0,2:1 1/5 = 2/3 : (6x + 7)

0,2:6/5 = 2/3 : (6x+7)

1/6 = 2/3 : (6x + 7 )

6x + 7 = 2/3 : 1/6

6x + 7 = 4

6x = 4 - 7 = - 3

x = -3 : 6

x = -1/2

2/3 : ( 6x + 7 ) = 1/6

\(0,2:1\frac{1}{5}=\frac{2}{3}:\left(6x+7\right)\)

\(\Leftrightarrow0,2\times\frac{5}{6}=\frac{2}{3}\times\frac{1}{6x+7}\)

\(\Leftrightarrow\frac{1}{6}:\frac{2}{3}=\frac{1}{6x+7}\)

\(\Leftrightarrow\frac{1}{4}=\frac{1}{6x+7}\)

\(\Leftrightarrow6x+7=4\)

\(\Leftrightarrow6x=-3\)

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

khó vậy

\(|x-\frac{1}{3}|=|\left(-3.2\right)+\frac{2}{5}|\)  

\(\Rightarrow|x-\frac{1}{3}|=|-3.2+0.4|\)

\(\Rightarrow|x-\frac{1}{3}|=|-2.8|\)

\(\Rightarrow|x-\frac{1}{3}|=2.8\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=2.8\\x-\frac{1}{3}=-2.8\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{43}{15}\\x=-\frac{41}{15}\end{cases}}\)

tính lại kết quả nhé

\(\left|x-\frac{1}{3}\right|=\left|-3.2+\frac{2}{5}\right|=\left|-\frac{14}{5}\right|\)\(=\frac{14}{5}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{14}{5}\\x-\frac{1}{3}=-\frac{14}{5}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{47}{15}\\x=-\frac{37}{15}\end{cases}}\)

Vậy............

b,

\(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}-\frac{x-4}{2008}=0\)

\(\Leftrightarrow\frac{x-1}{2011}-1+\frac{x-2}{2010}-1-\frac{x-3}{2009}+1-\frac{x-4}{2008}+1=0\)

\(\Leftrightarrow\frac{x-1-2011}{2011}+\frac{x-2-2010}{2010}-\frac{x-3-2009}{2009}-\frac{x-4-2008}{2008}=0\)

\(\Leftrightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)

\(\Leftrightarrow\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)

Mà \(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)

=> x-2012=0

=>x=2012

Vậy ..............

TK MK NHA

*****CHUC BN HOC GIỎI*****

\(\frac{1}{3}\left|\frac{1}{4}x-\frac{1}{5}\right|+\frac{3}{7}\left|\frac{x}{y}-0,2\right|=0\)

Nhận xét : \(\left|\frac{1}{4}x-\frac{1}{5}\right|\ge0\)với \(\forall x\)(vì giá trị tuyệt đối không âm) 

     \(\Rightarrow\frac{1}{3}\left|\frac{1}{4}x-\frac{1}{5}\right|\ge0\)(1) 

  \(\left|\frac{x}{y}-0,2\right|\ge0\)với \(\forall x\),\(\left(y\ne0\right)\)(vì giá trị tuyệt đối không âm)

    \(\Rightarrow\frac{3}{7}\left|\frac{x}{y}-0,2\right|\ge0\)  (2)

Từ (1) và (2) => \(\frac{1}{3}\left|\frac{1}{4}x-\frac{1}{5}\right|+\frac{3}{7}\left|\frac{x}{y}-0,2\right|\ge0\)

Để \(\frac{1}{3}\left|\frac{1}{4}x-\frac{1}{5}\right|+\frac{3}{7}\left|\frac{x}{y}-0,2\right|=0\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{3}\left|\frac{1}{4}x-\frac{1}{5}\right|=0\\\frac{3}{7}\left|\frac{x}{y}-0,2\right|=0\end{cases}\Leftrightarrow}\hept{\begin{cases}\left|\frac{1}{4}x-\frac{1}{5}\right|=\frac{1}{3}\\\left|\frac{x}{y}-0,2\right|=\frac{3}{7}\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{1}{4}x-\frac{1}{5}=0\\\frac{x}{y}-0,2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{4}x=\frac{1}{5}\\\frac{x}{y}=0,2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{4}{5}\\\frac{4}{5}\div y=0,2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{4}{5}\\y=4\left(tm\right)\end{cases}}\)

Vậy \(x=\frac{4}{5};y=4\) (Tm)

Ta có : \(\left(5x-3\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5x-3=0\\2x+5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x=3\\2x=-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{5}{2}\end{cases}}\)

a) x-1/3=5/14.(-7/6)

  x-1/3=-5/12

  x=-5/12+1/3

  x= -1/12

b) 3/4+1/4x=0,2

   1/4x=0,2-3/4

   1/4x=-11/20

    x=-11/20:1-4

    x=-11/5

c) 1/12.x^2=1.1/2

   1/12.x^2=1/2 

   x^2= 1/2:1/12

   x^2=6

=> x=căn bậc của 6

Chúc bn hok tốt

a)\(x-\frac{1}{3}=\frac{5}{14}\cdot\left(\frac{-7}{6}\right)\)

\(x-\frac{1}{3}=\frac{-5}{12}\)

\(x=\frac{-5}{12}+\frac{1}{3}\)

\(x=\frac{-5}{12}+\frac{4}{12}\)

\(x=\frac{-1}{12}\)

b)\(\frac{3}{4}+\frac{1}{4}\cdot x=0,2\)

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

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

\(\frac{1}{4}\cdot x=\frac{4}{20}-\frac{15}{20}\)

\(\frac{1}{4}\cdot x=\frac{-11}{20}\)

\(x=\frac{-11}{20}\cdot4\)

\(x=-\frac{11}{5}\)

c)\(\frac{1}{12}\cdot x^2=1\cdot\frac{1}{3}\)

\(\frac{1}{12}\cdot x^2=\frac{1}{3}\)

\(x^2=\frac{1}{3}\cdot12\)

\(x^2=4=\left(\pm2\right)^2\)

\(x=\pm2\)

\(\frac{x}{11,9}=\frac{1,5}{3,5}\Rightarrow x=\frac{1,5.11,9}{3,5}\Rightarrow x=5,1\)

\(\frac{x}{0,2}=\frac{80}{x}\Rightarrow x^2=80.0,2\Rightarrow x^2=16\Rightarrow x=4;-4\)

\(\frac{2}{3}x:\frac{1}{5}=\frac{4}{3}:\frac{1}{4}\Rightarrow\frac{2}{3}x:\frac{1}{5}=\frac{16}{3}\Rightarrow\frac{2}{3}x=\frac{16}{15}\Rightarrow x=\frac{8}{5}\)

\(3:\frac{2}{5}x=\frac{1}{0,001}\Rightarrow\frac{2}{5}x=\frac{3}{1000}\Rightarrow x=\frac{3}{400}\)

\(\frac{x}{11,9}=\frac{1,5}{3,5}\Leftrightarrow x=\frac{11,9.1,5}{3,5}=5,1\)

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