=-27/5x -4/15 + -4/15x2010 - -4/15x-27/5                                                                                                                                                            =-4/15x2010                                                                                                                                                                                                         =-52

\(\left|x+\frac{1}{3}\right|+\frac{4}{5}=\left|-3,2+\frac{2}{5}\right|+\left(27-\frac{3}{5}\right)\left(27-\frac{3^2}{6}\right)...\left(27-\frac{3^5}{9}\right)...\left(27-\frac{3^{2010}}{2014}\right)\)

\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}+\left(27-\frac{3^2}{6}\right)\left(27-\frac{3^3}{7}\right)...\left(27-27\right)...\left(27-\frac{3^{2010}}{2014}\right)\)

\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\)

\(\Leftrightarrow\left|x+\frac{1}{3}\right|=2\)

\(\Rightarrow\hept{\begin{cases}x+\frac{1}{3}=2\\x+\frac{1}{3}=-2\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\x=-\frac{7}{3}\end{cases}}}\)

bạn ơi, có một chỗ chưa chuẩn .bạn kiểm tra lại giú mình. chỗ vế trái bạn thiếu \(\left(27-\frac{3}{5}\right)\). bạn bổ sung vào cho đúng nhé. dù sao vẫn cảm ơn bạn.

Mọi người đánh giúp mình nhé! Hạn là tối nay!!

\(e,\frac{22}{15}-x=-\frac{8}{27}\)

=> \(x=\frac{22}{15}-\left[-\frac{8}{27}\right]\)

=> \(x=\frac{22}{15}+\frac{8}{27}\)

=> \(x=\frac{198}{135}+\frac{40}{135}=\frac{198+40}{135}=\frac{238}{135}\)

\(g,\left[\frac{2x}{5}-1\right]:\left[-5\right]=\frac{1}{4}\)

=> \(\left[\frac{2x}{5}-\frac{1}{1}\right]=\frac{1}{4}\cdot\left[-5\right]\)

=> \(\left[\frac{2x}{5}-\frac{5}{5}\right]=-\frac{5}{4}\)

=> \(\frac{2x-5}{5}=-\frac{5}{4}\)

=> \(2x-5=-\frac{5}{4}\cdot5=-\frac{25}{4}\)

=> \(2x=-\frac{5}{4}\)

=> \(x=-\frac{5}{8}\)

\(h,-2\frac{1}{4}x+9\frac{1}{4}=20\)

=> \(-\frac{9}{4}x+\frac{37}{4}=20\)

=> \(-\frac{9}{4}x=20-\frac{37}{4}=\frac{43}{4}\)

=> \(x=\frac{43}{4}:\left[-\frac{9}{4}\right]=\frac{43}{4}\cdot\left[-\frac{4}{9}\right]=\frac{43}{1}\cdot\left[-\frac{1}{9}\right]=-\frac{43}{9}\)

\(i,-4\frac{3}{5}\cdot2\frac{4}{23}\le x\le-2\frac{3}{5}:1\frac{6}{15}\)

=> \(-\frac{23}{5}\cdot\frac{50}{23}\le x\le-\frac{13}{5}:\frac{21}{15}\)

=> \(-\frac{1}{1}\cdot\frac{10}{1}\le x\le-\frac{13}{5}\cdot\frac{15}{21}\)

=> \(-10\le x\le-\frac{13}{1}\cdot\frac{3}{21}\)

=> \(-10\le x\le-\frac{13}{1}\cdot\frac{1}{7}\)

=> \(-10\le x\le-\frac{13}{7}\)

Đến đây tìm x

\(x^2=\frac{4}{5}.\frac{15}{27}=\frac{4.3.5}{5.3^3}=\frac{4}{3^2}=\frac{2^2}{3^2}=\left(\frac{2}{3}\right)^2\)

=> \(x=\pm\frac{2}{3}\)

Qui đồng rồi khử mẫu, ta được:

\(4x+12.\left(27-x\right)=15x+5.\left(27-x\right)\)

\(\Leftrightarrow4x+324-12x=15x+135-5x\)

\(\Leftrightarrow4x-12x-15x+5x=135-324\)

\(\Leftrightarrow-18x=-189\Leftrightarrow x=\frac{21}{2}=10,5\)

Vậy x = 10,5

\(\frac{x}{15}+\frac{27-x}{5}=\frac{x}{4}+\frac{27-x}{12}\)

\(\frac{x}{15}+\frac{3\left(27-x\right)}{15}=\frac{3x}{12}+\frac{27-x}{12}\)

\(\frac{x}{15}+\frac{81-3x}{15}=\frac{3x}{12}+\frac{27-x}{12}\)

\(\frac{x+81-3x}{15}=\frac{3x+27-x}{12}\)

\(\frac{-2x+81}{15}=\frac{2x+27}{12}\)

\(12\left(-2x+81\right)=15\left(2x+27\right)\)

\(-24x+972=30x+405\)

\(972-405=30x+24x\)

\(567=54x\)

\(x=567:54\)

\(x=10,5\)

\(\frac{x}{15}+\frac{27-x}{5}=\frac{x}{4}+\frac{27-x}{12}\)

\(\Leftrightarrow\frac{4x}{60}+\frac{12\left(27-x\right)}{5}=\frac{15x}{60}+\frac{5\left(27-x\right)}{12}\)

\(\Leftrightarrow4x+324-12x=15x+135-5x\)

\(\Leftrightarrow4x-12x-15x+5x=135-324\)

\(\Leftrightarrow-18x=-189\)

\(x=\frac{-189}{-18}=\frac{21}{2}=10,5\)