1/3-(-2/7+1/3)
= 1/3 + 2/7 - 1/3
= (1/3 - 1/3) + 2/7
= 2/7

\(\frac{1}{2}\cdot\frac{18}{14}+\frac{1}{2}\cdot\frac{3}{14}-\frac{1}{2}\cdot\frac{7}{14}\)

\(=\frac{1}{2}\left(\frac{18}{14}+\frac{3}{14}-\frac{7}{14}\right)\)

\(=\frac{1}{2}\cdot1=\frac{1}{2}\)

\(a:=\left(\dfrac{1}{4}+\dfrac{3}{4}\right)+\left(\dfrac{1}{5}+\dfrac{2}{5}\right)\\ =1+\dfrac{3}{5}=1\dfrac{3}{5}\\ b:=\left(\dfrac{2}{7}+\dfrac{1}{7}\right)+\left(\dfrac{5}{14}+\dfrac{3}{14}\right)\\ =\dfrac{3}{7}+\dfrac{8}{14}\\ =\dfrac{3}{7}+\dfrac{4}{7}=1\)

a: \(=\dfrac{2}{5}+\dfrac{3}{5}\cdot\dfrac{10}{3}\cdot\dfrac{1}{2}=\dfrac{2}{5}+\dfrac{10}{5}\cdot\dfrac{1}{2}=\dfrac{2}{5}+1=\dfrac{7}{5}\)

Đặt A=1+14+14²+14³+...+14¹⁴

A=1+14+14²+14³+...+14¹⁴=(1+14)+(14²+14³)+...+(14¹³+14¹⁴)

A=(1+14)+14²x(1+14)+...+14¹³x(1+14)

A=15+14²x15+...+14¹³x15

A=15x(1+14²+...+14¹³)

Vì 15 chia hết cho 3=>Vậy A chia hết cho 3(dpcm)

Đề bài có đúng không vậy nếu bỏ số 1 hoặc số 14¹⁴ thì mình làm được 

Này nhé. Nói cho mà biết muốn thì tự mà làm,liu liu?

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

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

|7 - \(\dfrac{3}{4}\)\(x\)| = 2 + \(\dfrac{3}{2}\)

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

\(\left[{}\begin{matrix}7-\dfrac{3}{4}x=\dfrac{7}{2}\\7-\dfrac{3}{4}x=-\dfrac{7}{2}\end{matrix}\right.\) 

\(\left[{}\begin{matrix}\dfrac{3}{4}x=7-\dfrac{7}{2}\\\dfrac{3}{4}=7+\dfrac{7}{2}\end{matrix}\right.\)

\(\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{7}{2}\\\dfrac{3}{4}x=\dfrac{21}{2}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{14}{3}\\x=14\end{matrix}\right.\)

 5  - |\(x-3\)| = 5

       |\(x-3\)| = 5 - 5

      |\(x-3\)| = 0

      \(x-3\) = 0 

      \(x\) = 3

|\(\dfrac{7}{2}\) - 42| + 14 = 14 ( vô lý xem lại đề bài nhé em)

\(\dfrac{2}{3}\) - |\(x+2\)| = 2

     |\(x+2\)| = \(\dfrac{2}{3}\) - 2

    |\(x+2\)| = - \(\dfrac{4}{3}\)  

   |\(x+2\)| ≥ 0 ∀ \(x\)

Vậy \(x\in\) \(\varnothing\)

a) Đặt \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}< \frac{1}{2}\)

Vậy A<\(\frac{1}{2}\).

b) Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)

                  \(\frac{1}{3^2}< \frac{1}{2.3}\)

                  \(\frac{1}{4^2}< \frac{1}{3.4}\)

                    ...

                   \(\frac{1}{100^2}< \frac{1}{99.100}\)

\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(B< 1-\frac{1}{100}< 1\)

Vậy \(B< 1\).