\(x:\dfrac{1}{2}+x:\dfrac{1}{4}+x:\dfrac{1}{8}+...+x:\dfrac{1}{512}=511\\ 2x+4x+8x+..+512x=511\\ x\left(2+4+8+...+512\right)=511\\ x\left(2^1+2^2+2^3+...+2^9\right)=511\\ \)

Gọi \(S=2^1+2^2+2^3+...+2^9\)

\(2S=2^2+2^3+2^4+...+2^{10}\\ 2S-S=\left(2^2+2^3+2^4+...+2^{10}\right)-\left(2^1+2^2+2^3+...+2^9\right)\\ S=2^{10}-2\)

\(x\left(2^{10}-2\right)=511\\ 2x\left(2^9-1\right)=511\\ 2x\left(512-1\right)=511\\ 2x\cdot511=511\\ 2x=1\\ x=\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2}\)

\(\Leftrightarrow\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)^2+\dfrac{1}{4}=\dfrac{1}{2}\)

\(\Leftrightarrow\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)^2=\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{1}{2}\\\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

Giải:

\(\dfrac{1}{2}.\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2-\dfrac{1}{5}=-\dfrac{3}{40}\)

\(\Leftrightarrow\dfrac{1}{2}.\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=-\dfrac{3}{40}+\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{1}{2}.\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{8}\)

\(\Leftrightarrow\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{8}:\dfrac{1}{2}\)

\(\Leftrightarrow\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{5}=\dfrac{1}{2}\\\dfrac{1}{3}x-\dfrac{1}{5}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{3}x=\dfrac{7}{10}\\\dfrac{1}{3}x=-\dfrac{3}{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{21}{10}\\x=-\dfrac{9}{10}\end{matrix}\right.\)

Vậy ...

Chúc bạn học tốt!

\(\dfrac{1}{2}.\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2-\dfrac{1}{5}=-\dfrac{3}{40}\\ \dfrac{1}{2}\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=-\dfrac{3}{40}+\dfrac{1}{5}\\ \dfrac{1}{2}\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{8}\\ \left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{8}:\dfrac{1}{2}\\\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{4}\\ \left(\dfrac{1}{3}x-\dfrac{1}{5}\right)=\left(\pm\dfrac{1}{2}\right)^2\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{5}=\dfrac{1}{2}\\\dfrac{1}{3}x-\dfrac{1}{5}=-\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{3}x=\dfrac{7}{10}\\\dfrac{1}{3}x=\dfrac{3}{10}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{21}{10}\\x=\dfrac{9}{10}\end{matrix}\right. \)

Vậy \(x=\dfrac{21}{10}\) hoặc \(x=\dfrac{9}{10}\)

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

=> \(\dfrac{1}{2}\)| \(\dfrac{1}{3}x\) - \(\dfrac{1}{4}\)| = \(\dfrac{11}{30}\)

=> | \(\dfrac{1}{3}x\)- \(\dfrac{1}{4}\)| = \(\dfrac{11}{15}\)

=> \(\left[{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{4}=\dfrac{11}{15}\\\dfrac{1}{3}x-\dfrac{1}{4}=\dfrac{-11}{15}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\dfrac{1}{3}x=\dfrac{59}{60}\\\dfrac{1}{3}x=\dfrac{-29}{60}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\dfrac{59}{20}\\x=\dfrac{-29}{20}\end{matrix}\right.\)

Chúc bạn học tốt !

Tích mình , mình làm nhé!

\(x+\left|\dfrac{1}{2}-\dfrac{1}{3}\right|=\left|\dfrac{-2}{3}-\dfrac{1}{4}\right|\)

\(x+\left|\dfrac{1}{6}\right|=\left|\dfrac{-11}{12}\right|\)

\(x+\dfrac{1}{6}=\dfrac{11}{12}\)

\(x=\dfrac{11}{12}-\dfrac{1}{6}\)

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

Vậy ...

1/2+1/3<x<=1+1/2+1/5

=>5/6<x<=1+7/10

=>5/6<x<17/10

mà x là số nguyên

nên x=1

\(\dfrac{x+1}{2017}+\dfrac{x+2}{2016}=\dfrac{x+3}{2015}-1\)

\(\Leftrightarrow\dfrac{x+1}{2017}+1+\dfrac{x+2}{2016}+1=\dfrac{x+3}{2015}-1+2\)

\(\Leftrightarrow\dfrac{x+100}{2017}+\dfrac{x+100}{2016}=\dfrac{x+100}{2015}\)

\(\Leftrightarrow\dfrac{x+100}{2017}+\dfrac{x+100}{2016}+\dfrac{x+100}{2015}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}+\dfrac{1}{2015}\right)=0\)

Do \(\dfrac{1}{2017}+\dfrac{1}{2016}+\dfrac{1}{2015}\ne0\) nên \(x+100=0\)

\(\Leftrightarrow x=\left(-100\right)\)

Vậy \(x=\left(-100\right)\)

(Thêm ở cả 2 vế cùng một số để tạo ra nhân tử chung ở tử (x + 2018))