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

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

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

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

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

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.

mai bn đăng lại nhé mk lm cho h đi ngủ

\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)+\frac{2}{5}\right|\\ \left|x-\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\\ \left|x-\frac{1}{3}\right|=2 \\ \Rightarrow\left[{}\begin{matrix}x-\frac{1}{3}=2\\x-\frac{1}{3}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{3}\\x=-\frac{5}{3}\end{matrix}\right.\)

Vậy...

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

=> \(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-\frac{16}{5}\right)+\frac{2}{5}\right|\)

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

=> \(\left[{}\begin{matrix}x-\frac{1}{3}=\frac{14}{5}\\x-\frac{1}{3}=-\frac{14}{5}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{14}{5}+\frac{1}{3}\\x=-\frac{14}{5}+\frac{1}{3}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{47}{15}\\x=-\frac{37}{15}\end{matrix}\right.\)

Vậy:..................................

P/s: Ko chắc!

a)\(\left|x+\frac{1}{5}\right|-4=-2\)

\(\Rightarrow\left|x+\frac{1}{5}\right|=2\)

\(\Rightarrow x+\frac{1}{5}=2\) hoặc \(-2\)

Xét \(x+\frac{1}{5}=2\Leftrightarrow x=\frac{9}{5}\)

Xét \(x+\frac{1}{5}=-2\Leftrightarrow x=-\frac{11}{5}\)

phần a dấu + fai là dấu =

a. \(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)+\frac{2}{5}\right|\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=\left|-\frac{16}{5}+\frac{2}{5}\right|-\frac{4}{5}\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=\left|-\frac{14}{5}\right|-\frac{4}{5}\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=\frac{14}{5}-\frac{4}{5}\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=2\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=2\\x-\frac{1}{3}=-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=-\frac{5}{3}\end{cases}.}\)

Vậy \(x\in\left\{-\frac{5}{3};\frac{7}{3}\right\}.\)

b. \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)\(\Leftrightarrow\left(x-7\right)^{x+1}-\left(x-7\right)^{x+1}\times\left(x-7\right)^{10}=0\)\(\Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\Leftrightarrow\orbr{\begin{cases}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{10}=0\end{cases}.}\)Xét 2 trường hợp:

Vậy \(x\in\left\{6;7;8\right\}.\)

ban nguyen nhat minh giang lai cho mk dong 2 cau b cai

mk cam on

Dòng 2 câu b là đặt nhân tử chung \(\left(x-7\right)^{x+1}\)ra ngoài

Ta có: 1/2 - (1/3 + 1/4) = 1/2 - 7/12 = -1/12 ;

           1/48 - (1/16 - 1/6) = 1/48 + 5/48 = 1/8

Vì \(-\frac{1}{12}< x< \frac{1}{8}\) nên x = 0

Ta có :

\(4\frac{5}{9}:2\frac{5}{18}-7=2-7=-5\)

\(\left(3\frac{1}{5}:3,2+4,5.1\frac{31}{45}\right):\left(-21\frac{2}{3}\right)=\left(1+\frac{38}{5}\right):\left(-21\frac{2}{3}\right)=\frac{43}{5}:\frac{-65}{3}=-\frac{129}{325}\)

Vì \(-5< x< -\frac{129}{325}\) nên \(x\in\left\{-4;-3;-2;-1\right\}\)

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

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

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

Vậy...

cau b nx

\(\Rightarrow\left(x-7\right)^{x+1}.\left(1-\left(x-7\right)^{10}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-7=0\\\left(x-7\right)^{10}=1=1^{10}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=7\\x=6;x=8\end{cases}}\)

\(\text{VT = }10,07-3,927+3,63-\frac{3}{20}=9,623\)

\(VP=13,8-\frac{3}{5}+\frac{7}{50}=13,34\)

\(\text{Vì }x\in Z\Rightarrow x\left\{10;11;12;13\right\}\)