Đặt x+1/x = t ta có t2 + 2t - 8 = 0 => (t +1)2 = 32
t= 2 hoặc t= -4
Với t= 2 => x2 - 2x +1=0 => x=1
Với t=-4=> x2 + 4x +1 =0 => ( x + 2)2 = 3 => x= -2 \(+-\sqrt{3}\)
Đặt x+1/x = t ta có t2 + 2t - 8 = 0 => (t +1)2 = 32
t= 2 hoặc t= -4
Với t= 2 => x2 - 2x +1=0 => x=1
Với t=-4=> x2 + 4x +1 =0 => ( x + 2)2 = 3 => x= -2 \(+-\sqrt{3}\)
giải pt
a) \(\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-3}=12\left(\frac{x-2}{x-3}\right)^2\)
b) \(\frac{2\left(x+1\right)}{3x^2+x}+\frac{13\left(x+1\right)}{3x^2+7x+6}=6\)
Giải hệ PT
\(\hept{\begin{cases}x+y^2+z^3=14\left(1\right)\\\left(\frac{1}{2x}+\frac{1}{3y}+\frac{1}{6z}\right)\left(\frac{x}{2}+\frac{y}{3}+\frac{z}{6}\right)=1\left(2\right)\end{cases}}\)
Giải pt: \(\left(\frac{x+2}{x+1}\right)^2+\left(\frac{x-2}{x-1}\right)^2-\frac{5x^2-4}{2x^2-1}=0\)
Giải pt \(\left(\frac{x+2}{x+1}\right)^2+\left(\frac{x-2}{x-1}\right)-\frac{5x^2-4}{2x^2-1}=0\)
1)giải pt \(x^3-9x^2+6x-6-3\sqrt[3]{6x^2+2}=0\)
2) giải hệ pt \(\int^{\sqrt{x}\left(1+\frac{3}{x+3y}\right)=2}_{\sqrt{7y}\left(1-\frac{3}{x+3y}\right)=4\sqrt{2}}\)
Giải pt : \(\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-3}=12\left(\frac{x-2}{x-3}\right)^2\)
Giải pt:
\(\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-3}=12.\left(\frac{x+2}{x-3}\right)^2\)
Giải các pt sau
a/\(\frac{3}{x+2}=1-\frac{2}{x-3}\)
b/ \(\frac{2x}{2x^2-5x+2}+\frac{13x}{3x^2+x+2}=6\)
c/ \(x^2+\frac{4x^2}{\left(x-2\right)^2}=4\)
d/\(\left(x+2\right).\left(x+4\right)+5.\left(x+2\right).\sqrt{\frac{x+4}{x+2}}=6\)
Giải nhanh dùm mk nha mk đang cần gấp
giải pt
\(\frac{2\left(x-\sqrt{2}\right)\left(x-\sqrt{3}\right)}{\left(1-\sqrt{2}\right)\left(1-\sqrt{3}\right)}+\frac{3\left(x-1\right)\left(x-\sqrt{3}\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}-\sqrt{3}\right)}+\frac{4\left(x-1\right)\left(x-\sqrt{2}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-2\right)}\)=3x-1