Đề phải vậy chứ nhỉ?
\(\frac{1}{x-1}+\frac{3x^2}{1-x^3}=\frac{2x}{x^2+x+1}\left(Đkxđ:x\ne1\right)\)
\(\Leftrightarrow x^2+x+1-3x^2=2x\left(x-1\right)\)
\(\Leftrightarrow x^2+x+1-3x^2=2x^2-2x\)
\(\Leftrightarrow4x^2-3x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(ktmđk\right)\\x=-\frac{1}{4}\left(tmđk\right)\end{matrix}\right.\)
Vậy ...........
Giải pt
\(\frac{x^2+x+1}{x+1}+\frac{x^2+2x+2}{x+2}=\frac{x^2+3x+3}{x+3}+\frac{x^2+4x+4}{x+4}\)
giải pt: a. (x - 2)(x+1)(x+3) = (x+3)(x+1)(2x-5)
b. \(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)
Giải pt: \(\frac{2x+1}{x^2}+\frac{x^2}{2\left(3x^2+4x+2\right)}=-\frac{1}{2}\)
giải pt
\(\frac{2x-1}{x-1}\) + \(\frac{3x-2}{x-2}\) = \(\frac{x^2+4x+5}{x^2-3x+2}\) + 4
Giải PT:
a)3-4x(25-2x)=8x\(^2\)+x-30
b)(2x+1)(3x-2)=(5x-8)(2x+1)
c)\(\frac{x}{2x+1}+\frac{x+1}{2x+3}=\frac{x+1}{2x+1}+\frac{x-1}{2x+3}\)
d)||x+1|-1|=5
giải pt
1,\(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
2,\(\frac{x+4}{2x^2-5x+2}+\frac{x+1}{2x^2-7x+3}=\frac{2x+5}{2x^2-7x+3}\)
3,\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=3x\left(1-\frac{x-1}{x+1}\right)\)
4,\(\frac{2x}{x-1}+\frac{4}{x^2+2x-3=}=\frac{2x-5}{x+3}\)
5,\(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{x^2+x-2}\)
6,\(\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{6x-8-x^2}\)
7,\(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{1-x^2}\)
Giair pt:
c, x ( 3x-1) (3x+1) (3x+2) =8
d, (x+1) (2x+3) (2x+5) (x+3)=45
e,x4+ 3x3 - 15x2 - 19x + 3 = 0
f, \(\frac{1}{x^2+x}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}=\frac{1}{3}\)
h,\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
giải pt sau: \(\frac{1}{x-1}+\frac{2}{x^2+x+1}=\frac{3x^2}{x^2-1}\)
1,Giải Pt
a,\(\frac{3x-7}{2}+\frac{x+1}{3}=-16\)
b,\(x-\frac{x+1}{3}=\frac{2x+1}{5}\)
c,\(\frac{7-3x}{12}+\frac{3}{4}=2\left(x-2\right)+\frac{5\left(5-2x\right)}{6}\)
e,\(\frac{3\left(x+3\right)}{4}+\frac{1}{2}=\frac{5x+9}{3}-\frac{7x-9}{4}\)
