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

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

\(-5x-\frac{1}{2}x-\frac{3}{2}x=1-\frac{1}{3}-\frac{5}{6}\)

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

\(x.\left(-7\right)=\frac{-1}{6}\)

x=\(\frac{1}{42}\)

\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

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

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)

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

_Tần vũ_

\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)

\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)

\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)

\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)

\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)

\(\Leftrightarrow3x=\frac{1}{6}\)

\(\Leftrightarrow x=\frac{1}{18}\)

_Tần Vũ_

a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=0+\frac{9}{25}\)

\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\frac{3^2}{5^2}\)

\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

\(\Rightarrow\hept{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2x=\frac{3}{5}-\frac{3}{5}\\2x=-\frac{3}{5}-\frac{3}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2x=0\\2x=\frac{-6}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0:2\\x=-\frac{6}{5}:2\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)

b) \(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)

\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=0-\frac{1}{9}\)

\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)

\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}:3\)

\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)

\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\left(\frac{-1^3}{3^3}\right)\)

\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)

\(\Rightarrow3x-\frac{1}{2}=-\frac{1}{3}\)

\(\Rightarrow3x=-\frac{1}{3}+\frac{1}{2}\)

\(\Rightarrow3x=\frac{1}{6}\)

\(\Rightarrow x=\frac{1}{6}:3\)

\(\Rightarrow x=\frac{1}{18}\)

a) \(\left(\frac{2}{3}x-\frac{4}{9}\right)\left(\frac{1}{2}-\frac{3}{7}:x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}-\frac{3}{7}:x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{6}{7}\end{cases}}\)

Vậy \(x\in\left\{\frac{2}{3};\frac{6}{7}\right\}\)

b) 

 \(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

        \(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+329}{5}+4=4\)

         \(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

          \(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

Mà \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\ne0\)

\(\Rightarrow x+329=0\)

\(\Rightarrow x=-329\)

Vậy \(x=-329\)

\(x=1\)nha bạn

x-0,27=(73/77+73/165+73/185):[25.12/133)

x-0,27=(73/77+73/165+73/185):[25/1.12/133)

x-0,27=(73/77+73/165+73/185):300/133

x-0,27=1387/777:300/133

x-0,27=0,791381381

x=0,791381381-0,27

x=0,521381381

                                                                                Bài làm 

x-0.27=\(\frac{73}{100}\)

x= \(\frac{73}{100}\)+\(\frac{27}{100}\)

x=\(\frac{100}{100}\)=1

      Vậy x = 1 

a) (x-3)+(x-2)+(x-1)+....+10+11=11

(x-3)+(x-2)+(x-1)+....+10      =0

gọi số hạng của tổng vế trái là  n

(x-3+10).\(\frac{n}{2}\)=0

(x+7).n:2=0

(x+7)  =0

\(\Rightarrow\)x+7=0           (do n\(\ne\)0)

       x=0-7

       x=-7

b) \(\frac{2}{3}\left[\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right]<=x<=4\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{6}\right]\)

     \(\frac{2}{3}.\frac{11}{12}<=x<=\frac{13}{3}.\frac{1}{3}\)

      \(\frac{11}{18}<=x<=\frac{13}{9}\)

do x\(\in\)z nên x=1

vậy x=1

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)

\(\Leftrightarrow x+x+x+x+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=1\)

\(\Leftrightarrow4x+\frac{15}{16}=1\Leftrightarrow4x=\frac{1}{16}\Leftrightarrow x=\frac{1}{16}:4=\frac{1}{64}\)

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)

\(\Rightarrow x+\frac{1}{2}+x+\frac{1}{4}+x+\frac{1}{8}+x+\frac{1}{16}=1\)

\(\Rightarrow4x+\frac{15}{16}=1\)

\(\Rightarrow4x=1-\frac{15}{16}=\frac{1}{16}\)

\(\Rightarrow x=\frac{1}{16}:4=\frac{1}{16}.\frac{1}{4}=\frac{1}{64}\)

 bài nay mình bít cách làm nhưng hk bít đúng hông. cảm ơn các cậu nhiều.