a)4/3:x=-4/7                         b)5/3.x=7/12

⇒x=-16/21                             ⇒x=7/20

a. \(\dfrac{5}{7}+\dfrac{4}{3}:x=\dfrac{1}{7}\)

<=> \(\dfrac{5}{7}+\dfrac{4}{3}.\dfrac{1}{x}=\dfrac{1}{7}\)

<=> \(\dfrac{5}{7}+\dfrac{4}{3x}=\dfrac{1}{7}\)         ĐKXĐ: x \(\ne\) 0

<=> \(\dfrac{15x}{21x}+\dfrac{28}{21x}=\dfrac{3x}{21x}\)

<=> 15x + 28 = 3x

<=> 15x - 3x = -28

<=> 12x = -28

<=> x = \(\dfrac{-28}{12}=-\dfrac{7}{3}\)

b. \(\dfrac{5}{3}x.\dfrac{-1}{4}=\dfrac{2}{6}\)

<=> \(\dfrac{-5x}{12}=\dfrac{2}{6}\)

<=> -5x . 6 = 12 . 2

<=> -30x = 24

<=> x = \(-\dfrac{4}{5}\)

a)\(\frac{2}{3}\cdot x+\frac{1}{4}=\frac{5}{6}\)

\(\frac{2}{3}\cdot x=\frac{5}{6}-\frac{1}{4}=\frac{7}{12}\)

\(x=\frac{7}{12}:\frac{2}{3}=\frac{7}{8}\)

b) \(\frac{7}{12}-\frac{5}{6}\cdot x=\frac{1}{4}:\frac{2}{3}\)

\(\frac{7}{12}-\frac{5}{6}\cdot x=\frac{3}{8}\)

\(\frac{5}{6}\cdot x=\frac{7}{12}-\frac{3}{8}=\frac{5}{24}\)

\(x=\frac{5}{24}:\frac{5}{6}=\frac{1}{4}\)

c) \(2\frac{1}{3}-\left(x+1\right)=\frac{5}{9}\)

\(\frac{7}{3}-\left(x+1\right)=\frac{5}{9}\)

\(x+1=\frac{7}{3}-\frac{5}{9}=\frac{16}{9}\)

\(x=\frac{16}{9}-1=\frac{7}{9}\)

d) \(\frac{2\cdot x+1}{15}=\frac{3}{5}\)

\(\left(2\cdot x+1\right):15=\frac{3}{5}\)

\(2\cdot x+1=\frac{3}{5}\cdot15=9\)

\(2\cdot x=9-1=8\)

\(x=8:2=4\)

a, 2/3 . x +1/4=5/6

      2/3 . x=5/6-1/4

      2/3 . x=10/12 -3/12

       2/3 . x=7/12

       x= 7/12 : 2/3

      x=7/8 

Vậy x=7/8

e) (x + 1) + (x + 2) + .... + (x + 211) = 23632 

   (x + x + .... + x) + (1 + 2 + ... + 211) = 23632 

  211x + 22366 = 23632 

  211x = 23632 - 22366 

  211x = 1266

=> x= 6 

f) (1 + 3 + 5 + 7 + .... + 2009) + x = 1010026 

Số số hạng của 1 + 3 + 5 + 7 + .... + 2009 là: (2009 - 1) : 2 + 1 = 1005

Tổng của 1 + 3 + 5 +... + 2009 là: (1 + 2009) x 1005 : 2 = 1010025

Ta được : 1010025 + x = 1010026 

             => x = 1010026 - 1010025 = 1 

các bạn giúp mình làm bài 8 đi

Bài 8:

( 752248-752248) : ( 2+4+6+...+200)

0 : ( 2+4+6+200)=0

Chúc bạn hok tốt nha!

A.\(\left(\frac{4}{7}.x-1\right).3=\frac{1}{21}\)

\(\frac{4}{7}.x-1=\frac{1}{21}:3\)

\(\frac{4}{7}.x-1=\frac{1}{21}.3\)

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

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

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

\(x=\frac{8}{7}:\frac{4}{7}\)

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

\(x=2\)

Vậy x=2

C.\(\frac{2}{3}.x-\frac{1}{3}.x=\frac{5}{9}\)

\(\left(\frac{2}{3}-\frac{1}{3}\right).x=\frac{5}{9}\)

\(\frac{1}{3}.x=\frac{5}{9}\)

\(x=\frac{5}{9}:\frac{1}{3}\)

\(x=\frac{5}{9}.3\)

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

Vậy \(x=\frac{5}{3}\)

D. \(\left(x-2\frac{1}{4}\right).\left(\frac{-2}{3}\right).50\%=2\frac{5}{6}\)

\(x.\frac{-2}{3}-\frac{9}{4}.\frac{-2}{3}=\frac{17}{6}:\frac{1}{2}\)

\(x.\frac{-2}{3}-\frac{-3}{2}=\frac{17}{6}.2\)

\(x.\frac{-2}{3}+\frac{3}{2}=\frac{17}{3}\)

\(x.\frac{-2}{3}=\frac{17}{3}-\frac{3}{2}\)

\(x.\frac{-2}{3}=\frac{34}{6}-\frac{9}{6}\)

\(x.\frac{-2}{3}=\frac{25}{6}\)

\(x=\frac{25}{6}:\frac{-2}{3}\)

\(x=\frac{25}{6}.\frac{3}{-2}\)

\(x=\frac{25}{-4}\)

Vậy \(x=\frac{25}{-4}\)

\(\left|x+\frac{3}{4}\right|-\frac{1}{3}=0\)

\(\left|x+\frac{3}{4}\right|=\frac{1}{3}\)

\(\Rightarrow x+\frac{3}{4}=\pm\frac{1}{3}\)

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

\(x=-\frac{5}{12}\)

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

\(x=-\frac{13}{12}\)

A = 7 ⇔ 4 x - 1 = 7 ⇔ x = 11 7 t m đ k

Vậy  x = 11 7 thì A=7