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

=>\(\left|x-\dfrac{1}{3}\right|+0,8=\left|-3,2+0,4\right|=2,8\)

=>\(\left|x-\dfrac{1}{3}\right|=2,8-0,8=2\)

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

đề sai

a: \(\left(\sqrt{3}\right)^x=243\)

=>\(3^{\dfrac{1}{2}\cdot x}=3^5\)

=>\(\dfrac{1}{2}\cdot x=5\)

=>x=10

b: \(0,1^x=1000\)

=>\(\left(\dfrac{1}{10}\right)^x=1000\)

=>\(10^{-x}=10^3\)

=>-x=3

=>x=-3

c: \(\left(0,2\right)^{x+3}< \dfrac{1}{5}\)

=>\(\left(0,2\right)^{x+3}< 0,2\)

=>x+3>1

=>x>-2

d: \(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{5}{3}\right)^2\)

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

=>2x+1<-2

=>2x<-3

=>\(x< -\dfrac{3}{2}\)

e: \(5^{x-1}+5^{x+2}=3\)

=>\(5^x\cdot\dfrac{1}{5}+5^x\cdot25=3\)

=>\(5^x=\dfrac{3}{25,2}=\dfrac{1}{8,4}=\dfrac{10}{84}=\dfrac{5}{42}\)

=>\(x=log_5\left(\dfrac{5}{42}\right)=1-log_542\)

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

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

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

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

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

Vậy..............

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

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

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

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

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

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

\(x = \frac{7}{3}\)

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

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

\(x = - \frac{5}{3}\)

Vậy x = \(x = \frac{7}{3}; x = - \frac{5}{3}\)

a: \(P=\left(\dfrac{3x+6}{2\left(x^2+4\right)}-\dfrac{2x^2-x-10}{\left(x+1\right)\left(x^2+1\right)}\right):\left(\dfrac{10\left(x^2-1\right)+3\left(x^2+1\right)\left(x-1\right)-6\left(x+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\cdot2}\right)\cdot\dfrac{2}{x-1}\)

\(=\left(\dfrac{\left(3x+6\right)\left(x^3+x^2+x+1\right)-\left(2x^2+8\right)\left(2x^2-x-10\right)}{2\left(x^2+4\right)\left(x+1\right)\left(x^2+1\right)}\right)\cdot\dfrac{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)\cdot2}{-3x^3+x^2-3x-13}\cdot\dfrac{2}{x-1}\)

\(=\dfrac{-x^4+11x^3+13x^2+17x+16}{\left(x^2+4\right)}\cdot\dfrac{2}{-3x^3+x^2-3x-13}\)

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

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

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

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

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

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

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

làm tiếp câu a) nhé

b) \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}x=7\\x-7=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=8\end{cases}}\)

a: =>11(x-3)=6(x-5)

=>11x-33=6x-30

=>5x=3

=>x=3/5

b: =>(4/3-1/4x-5/12)-2x=8/5*5/3=8/3

=>-9/4x+11/12=8/3

=>-9/4x=32/12-11/12=21/12=7/4

=>x=-7/9

c: =>1/2x-1/3-2/3x-1=x

=>-1/6x-4/3=x

=>-7/6x=4/3

=>x=-4/3:7/6=-4/3*6/7=-24/21=-8/7

d: =>1-2x-3x+1=7/2

=>-5x=3/2

=>x=-3/10