a, \(\dfrac{3}{7}\)\(x\) - 0,4 = - \(\dfrac{17}{35}\)

    \(\dfrac{3}{7}\)\(x\)         = - \(\dfrac{17}{35}\) + 0,4

     \(\dfrac{3}{7}\)\(x\)       = - \(\dfrac{3}{35}\)

        \(x\)       = - \(\dfrac{3}{35}\): \(\dfrac{3}{7}\)

        \(x\)       = - \(\dfrac{1}{5}\)

b, 0,2.(\(x\) - 3) +2,4 = 10

    0,2.(\(x\) - 3)          = 10 - 2,4

    0,2.(\(x\) - 3)          = 7,6

           \(x\) - 3            = 7,6:0,2

           \(x\) - 3            = 38

            \(x\)                = 38 + 3

             \(x\)                = 41

\(\dfrac{3}{7}x-0,4=\dfrac{-17}{35}\)

\(\dfrac{3}{7}x=\dfrac{-17}{35}+0,4=\dfrac{-34}{70}+\dfrac{28}{70}\)

\(\dfrac{3}{7}x=\dfrac{-6}{70}=\dfrac{-3}{35}\)

\(x=\dfrac{-3}{35}:\dfrac{3}{7}=\dfrac{-3}{35}\cdot\dfrac{7}{3}\)

\(x=\dfrac{3\cdot\left(-1\right)\cdot7}{5\cdot7\cdot3}=-\dfrac{1}{5}\)

b) \(0,2\left(x-3\right)+2,4=10\)

\(0,2\left(x-3\right)=10-2,4=7,6\)

\(x-3=7,6:0,2=38\)

\(x=38+3=41\)

c) \(\left(x:2,2\right)\cdot\dfrac{1}{6}=\dfrac{-3}{8}\cdot\left(0,5-1\dfrac{3}{5}\right)\)

\(\left(x:2,2\right)\cdot\dfrac{1}{6}=\dfrac{-3}{8}\cdot\dfrac{-11}{10}\)

\(x:2,2=\dfrac{-3}{8}\cdot\dfrac{-11}{10}\cdot\dfrac{6}{1}\)

\(x:2,2=\dfrac{-3\cdot\left(-11\right)\cdot2\cdot3}{2\cdot4\cdot10\cdot1}=\dfrac{99}{40}\)

\(x=\dfrac{99}{40}\cdot2,2=\dfrac{99}{40}\cdot\dfrac{22}{10}\)

\(x=\dfrac{99\cdot2\cdot11}{40\cdot2\cdot5}=\dfrac{1089}{200}\)

chị phải giúp em 1 bài

hên xui

ê Duyên tui cũng hok bik làm bài đó lun

\(a,\dfrac{6}{5}=\dfrac{18}{x}\\ \Rightarrow x=18:\dfrac{6}{5}\\ \Rightarrow x=15\\ b,\dfrac{3}{4}=\dfrac{-21}{x}\\ \Rightarrow x=-21:\dfrac{3}{4}\\ \Rightarrow x=-28\\ c,\dfrac{2}{-7}=\dfrac{18}{x}\\ \Rightarrow x=18:\dfrac{2}{-7}\\ \Rightarrow x=-63\\ d,\dfrac{-5}{2}=\dfrac{10}{-x}\\ \Rightarrow x=-10:\dfrac{-5}{2}\\ \Rightarrow x=4\)

\(a,\dfrac{6}{5}=\dfrac{18}{x}\Rightarrow6.x=5.18=90\\ \Rightarrow6.x=90\\ \Rightarrow x=15\\ b,\dfrac{3}{4}=\dfrac{-21}{x}\Rightarrow3.x=4.21=84\\ \Rightarrow x=28\)

\(a,\dfrac{6}{5}=\dfrac{18}{x}=>x=\dfrac{18.5}{6}=\dfrac{90}{6}=15\)

