\(\frac{3x-11}{2}-\frac{x-3}{3}=\frac{1}{6}\)

\(\frac{3\times\left(3x-11\right)}{3\times2}-\frac{2\times\left(x-3\right)}{2\times3}=\frac{1}{6}\)

\(\frac{9x-33}{6}-\frac{2x-6}{6}=\frac{1}{6}\)

\(\frac{\left(9x-33\right)-\left(2x-6\right)}{6}=\frac{1}{6}\)

\(9x-33-2x+6=1\)

\(\left(9x-2x\right)-\left(33-6\right)=1\)

\(7x-27=1\)

\(7x=1+27\)

\(7x=28\)

\(x=\frac{28}{7}\)

\(x=4\)

Chúc bạn học tốt

\(PT\Leftrightarrow\frac{3.\left(3x-11\right)-2.\left(x-3\right)}{6}=\frac{1}{6}\)

<=> 3.(3x - 11) - 2.(x - 3) = 1

<=> 9x - 33 - 2x + 6 = 1

<=> 7x = 28

<=> x = 4

\(\frac{3x-11}{2}-\frac{x-3}{3}=\frac{1}{6}\) MC: 6

\(\Leftrightarrow\frac{3\left(3x-11\right)}{6}-\frac{2\left(x-3\right)}{6}=\frac{1}{6}\)

\(\Leftrightarrow3\left(3x-11\right)-2\left(x-3\right)=1\)

\(\Leftrightarrow9x-33-2x+6=1\)

\(\Leftrightarrow9x-2=33-6+1\)

\(\Leftrightarrow7x=28\)

\(\Leftrightarrow x=4\)

Vậy x = 4

|x| = 3/4 

=>  x= -3/4 ; 3/4

Mà x < 0

=>  x= -3/4

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

\(\Rightarrow x\in\varphi\)

|x| = 0,35 

=>  x = -0,35; 0,35

Mà x >0 

=> x= 0,35

b) \(\left(\frac{2}{3}x-1\right).\left(\frac{3}{4}x+\frac{1}{2}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x-1=0\\\frac{3}{4}x+\frac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x=1\\\frac{3}{4}x=-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1:\frac{2}{3}\\x=\left(-\frac{1}{2}\right):\frac{3}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=-\frac{2}{3}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{3}{2};-\frac{2}{3}\right\}.\)

c) \(x:\frac{9}{14}=\frac{7}{3}:x\)

\(\Rightarrow\frac{x}{\frac{19}{4}}=\frac{\frac{7}{3}}{x}\)

\(\Rightarrow x.x=\frac{7}{3}.\frac{19}{4}\)

\(\Rightarrow x.x=\frac{133}{12}\)

\(\Rightarrow x^2=\frac{133}{12}\)

\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\frac{133}{12}}\\x=-\sqrt{\frac{133}{12}}\end{matrix}\right.\)

Vậy \(x\in\left\{\sqrt{\frac{133}{12}};-\sqrt{\frac{133}{12}}\right\}.\)

d) \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\)

\(\Rightarrow\left(3x-1\right)^{10}-\left(3x-1\right)^{20}=0\)

\(\Rightarrow\left(3x-1\right)^{10}.\left[1-\left(3x-1\right)^{10}\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(3x-1\right)^{10}=0\\1-\left(3x-1\right)^{10}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-1=0\\\left(3x-1\right)^{10}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x-1=\pm1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1:3\\3x-1=1\\3x-1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\3x=2\\3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=\frac{2}{3}\\x=0\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{1}{3};\frac{2}{3};0\right\}.\)

Chúc bạn học tốt!

M=9.1000-[169-(25+4+4+15)]1000

M=9000-[169-48]1000

M=9000-121.1000

M=9000-121000

M=-112000

234rtf

Ta có \(3^2.10^3-\left[13^2-\left(5^2+4+2^2+15\right)\right]10^3\)

          \(=9.1000-\left[169-\left(25+4+4+15\right)\right]1000\)

           \(=9.1000-\left(169-48\right)1000\)

           \(=1000\left[9-\left(169-48\right)\right]\)

            \(=1000.\left(-112\right)\) 

              \(=-112000\)

\(\frac{x+10}{2000}+\frac{x+20}{1990}+\frac{x+30}{1980}+\frac{x+40}{1970}=-4\)

\(\Leftrightarrow\frac{x+10}{2000}+1+\frac{x+20}{1990}+1+\frac{x+30}{1980}+1+\frac{x+40}{1970}+1=0\)

\(\Leftrightarrow\frac{x+2010}{2000}+\frac{x+2010}{1990}+\frac{x+2010}{1980}+\frac{x+2010}{1970}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}\right)=0\)

Vì  \(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}>0\)

\(\Rightarrow x+2010=0\)

\(\Leftrightarrow x=-2010\)

\(\Leftrightarrow\frac{x+10}{2000}+1+\frac{x+20}{1990}+1+\frac{x+30}{1980}+1+\frac{x+40}{1970}+1=0\)

\(\Leftrightarrow\frac{x+2010}{2000}+\frac{x+2010}{1990}+\frac{x+2010}{1980}+\frac{x+2010}{1970}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}\right)=0\)

mà\(\left(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}\right)\ne0\Rightarrow\left(x+2010\right)=0\\ \Rightarrow x=-2010\)

x-4-1 = 10-2x

x = 5

nhanh chưa bn

x=5

k mik nha

mik k lại cho bạn heng

bài dễ z mà k hỉu hả, mẫu số = nhau thi tử số = nhau,