a)\(\Leftrightarrow\)\(x(27+30+43)=210500\)

\(\Leftrightarrow\)\(100x=210500\)

\(\Leftrightarrow\)\(x=2105\)

b)\(\Leftrightarrow\)\(x(128-12-16)=5208000\)

\(\Leftrightarrow\)\(100x=5208000\)

\(\Leftrightarrow\)\(x=52080\)

\(x\times27+x\times30+x\times43=210500\)

\(\Leftrightarrow x\times\left(27+30+43\right)=210500\)

\(\Leftrightarrow x\times100=210500\)

\(\Leftrightarrow x=2105\)

a: 27x+30x+43x=210500

nên 100x=210500

hay x=2105

b: 128x-12x-16x=5208000

nên 100x=5208000

hay x=52080

Giải:

\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{x.\left(x+2\right)}=\dfrac{16}{99}\) 

\(\dfrac{1}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{x.\left(x+2\right)}\right)=\dfrac{16}{99}\) 

\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{16}{99}\) 

                           \(\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{x+2}\right)=\dfrac{16}{99}\) 

                                     \(\dfrac{1}{3}-\dfrac{1}{x+2}=\dfrac{16}{99}:\dfrac{1}{2}\) 

                                     \(\dfrac{1}{3}-\dfrac{1}{x+2}=\dfrac{32}{99}\) 

                                             \(\dfrac{1}{x+2}=\dfrac{1}{3}-\dfrac{32}{99}\) 

                                             \(\dfrac{1}{x+2}=\dfrac{1}{99}\) 

\(\Rightarrow x+2=99\) 

           \(x=99-2\) 

           \(x=97\) 

Chúc em học tốt!

\(\dfrac{1}{3x5}+\dfrac{1}{5x7}+\dfrac{1}{7x9}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{16}{99}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{3x5}+\dfrac{2}{5x7}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{16}{99}\)

\(=\dfrac{2}{3x5}\)\(+\dfrac{2}{5x7}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{32}{99}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}+.....+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{32}{99}\)

\(=\dfrac{1}{3}-\dfrac{1}{x+2}=\dfrac{32}{99}=>x=97\)

\(\dfrac{x}{12}=x+\dfrac{15}{16}\)

\(\Leftrightarrow\dfrac{x}{12}=\dfrac{16x+15}{16}\)

\(\Rightarrow12\left(16x+15\right)=16x\)

\(\Leftrightarrow192x+180=16x\)

\(\Leftrightarrow176x=-180\)

\(\Leftrightarrow x=-\dfrac{45}{44}\)

Vậy \(x=-\dfrac{45}{44}\)

\(\dfrac{x}{12}=\dfrac{x+15}{16}\)

\(\Rightarrow12\left(x+15\right)=16x\)

\(\Leftrightarrow12x+180=16x\)

\(\Leftrightarrow4x=180\)

\(\Leftrightarrow x=45\)

Vậy x = 45

\(\dfrac{x}{12}=\dfrac{x+15}{16}\)

\(16x=12\left(x+15\right)\)

\(16x=12x+180\)

\(4x=180\)

\(x=45\)

\(x=-\dfrac{3}{16}+\dfrac{1}{4}=\dfrac{1}{16}\)

bạn viết rõ đề nhé 

=> cái vế  trong ngoặc là 0.

rồi bạn làm tiếp nha

\(\dfrac{-12}{25}.\left(\dfrac{3}{4}-x+\dfrac{6}{-11}-\dfrac{5}{6}\right)=0\)

\(\dfrac{3}{4}-x+\dfrac{-6}{11}-\dfrac{5}{6}=0\)

\(\dfrac{3}{4}-x+\dfrac{-91}{66}=0\)

\(\dfrac{3}{4}-x=0-\left(\dfrac{-91}{66}\right)\)

\(\dfrac{3}{4}-x=\dfrac{91}{66}\)

\(x=\dfrac{-83}{132}\)

a: Ta có: \(-\dfrac{12}{25}\cdot\left(\dfrac{3}{4}-x+\dfrac{6}{-11}-\dfrac{5}{6}\right)=0\)

\(\Leftrightarrow x+\dfrac{83}{132}=0\)

hay \(x=-\dfrac{83}{132}\)

`x+4/5=45/5-36/35`

`x+4/5=279/35`

`x=251/34`

a. (80x - 801) . 12 = 0

<=> 80x - 801 = 0

<=> 80x = 801

<=> x = \(\dfrac{801}{80}\)

(Mấy câu tiếp mik ko hiểu đề, bn viết lại để dễ hiểu hơn nhé)

c: Ta có: \(\overline{xxx}=16\)

\(\Leftrightarrow100x+10x+1=16\)

\(\Leftrightarrow101x=16\)

hay \(x=\dfrac{16}{101}\)

=>x-3/4=2

hay x=11/4

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

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

\(x=\dfrac{11}{4}\)

x - 3/4 = 8/5 x 5/4

x - 3/4 =  2

       x  = 2 + 3/4

       x  = 11/4

vậy x =...

\(\dfrac{x}{10}=\dfrac{-3}{5}\\ \Rightarrow x=\dfrac{10.\left(-3\right)}{5}=-6\\ ------------\\ x+\dfrac{5}{12}=\dfrac{-2}{3}\\ x=\dfrac{-2}{3}-\dfrac{5}{12}\\ x=\dfrac{-8}{12}-\dfrac{5}{12}\\ x=\dfrac{-13}{12}\)

`x`=`-6`

`x` = `-13/12`

x/10=-3/5

x/10=-3.2/5.2

x/10=-6/10

x=-6

x+5/12=-2/3

x=-2/3-5/12

x=8/12-5/12

x=3/12

x=1/4