\(b,\dfrac{3}{4}=-\dfrac{21}{x}=>x=-\dfrac{21.4}{3}=-\dfrac{84}{3}=24\)

\(c,\dfrac{2}{-7}=\dfrac{18}{x}=>x=\dfrac{18.-7}{2}=-\dfrac{126}{2}=-63\)

\(d,-\dfrac{5}{2}=\dfrac{10}{-x}=>-x=\dfrac{10.2}{-5}=\dfrac{20}{-5}=-4=>x=4\)

a) \(\dfrac{x}{4}\)=\(\dfrac{1}{x}\)→x²=4→x=2 hoặc x=-2
b) \(\dfrac{1}{5}\) = x:4-\(\dfrac{1}{10}\)→x:4=\(\dfrac{1}{5}\)+\(\dfrac{1}{10}\)→x:4=\(\dfrac{2+1}{10}\)→x:4=\(\dfrac{3}{10}\)→x=\(\dfrac{3}{10}\)x4→x=\(\dfrac{12}{10}\)→x=\(\dfrac{6}{5}\)

cmr là cái j

Lê Thanh Thùy Ngân 

cmr là chứng minh rằng bạn nhé 

a) \(A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{x+3}\right)\) (ĐK: \(x\ne\pm3\))

\(A=\left[\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x^2-1}{\left(x+3\right)\left(x-3\right)}\right]:\left(2+\dfrac{x+5}{x+3}\right)\)

\(A=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x+3\right)\left(x-3\right)}:\dfrac{2\left(x+3\right)-\left(x+5\right)}{x+3}\)

\(A=\dfrac{-5x-5}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x+3}{x+1}\)

\(A=\dfrac{-5\left(x+1\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)\left(x+1\right)}\)

\(A=\dfrac{-5}{x-3}\)

b) Ta có: \(\left|x\right|=1\)

TH1: \(\left|x\right|=-x\) với \(x< 0\)

Pt trở thành:

\(-x=1\) (ĐK: \(x< 0\)) 

\(\Leftrightarrow x=-1\left(tm\right)\)

Thay \(x=-1\) vào A ta có:

\(A=\dfrac{-5}{x-3}=\dfrac{-5}{-1-3}=\dfrac{5}{4}\)

TH2: \(\left|x\right|=x\) với \(x\ge0\)

Pt trở thành:

\(x=1\left(tm\right)\) (ĐK: \(x\ge0\)) 

Thay \(x=1\) vào A ta có:

\(A=\dfrac{-5}{x-3}=\dfrac{-5}{1-2}=\dfrac{5}{2}\)

c) \(A=\dfrac{1}{2}\) khi:

\(\dfrac{-5}{x-3}=\dfrac{1}{2}\)

\(\Leftrightarrow-10=x-3\)

\(\Leftrightarrow x=-10+3\)

\(\Leftrightarrow x=-7\left(tm\right)\)

d) \(A\) nguyên khi:

\(\dfrac{-5}{x-3}\) nguyên

\(\Rightarrow x-3\inƯ\left(-5\right)\)

\(\Rightarrow x\in\left\{8;-2;2;4\right\}\)

a: \(A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{x+3}\right)\)

\(=\dfrac{x\left(x-3\right)-2\left(x+3\right)-x^2+1}{\left(x-3\right)\left(x+3\right)}:\dfrac{2x+6-x-5}{x+3}\)

\(=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x+1}\)

\(=\dfrac{-5x-5}{\left(x-3\right)}\cdot\dfrac{1}{x+1}=\dfrac{-5}{x-3}\)

b: |x|=1

=>x=-1(loại) hoặc x=1(nhận)

Khi x=1 thì \(A=\dfrac{-5}{1-3}=-\dfrac{5}{-2}=\dfrac{5}{2}\)

c: A=1/2

=>x-3=-10

=>x=-7

d: A nguyên

=>-5 chia hết cho x-3

=>x-3 thuộc {1;-1;5;-5}

=>x thuộc {4;2;8;-2}