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

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

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

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

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

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

\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|-2,8\right|\)

\(\left|x-\frac{1}{3}\right|=2,8-\frac{4}{5}\)

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

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

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

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

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

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

\(\left|x-\frac{1}{3}\right|=\frac{19}{5}\)

\(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{3}=\frac{19}{5}\Leftrightarrow x=\frac{62}{15}\\x-\frac{1}{3}=-\frac{19}{5}\Leftrightarrow x=-\frac{52}{15}\end{cases}}\)

vây x= \(\frac{62}{15}\)hoặc x=\(-\frac{52}{15}\)

Chúc bn hok tốt

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

\(\Rightarrow|x-\frac{1}{3}|+\frac{4}{5}=|-\frac{7}{10}|\)

\(\Rightarrow|x-\frac{1}{3}|+\frac{4}{5}=\frac{7}{10}\)

\(\Rightarrow|x-\frac{1}{3}|=\frac{7}{10}-\frac{4}{5}\)

\(\Rightarrow|x-\frac{1}{3}|=-\frac{1}{10}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{10}\\x-\frac{1}{3}=-\frac{1}{10}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{13}{30}\\x=\frac{7}{30}\end{cases}}\)

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*****

d,

\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)

e,

\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)

\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)

\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)

Vậy không tồn tại $x$ thỏa mãn đề bài.

f, 

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

\(\Leftrightarrow 6x-3=10+6x\)

\(\Leftrightarrow 13=0\) (vô lý)

Vậy không tồn tại $x$ thỏa mãn đề bài.

a,

$0-|x+1|=5$

$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)

Do đó không tồn tại $x$ thỏa mãn điều kiện đề.

b,

\(2-|\frac{3}{4}-x|=\frac{7}{12}\)

\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)

\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)

c, 

\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)

\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)

\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)

\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)

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 \(x\in\left\{-4;-3;-2;-1\right\}\)

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