Đk: `x ne +-3`.
Với `x` thỏa mãn đkxđ thì:
`<=> (x(2x-6))/((x+3)(2x-6)) - ((2x-1)(x+3))/((x+3)(2x-6)) = (2(x+2))/(2x-6)(x+3))`
`=> 2x^2 - 6x - 2x^2 - 5x + 3 = 2x + 4`
`=> -11x + 3 = 2x + 4`
`=> 13x + 1 = 0`
`=> x = -1/13`.
Vậy `x = -1/13`.
Đk: `x ne +-3`.
Với `x` thỏa mãn đkxđ thì:
`<=> (x(2x-6))/((x+3)(2x-6)) - ((2x-1)(x+3))/((x+3)(2x-6)) = (2(x+2))/(2x-6)(x+3))`
`=> 2x^2 - 6x - 2x^2 - 5x + 3 = 2x + 4`
`=> -11x + 3 = 2x + 4`
`=> 13x + 1 = 0`
`=> x = -1/13`.
Vậy `x = -1/13`.
1/ \(\dfrac{x-4}{3}+2x=\dfrac{4x-2}{6}\)
2/ \(\dfrac{5x-2}{5}-2=\dfrac{1-2x}{3}\)
3/ \(\dfrac{x-2}{2}-\dfrac{2}{3}=x-1\)
4/ \(\dfrac{2x-1}{3}+\dfrac{3x-2}{4}=\dfrac{4x-3}{5}\)
5/ \(\dfrac{x-3}{9}-\dfrac{x+2}{6}=\dfrac{x+4}{18}-\dfrac{1}{2}\)
1/ \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
2/ \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
3/ \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
4/ \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
5/ \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
thực hiện phép tính
\(\dfrac{11x}{2x-3}+\dfrac{x-18}{2x-3}\)
\(\dfrac{2x+12}{4x^2-9}+\dfrac{2x+5}{4x-6}\)
\(\dfrac{x}{2x+1}+\dfrac{-1}{4x^2-1}+\dfrac{2-x}{2x-1}\)
Giải các phương trình sau: (TM ĐK)
1) \(\dfrac{11}{x}=\dfrac{9}{x+1}+\dfrac{2}{x-4}\)
2) \(\dfrac{14}{3x-12}-\dfrac{2+x}{x-4}=\dfrac{3}{8-2x}-\dfrac{5}{6}\)
3) \(\dfrac{x+5}{x^2-5x}-\dfrac{x+25}{2x^2-50}=\dfrac{x-5}{2x^2+10}\)
4) \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{x^2-1}\)
5) \(\left(1-\dfrac{x-1}{x+1}\right)\left(x+2\right)=\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}\)
mng giúp mk bài này nha. Cảm ơn bạn nhiều
P= \(\dfrac{2x-9}{x^2-5x+6}-\dfrac{x+3}{x-2}-\dfrac{2x+1}{3-x}\)(x≠2;x≠3)
giải các phương trinh sau
1/ \(\dfrac{4x-4}{3}-\dfrac{7-x}{5}\)
2/ \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)
3/ \(\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\)
4/ \(\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\)
5/ \(\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\)
giải các phương trình sau
a, 3(x-1) -3=2(x+3)
b, \(\dfrac{x+4}{4}-\dfrac{x+3}{3}=\dfrac{x+6}{6}\)
c,\(\left(2x-1\right)^2-x^2=0\)
d,\(\dfrac{x}{x+3}-\dfrac{2x}{x-3}-\dfrac{3x}{9-x^2}=0\)
A = \(\dfrac{2x-9}{x^2-5x+6}-\dfrac{x+3}{x-2}\)và B = \(\dfrac{2x+1}{3-x}\)(x≠2,x≠3)
Tìm x để P ≤ 1
Tính:
a) \(\dfrac{x+1}{2x-6}+\dfrac{2x+3}{x^2+3x}\)
b) \(\dfrac{3}{2x+6}-\dfrac{x-6}{3x^2+6}\)
c) \(\dfrac{2x+6}{3x^2-x}:\dfrac{x^2+3x}{1-3x}\)