`(x+1)/9+(x+2)/8=(x+3)/7+(x+4)/6`
`<=>(x+1)/9+1+(x+2)/8+1=(x+3)/7+1+(x+4)/6+1`
`<=>(x+10)/9+(x+10)/8=(x+10)/7+(x+10)/6`
`<=>(x+10)(1/9+1/8-1/7-1/6)=0`
`<=>x+10=0`do `1/9+1/8-1/7-1/6<0`
`<=>x=-10`
Vậy `x=-10`
\(\dfrac{x+1}{9}+\dfrac{x+2}{8}=\dfrac{x+3}{7}+\dfrac{x+4}{6}\)
a) 6-7x+7=-8x+12
b) \(\dfrac{\text{4x-1}}{8}\)\(\)=\(\dfrac{6-x}{2}\)-\(\dfrac{1}{4}\)
c) 2x(x-3)+7x-21+0
d) \(\dfrac{x}{x-3}\)+\(\dfrac{x-2}{x+3}\)\(\dfrac{2\left(x^2+6\right)}{x^2-9}\)
Giải các phương trình sau:
a) 2,3 - 2(0,7 + 2) = 3,6 - 1,7x
b) \(\dfrac{5x+7}{4}-\dfrac{3x+5}{8}=\dfrac{4x+9}{5}-\dfrac{x-9}{3}\)
c) \(\dfrac{2x-1}{4}+\dfrac{x-3}{3}=\dfrac{4x-2}{3}-\dfrac{6x+7}{12}\)
d) (x - 1)(x + 2) - x(x + 3) = 8
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/ \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
d.\(\dfrac{12x+1}{11x-4}+\dfrac{10x-4}{9}=\dfrac{20x+17}{18}\) e.\(\dfrac{11}{x}=\dfrac{9}{x+1}+\dfrac{2}{x-4}\) f.\(\dfrac{14}{3x-12}-\dfrac{2+x}{x-4}=\dfrac{3}{8-2x}-\dfrac{5}{6}\)
d.\(\dfrac{12x+1}{11x-4}+\dfrac{10x-4}{9}=\dfrac{20x+17}{18}\) e.\(\dfrac{11}{x}=\dfrac{9}{x+1}+\dfrac{2}{x-4}\) f.\(\dfrac{14}{3x-12}-\dfrac{2+x}{x-4}=\dfrac{3}{8-2x}-\dfrac{5}{6}\)
thực hiện phép tính
\(\dfrac{x}{x-3}-\dfrac{6}{x}-\dfrac{9}{x^2-3x}\)
\(\dfrac{7}{x}-\dfrac{x}{x+6}+\dfrac{36}{x^2+6x}\)
\(\dfrac{6}{x-3}-\dfrac{2x-16}{x^2-9}-\dfrac{4}{x+3}\)
a) (X-2)(x+3)-3(4x-2)=(x-4)\(^{^{ }2}\)
b) \(\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\)
c) \(x-\dfrac{2x-2}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)
d) \(\left(2x+5\right)^2=\left(x+2\right)^2\)
e) \(x^2-5+6=0\)
g) \(2x^3+6x^2=x^2+3x\)
h) \(\left(x+\dfrac{1}{2}\right)^2+2\left(x+\dfrac{1}{x}\right)-8=0\)
mọi người giúp e với ạ
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
