Bài 1: 

a: x/-2=-18/x

=>x²=36

=>x=6 hoặc x=-6

b: x/2+x/5=17/10

=>7/10x=17/10

hay x=17/7

\(x-\dfrac{2}{3}.\left(x+9\right)=1\)

\(\Rightarrow x-\dfrac{2}{3}x-6=1\)

\(\Rightarrow\dfrac{1}{3}x=7\)

\(\Rightarrow x=21\)

\(x-\dfrac{2}{3}\times\left(x+9\right)=1\)

\(x-\dfrac{2}{3}x-6=1\)

\(\dfrac{1}{3}x-6=1\)

\(\dfrac{1}{3}x=1+6\)

\(\dfrac{1}{3}x=7\)

\(x=7:\dfrac{1}{3}\)

\(x=21\)

`x xx 6/7=5/14`

`=>x=5/14:6/7`

`=>x=5/14xx7/6`

`=>x=35/84`

`=>x=5/12`

Vậy `x=5/12`

__

`x:2/3=4/9`

`=>x=4/9xx2/3`

`=>x=8/27`

Vậy `x=8/27`

__

`x-1/4=3/2`

`=>x=3/2+1/4`

`=>x=6/4+1/4`

`=>x=7/4`

Vậy `x=7/4`

__

`x+4/5=8/9`

`=>x=8/9-4/5`

`=>x=40/45-36/45`

`=>x=4/45`

Vậy `x=4/45`

\(x\cdot\dfrac{6}{7}=\dfrac{5}{14}\)

\(x\)         \(=\dfrac{5}{14}:\dfrac{6}{7}\)

\(x\)           \(=\dfrac{5}{12}\)

\(x:\dfrac{2}{3}=\dfrac{4}{9}\)

\(x\)        \(=\dfrac{4}{9}\cdot\dfrac{2}{3}\)

\(x\)          \(=\dfrac{8}{27}\)

\(x-\dfrac{1}{4}=\dfrac{3}{2}\)

\(x\)          \(=\dfrac{3}{2}+\dfrac{1}{4}\)

\(x\)            \(=\dfrac{7}{4}\)

\(x+\dfrac{4}{5}=\dfrac{8}{9}\)

\(x\)          \(=\dfrac{8}{9}-\dfrac{4}{5}\)

\(x\)            \(=\dfrac{4}{45}\)

a, \(x + 1/6 = -3/8 \)

\(x = -3/8 - 1/6\)

\(x = -13/24\)

Vậy \(x = -13/24\)

b, \(-3/7 - x = 4/5 - 2/3\)

\(-3/7 - x = 2/15\)

\(x = -3/7 - 2/15\)

\(x = -59/105.\)

Vậy \(x = -59/105\)

x+1/6=-3/8

x=-3/8-1/6

x=-13/24

-3/7-x=4/5+-2/3
-3/7-x=2/15

x=-3/7-2/15

x=-59/105

tham khảo

Chọn D.

Ta có:

 khi x → +∞