{1}{x+2}=\dfrac{12}{8+x^3}\)
d) \(\dfrac{13}{\left(x-3\right)\left(2x+7\right)}+\dfrac{1}{2x+7}=\dfrac{6}{\left(x-3\right)\left(x+3\right)}\)
\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)
\(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{7}{6\left(x+5\right)}\)
\(\frac{x}{2x+6}-\frac{x}{2x+2}=\frac{3x+2}{\left(x+1\right)\left(x+3\right)}\)
Giải phương trình sau:
\(\left(\dfrac{x+4}{x-1}\right)^2+6\left(\dfrac{x-2}{x+3}\right)^2=\dfrac{7\left(x^2+2x-8\right)}{x^2+2x-3}\)
