Giải:
Ta có: \(\frac{x^2-6}{x}=x+\frac{3}{2}\) (ĐKXĐ: x ≠ 0)
\(\text{ }\frac{x^2-6}{x}-x=\frac{3}{2}\text{ ⇔ }\frac{x^2-6-x^2}{x}=\frac{3}{2}\text{ ⇔ }\frac{6}{x}=\frac{3}{2}\text{ ⇔ }x=\frac{6.2}{3}\text{ ⇔ }x=4\)
Vậy phương trình có nghiệm là x = 4
\(\frac{4}{x^2-3x+2}-\frac{3}{2x^2-6x+1}+1=0\)
\(\frac{1}{x-1}+\frac{2}{x-2}+\frac{3}{x-3}=\frac{6}{x-6}\)
\(1.\frac{1}{x^2-2x+2}+\frac{2}{x^2-2x+3}=\frac{6}{x^2-2x+4}
\)
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.\(\frac{2x}{3x^2-x+2}-\frac{7x}{3x^2+5x+2}=1\)
\(\frac{1}{x-1}+\frac{2}{x-2}+\frac{3}{x-3}=\frac{6}{x-6}\)
Em cảm ơn
f.\(\frac{3}{x^2+x-2}-\frac{1}{x-1}=\frac{-7}{x+2}\)
g.\(\frac{2}{-x^2+6x-8}-\frac{x-1}{x-2}=\frac{x+3}{x-4}\)
h.\(\frac{2}{x^3-x^2-x+1}=\frac{3}{1-x^2}-\frac{1}{x-1}\)
i.\(\frac{x+2}{x-2}-\frac{2}{x^2-2x}=\frac{1}{x}\)
j.\(\frac{5}{-x^2+5x-6}+\frac{x+3}{2-x}=0\)
Giari các phương trình sau.
a. \(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\)
b. \(\frac{x+3}{x-3}-\frac{1}{x}=\frac{3}{x\left(x-3\right)}\)
c. \(\frac{3}{x+2}-\frac{2}{x-2}+\frac{8}{x^2-4}=0\)
d. \(\frac{3}{x+2}-\frac{2}{x-3}=\frac{8}{\left(x-3\right)\left(x+2\right)}\)
e. \(\frac{x}{2x+6}-\frac{x}{2x+2}=\frac{3x+2}{\left(x+1\right)\left(x+3\right)}\)
f. \(\frac{x}{x+1}-\frac{2x-3}{1-x}=\frac{3x^2+5}{x^2-1}\)
g. \(\frac{5}{x+7}+\frac{8}{2x+14}=\frac{3}{2}\)
h. \(\frac{x-1}{x}-\frac{1}{x+1}=\frac{2x-1}{x^2+x}\)
\(\frac{4}{-25x^2+20x-3}=\frac{3}{5x-1}-\frac{2}{5x-3}\)
\(\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}-\frac{2}{x^2-4x+3}=0\)
\(\frac{x-1}{2x^2-4x}-\frac{7}{8x}=\frac{5-x}{4x^2-8x}-\frac{1}{8x-16}\)
\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
giải phương trình sau:
a) \(\frac{x^2-6}{x}=x+\frac{3}{2}\)
b) \(\frac{x+3}{x+1}+\frac{x-2}{x}=2\)
Mọi người giúp mk
\(\frac{1}{x-1}+\frac{2}{x-2}+\frac{3}{x-3}=\frac{6}{x-6}\)
Cảm ơn trước nha!
GIÚP MÌNH VỚI!!
a) \(\frac{x}{2x-6}+\frac{x}{2x+2}+\frac{2x}{\left(x+1\right)\left(3-x\right)}=0\)
b) \(\frac{x-3}{x+1}-\frac{x+1}{x+3}=\frac{x^2-x-10}{x^2+4x+3}\)
Giải phương trình:\(\left(\frac{x+3}{x-2}\right)^2+6\left(\frac{x-3}{x+2}\right)^2=\frac{7\left(x^2-9\right)}{x^2-4}\)
