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

     \(\orbr{\begin{cases}2x+\frac{3}{4}=\frac{1}{2}\\2x+\frac{3}{4}=\frac{-1}{2}\end{cases}}\) =>   \(\orbr{\begin{cases}2x=\frac{1}{2}-\frac{3}{4}\\2x=\frac{-1}{2}-\frac{3}{4}\end{cases}}\)  =>   \(\orbr{\begin{cases}2x=\frac{-1}{4}\\2x=\frac{-5}{4}\end{cases}}\) =>   \(\orbr{\begin{cases}x=\frac{-1}{8}\\x=\frac{-5}{8}\end{cases}}\)

Vậy \(x=\left\{\frac{-1}{8},\frac{-5}{8}\right\}\)

b) \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{2\frac{1}{4}}\)= \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{\frac{9}{4}}\)

=> \(3x.\frac{9}{4}=2,7.\frac{1}{4}\)=>  \(\frac{27x}{4}=\frac{27}{40}\)

\(27x.40=27.4\)

\(1080.x=108\)

             \(x=\frac{1}{10}\)

Vậy \(x=\frac{1}{10}\)

c) \(\left|x-1\right|+4=6\)

\(\left|x-1\right|=6-4\)

\(\left|x-1\right|=2\)

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

Vậy \(x=\left[3,-1\right]\)

d) \(\frac{x}{3}=\frac{y}{5}=>\frac{y}{5}=\frac{x}{3}=>\frac{y-x}{5-3}=\frac{24}{2}=12\)

e) \(\left(x^2-3\right)^2=16\)

\(\left(x^2-3\right)^2=4^2\)\(=>x^2-3=4\)

\(x^2=7=>x=\sqrt{7}\)

Vậy \(x=\sqrt{7}\)

f) \(\frac{3}{4}+\frac{2}{5}x=\frac{29}{60}\)

               \(\frac{2}{5}x=\frac{29}{60}-\frac{3}{4}\) 

               \(\frac{2}{5}x=-\frac{4}{15}\)

          \(x=-\frac{4}{15}:\frac{2}{5}=-\frac{4}{15}.\frac{5}{2}=-\frac{2}{3}\)

Vậy \(x=-\frac{2}{3}\)

g) \(\left(-\frac{1}{3}\right)^3.x=\frac{1}{81}\)

\(\left(-\frac{1}{27}\right).x=\frac{1}{81}\)

\(x=\left(-\frac{1}{27}\right):\frac{1}{81}=\left(-\frac{1}{27}\right).81=-3\)

Vậy \(x=-3\)

k)\(\frac{3}{4}-\frac{2}{5}x=\frac{29}{60}\)

\(\frac{2}{5}x=\frac{3}{4}-\frac{29}{60}\)

\(\frac{2}{5}x=\frac{4}{15}\)

      \(x=\frac{2}{5}-\frac{4}{15}=>x=\frac{2}{15}\)

Vậy \(x=\frac{2}{15}\)

I) \(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)

\(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}\)

\(\frac{3}{5}x=\frac{5}{14}\)

\(x=\frac{5}{14}:\frac{3}{5}=\frac{5}{14}.\frac{5}{3}=\frac{25}{42}\)

Vậy \(x=\frac{25}{42}\)

Câu 1 có sai đề bài không đấy?

Câu 2: Ta có \(S=6^2+18^2+30^2+...+126^2\)

                   \(S=6^2\left(1^2+3^2+5^2+...+21^2\right)\)

                       \(=6^2.1771=36.1771=63756\)

2 cách nhé các bạn

\(A=1+3+3^2+3^3+...+3^{2016}\)

\(A=1+3\left(1+3^2+...+3^{2015}\right)\)

\(A=1+3\left(A-3^{2016}\right)\)

\(A=1+3A-3^{2017}\)

\(2A=3^{2017}-1\Rightarrow A=\frac{3^{2017}-1}{2}\)

\(A< B\)

\(a)\frac{3}{4}+\frac{2}{5}x=\frac{29}{60}\)

  \(\)TỰ LÀM NHA HIHI

 MI SUỐT NGÀY NGỒI MÁY TÍNH LƯỚT FACE, LÚC NÀO ĐI QUA CŨNG THẤY

a) \(\dfrac{5}{6}:x=30:3\)

\(\Leftrightarrow\dfrac{5}{6}:x=10\)

\(\Leftrightarrow x=\dfrac{5}{6}:10\)

\(\Leftrightarrow x=\dfrac{1}{12}\)

Vậy .......

b) \(x:2,5=0,003:0,75\)

\(\Leftrightarrow x:2,5=0,004\)

\(\Leftrightarrow x=0,004.2,5\)

\(\Leftrightarrow x=0,01\)

Vậy .......

c) \(3,8:\left(2x\right)=\dfrac{1}{4}:2\dfrac{2}{3}\)

\(\Leftrightarrow3,8:\left(2x\right)=\dfrac{1}{4}:\dfrac{8}{3}=\dfrac{3}{32}\)

\(\Leftrightarrow2x=3,8:\dfrac{3}{32}\)

\(\Leftrightarrow2x=\dfrac{698}{25}\)

\(\Leftrightarrow x=\dfrac{304}{15}\)

Vậy ...

d) \(\dfrac{2}{3}:0,4=x:\dfrac{4}{5}\)

\(\Leftrightarrow x:\dfrac{4}{5}=\dfrac{2}{3}\)

\(\Leftrightarrow x=\dfrac{8}{15}\)

Vậy ....

e) \(3\dfrac{4}{5}:40\dfrac{8}{15}=0,25:x\)

\(\Leftrightarrow0,25:x=\dfrac{19}{5}:\dfrac{608}{15}\)

\(\Leftrightarrow0,25x=\dfrac{57}{608}\)

\(\Leftrightarrow x=\dfrac{228}{608}\)

Vậy ...

e) \(\dfrac{x}{-15}=\dfrac{-60}{x}\)

\(\Leftrightarrow xx=\left(-60\right)\left(-15\right)\)

\(\Leftrightarrow x^2=900\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=30^2\\x^2=\left(-30\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\)

Vậy ...

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