\(\frac{4}{7}.x+0,125=1\frac{1}{8}+x\)

=>\(\frac{4}{7}.x+\frac{1}{8}=1+x+\frac{1}{8}\)

=>\(\frac{4}{7}.x=1+x\)

=>\(\frac{4}{7}.x-x=1\)

=>\(\left(\frac{4}{7}-1\right)x=1\)

=>\(\frac{-3}{7}.x=1\)

=>\(x=1:\frac{-3}{7}=\frac{1.7}{-3}=-\frac{7}{3}\)

a) \(\dfrac{3}{4}x=1\)

\(x=1:\dfrac{3}{4}\)

\(x=1.\dfrac{4}{3}\)

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

b)\(\dfrac{4}{7}x=\dfrac{9}{8}-0,125\)

\(\dfrac{4}{7}x=\dfrac{9}{8}-\dfrac{1}{8}\)

\(\dfrac{4}{7}x=1\)

\(x=1:\dfrac{4}{7}\)

\(x=1.\dfrac{7}{4}\)

\(x=\dfrac{7}{4}\)

a) \(\dfrac{3}{4}x=1\)

\(x=1:\dfrac{3}{4}=1.\dfrac{4}{3}\)

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

Vậy \(x=\dfrac{4}{3}\)

b) \(\dfrac{4}{7}x=\dfrac{9}{8}-0,125\)

\(\Leftrightarrow\dfrac{4}{7}x=\dfrac{9}{8}-\dfrac{1}{8}\)

\(\Rightarrow\dfrac{4}{7}x=1\)

\(x=1:\dfrac{4}{7}=1.\dfrac{7}{4}\)

\(x=\dfrac{7}{4}\)

Vậy \(x=\dfrac{7}{4}\)

a ) 3/4 . x = 1 

=>        x  = 1 : 3/4

=>        x  = 1 . 4/3

=>        x  = 4/3

Vậy x = 4/3 

b )  4/7x    =    9/8 - 0,125

=> 4/7x     =   9/8  - 1/8

=> 4/7x    =      8/8 = 1

=>      x    =  1 : 4/7

=>      x    =   1. 7/4

=>      x    =    7/4

Vậy     x    =   7/4

Chúc học giỏi 

\(\frac{3}{4}x=1\)

\(x=1:\frac{3}{4}\)

\(x=\frac{4}{3}\)

1,\(\left(\frac{7}{2}-2x\right).\frac{4}{3}=\frac{22}{3}\)

\(x.\left(\frac{7}{2}-2\right)=\frac{22}{3}:\frac{4}{3}=\frac{22}{3}.\frac{3}{4}=\frac{11}{2}\)

\(x.\frac{3}{2}=\frac{11}{2}\)

\(x=\frac{11}{2}:\frac{3}{2}=\frac{11}{2}.\frac{2}{3}=\frac{11}{3}\)

Số số hạng là :

      (2x - 2) : 2 + 1 = x - 1 + 1 = x (số)

Tổng là : 

       (2x + 2).x : 2 = 210

=> (2x² + 2x) : 2 = 210

=> x² + x = 210

=> x(x + 1) = 210

=> x(x + 1) = 20.21

=> x = 20

Vậy x = 20 

Ta có : \(\frac{x}{2}=\frac{10}{x+1}\)

=> x(x + 1) = 10.2

=> x(x + 1) = 20

=> sai đề 

a) \(\left(3\frac{1}{2}-2x\right).3\frac{1}{3}=7\frac{1}{3}\)

 \(\left(\frac{7}{2}-2x\right).\frac{10}{3}=\frac{22}{3}\)

  \(\frac{7}{2}-2x=\frac{11}{5}\)

              \(2x=\frac{13}{10}\)

                \(x=\frac{13}{20}\)

Vậy ...

b) \(\frac{4}{9}x=\frac{9}{8}-0,125\)

   \(\frac{4}{9}x=1\)

         \(x=\frac{9}{4}\)

Vậy...

\(a,\)\(-\frac{3}{5}\cdot x=\frac{1}{4}+0,75\)

\(-\frac{3}{5}\cdot x=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\)

\(x=1\div\left(-\frac{3}{5}\right)\)

\(x=-\frac{5}{3}\)

\(b,\)\(\left(\frac{1}{7}-\frac{1}{3}\right)\cdot x=\frac{28}{5}\times\left(\frac{1}{4}-\frac{1}{7}\right)\)

\(\left(\frac{3}{21}-\frac{7}{21}\right)\cdot x=\frac{28}{5}\cdot\left(\frac{7}{28}-\frac{4}{28}\right)\)

\(-\frac{4}{21}\cdot x=\frac{28}{5}\cdot\frac{3}{28}\)

\(-\frac{4}{21}\cdot x=\frac{3}{5}\)

\(x=\frac{3}{5}\div\left(-\frac{4}{21}\right)\)

\(x=-\frac{63}{20}\)

