\(a,1-3\left|2x-3\right|=-\dfrac{1}{2}\\ 3\left|2x-3\right|=1+\dfrac{1}{2}\\ 3\left|2x-3\right|=\dfrac{3}{2}\\ \left|2x-3\right|=\dfrac{3}{2}:3\\ \left|2x-3\right|=\dfrac{9}{2}\\ \Rightarrow\left[{}\begin{matrix}2x-3=\dfrac{9}{2}\\2x-3=-\dfrac{9}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=\dfrac{15}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

Vậy `x in {15/4;-3/4}`

\(b,\left(\left|x\right|-0,2\right)\left(x^3-8\right)=0\\ \left(\left|x\right|-0,2\right)\left(x-2\right)\left(x^2+2x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}\left|x\right|-0,2=0\\x-2=0\\x^2+2x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\left|x\right|=0,2\\x=2\\\left(x+1\right)^2+3=0\left(lọai\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0,2\\x=-0,2\\x=2\end{matrix}\right.\)

Vậy `x in {+-0,2;2}`

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: \(5^{x-1}+5^{x+2}=3\)

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

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

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

\(\left(2x-5\right)^2=0,81\)

\(\left(2x-5\right)^2=0,9^2\)

\(\Rightarrow2x-5=0,9\)

\(2x=0,9+5\)

\(2x=5,9\)

\(x=5,9:2\)

\(x=2,95\)

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\(\left(x-\frac{1}{3}\right)^3=0,027\)

\(\left(x-\frac{1}{3}\right)^3=0,3^3\)

\(\Rightarrow x-\frac{1}{3}=0,3\)

\(x=0,3+\frac{1}{3}\)

\(x=\frac{19}{30}\)

\(\left(2x-5\right)^2=0,81\)

\(\Rightarrow2x-5=0,9\)

\(\Rightarrow2x=5,9\)

\(\Rightarrow x=2,95\)

\(\left(x-\frac{1}{3}\right)^3=0,027\)

\(\Rightarrow x-\frac{1}{3}=0,3\)

\(\Rightarrow x=\frac{19}{30}\)

\(\left(0,09\right)^3=\left(\dfrac{9}{100}\right)^3=\left[\left(\dfrac{3}{10}\right)^2\right]^3=\left(\dfrac{3}{10}\right)^6\\ \left(\dfrac{3}{10}\right)^8=\left(\dfrac{3}{10}\right)^8\\ \left(0,027\right)^2=\left(\dfrac{27}{1000}\right)^2=\left[\left(\dfrac{3}{10}\right)^3\right]^2=\left(\dfrac{3}{10}\right)^6\)

a)\(\left[{}\begin{matrix}\dfrac{5}{2}-x=\dfrac{1}{3}\\\dfrac{5}{2}-x=-\dfrac{1}{3}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{13}{6}\\x=\dfrac{17}{6}\end{matrix}\right.\)

b) 8/6-x-1/5=0

9/6-x=1/5

x=13/10

a = 1,2

b = 1,8

a) \(\left|2,5-x\right|=1,3\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1,2\\x=3,8\end{matrix}\right.\)

b) \(1,6-\left|x-0,2\right|=0\)

\(\Leftrightarrow\left|x-0,2\right|=1,6\)

\(\Leftrightarrow\left[{}\begin{matrix}x-0,2=1,6\\x-0,2=-1,6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1,8\\x=-1,4\end{matrix}\right.\)

\(\frac{1}{3}\left|\frac{1}{4}x-\frac{1}{5}\right|+\frac{3}{7}\left|\frac{x}{y}-0,2\right|=0\)

Nhận xét : \(\left|\frac{1}{4}x-\frac{1}{5}\right|\ge0\)với \(\forall x\)(vì giá trị tuyệt đối không âm) 

     \(\Rightarrow\frac{1}{3}\left|\frac{1}{4}x-\frac{1}{5}\right|\ge0\)(1) 

  \(\left|\frac{x}{y}-0,2\right|\ge0\)với \(\forall x\),\(\left(y\ne0\right)\)(vì giá trị tuyệt đối không âm)

    \(\Rightarrow\frac{3}{7}\left|\frac{x}{y}-0,2\right|\ge0\)  (2)

Từ (1) và (2) => \(\frac{1}{3}\left|\frac{1}{4}x-\frac{1}{5}\right|+\frac{3}{7}\left|\frac{x}{y}-0,2\right|\ge0\)

Để \(\frac{1}{3}\left|\frac{1}{4}x-\frac{1}{5}\right|+\frac{3}{7}\left|\frac{x}{y}-0,2\right|=0\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{3}\left|\frac{1}{4}x-\frac{1}{5}\right|=0\\\frac{3}{7}\left|\frac{x}{y}-0,2\right|=0\end{cases}\Leftrightarrow}\hept{\begin{cases}\left|\frac{1}{4}x-\frac{1}{5}\right|=\frac{1}{3}\\\left|\frac{x}{y}-0,2\right|=\frac{3}{7}\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{1}{4}x-\frac{1}{5}=0\\\frac{x}{y}-0,2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{4}x=\frac{1}{5}\\\frac{x}{y}=0,2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{4}{5}\\\frac{4}{5}\div y=0,2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{4}{5}\\y=4\left(tm\right)\end{cases}}\)

Vậy \(x=\frac{4}{5};y=4\) (Tm)

\(\Rightarrow x-2,2=0,2+x \)

\(\Leftrightarrow x\in rổng\)

Đây là toán chuyên đề giá trị tuyệt đối. Hôm nay, Olm sẽ hướng dẫn các em giải chi tiết dạng này như sau:

               |\(x-2,2\)| = |0,2  + \(x\)|

               \(\left[{}\begin{matrix}x-2,2=0,2+x\\x-2,2=-0,2-x\end{matrix}\right.\)

               \(\left[{}\begin{matrix}-2,2=0,2\left(loại\right)\\x+x=-0,2+2,2\end{matrix}\right.\)

                 2\(x\) = -0,2 + 2,2

                  2\(x\) = 2 

                  \(x=2:2\)

                 \(x=1\)

        Vậy \(x\) = 1

1.

\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{2}{3}.\left[\dfrac{2}{3}-\dfrac{2}{3}.\left(\dfrac{2}{3}-\dfrac{1}{2}\right)\right]\right\}\)

=\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{2}{3}.\left[\dfrac{2}{3}-\dfrac{2}{3}.\dfrac{1}{6}\right]\right\}\)

=\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{2}{3}.\left[\dfrac{2}{3}-\dfrac{1}{9}\right]\right\}\)

=\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{2}{3}.\dfrac{5}{9}\right\}\)

=\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{10}{27}\right\}\)

=\(\dfrac{2}{3}.\dfrac{8}{27}\)

=...