\(c,\)\(\frac{5}{7}\cdot x=\frac{9}{8}-0,125\)

\(\frac{5}{7}\cdot x=\frac{9}{8}-\frac{1}{8}\)

\(\frac{5}{7}\cdot x=1\)

\(x=1\div\frac{5}{7}\)

\(x=\frac{7}{5}\)

\(d,\)\(\left(\frac{2}{11}+\frac{1}{3}\right)\cdot x=\left(\frac{1}{7}-\frac{1}{8}\right)\cdot36\)

\(\left(\frac{6}{33}+\frac{11}{33}\right)\cdot x=\left(\frac{8}{56}-\frac{7}{56}\right)\cdot36\)

\(\frac{17}{33}\cdot x=\frac{1}{56}\cdot36\)

\(\frac{17}{33}\cdot x=\frac{9}{14}\)

\(x=\frac{9}{14}\div\frac{17}{33}\)

\(x=\frac{9}{14}\cdot\frac{33}{17}=\frac{297}{238}\)

\(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+...+\frac{1}{\left(2x-2\right)\cdot2x}=\frac{1}{8}\left(x\inℕ;x\ge2\right)\)

Đặt \(A=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+...+\frac{1}{\left(2x-2\right)2x}\)

\(2A=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+...+\frac{2}{\left(2x-2\right)2x}\)

\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2x-2}-\frac{1}{2x}\)

\(2A=\frac{1}{2}-\frac{1}{2x}=\frac{x-1}{2x}\)

\(\Rightarrow A=\frac{x-1}{2x}:2=\frac{x-1}{2x}\cdot\frac{1}{2}=\frac{x-1}{4x}\)

Mà \(A=\frac{1}{8}\Rightarrow\frac{x-1}{4}=\frac{1}{8}\)

\(\Leftrightarrow8x-8=4\)

\(\Leftrightarrow8x=12\)

\(\Leftrightarrow x=\frac{12}{8}=\frac{3}{2}\left(ktm\right)\)

Vậy không có x thỏa mãn yêu cầu đề bài

a./ \(\Leftrightarrow x^{10}=1\Leftrightarrow x=\pm1\)

b./ \(\Leftrightarrow x^{10}-x=0\Leftrightarrow x\left(x^9-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^9=1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)

c./ \(\Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\Leftrightarrow\left(2x-15\right)^3\left(\left(2x-15\right)^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}2x-15=0\\\left(2x-15\right)^2=1\end{cases}}\)

• 2x - 15 = 0 \(\Leftrightarrow x=\frac{15}{2}\)
• 2x - 15 = 1 \(\Leftrightarrow x=\frac{16}{2}=8\)
• 2x - 15 = -1 \(\Leftrightarrow x=\frac{14}{2}=7\)

1^10=1^1

1^10=1

(2*8-15)^3=(2*8)^3

Ý c,x có thể bằng 7

f) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)( ĐKXĐ : \(x\ne-\frac{1}{2}\))

\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=21\cdot3\)

\(\Leftrightarrow4x^2-1=63\)

\(\Leftrightarrow4x^2=64\)

\(\Leftrightarrow x^2=16\)

\(\Leftrightarrow x^2=\left(\pm4\right)^2\)

\(\Leftrightarrow x=\pm4\)(tmđk)

h) \(\frac{10x+5}{6}=\frac{5}{x+1}\)( ĐKXĐ : \(x\ne-1\))

\(\Leftrightarrow\left(10x+5\right)\left(x+1\right)=6\cdot5\)

\(\Leftrightarrow10x^2+15x+5=30\)

\(\Leftrightarrow10x^2+15x+5-30=0\)

\(\Leftrightarrow10x^2+15x-25=0\)

\(\Leftrightarrow5\left(2x^2+3x-5\right)=0\)

\(\Leftrightarrow2x^2+3x-5=0\)

\(\Leftrightarrow2x^2-2x+5x-5=0\)

\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+5\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)(tmđk)

f) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=21.3\)

\(\Leftrightarrow4x^2-1=63\)

\(\Leftrightarrow4x^2=64\)

\(\Leftrightarrow x^2=16\)\(\Leftrightarrow x^2=4^2\)\(\Leftrightarrow x=4\)

Vậy \(x=4\)

h) \(\frac{10x+5}{6}=\frac{5}{x+1}\)

\(\Leftrightarrow\left(10x+5\right)\left(x+1\right)=5.6\)

\(\Leftrightarrow5\left(2x+1\right)\left(x+1\right)=30\)

\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=6\)

\(\Leftrightarrow2x^2+3x+1=6\)

\(\Leftrightarrow2x^2+3x-5=0\)

\(\Leftrightarrow\left(2x^2-2x\right)+\left(5x-5\right)=0\)

\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\2x=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{-5}{2}\end{cases}}\)

Vậy \(x\in\left\{\frac{-5}{2};1\right\}\)