Lời giải:
\(\frac{-3}{14}+\frac{-12}{25}.\frac{-3}{14}+(\frac{-3}{14})^1\\ =\frac{-3}{14}+\frac{-12}{25}.\frac{-3}{14}+\frac{-3}{14}\\ =\frac{-3}{14}(1+\frac{-12}{25}+1)=\frac{-3}{14}.\frac{38}{25}=\frac{-57}{175}\)

Ta có: \(\frac{15}{14}\left(\frac{14}{9}-\frac{3}{7}\right)-\frac{15}{14}\left(\frac{14}{6}-\frac{3}{7}\right)\)

\(=\frac{15}{14}\left[\left(\frac{14}{9}-\frac{3}{7}\right)-\left(\frac{14}{6}-\frac{3}{7}\right)\right]\)

\(=\frac{15}{14}\left(\frac{14}{9}-\frac{3}{7}-\frac{14}{6}+\frac{3}{7}\right)\)

\(=\frac{15}{14}\left(\frac{14}{9}-\frac{14}{6}\right)\)

\(=\frac{15}{14}\cdot\left(\frac{28}{18}-\frac{42}{18}\right)\)

\(=\frac{15}{14}\cdot\frac{-14}{18}\)

\(=\frac{-15}{18}=-\frac{5}{6}\)

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?

thiếu

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

Bài 2: 

x=13 nên x+1=14

\(f\left(x\right)=x^{14}-x^{13}\left(x+1\right)+x^{12}\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+14\)

\(=x^{14}-x^{14}-x^{13}+x^{13}-...+x^3+x^2-x^2-x+14\)

=14-x=1

x=13 nên x+1=14

f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14

=x14−x14−x13+x13−...+x3+x2−x2−x+14=x14−x14−x13+x13−...+x3+x2−x2−x+14

=14-x=1

ko biết

 `-3/5xx9/14-3/5xx5/14+3/5`

`=(-3/5+3/5)xx(9/14+5/14)xx3/5`

`=0xx14xx3/5`

`=0xx1xx3/5`

`=0xx3/5`

`=0`

`@Nae`

\(-\dfrac{3}{5}\times\dfrac{9}{14}-\dfrac{3}{5}\times\dfrac{5}{14}+\dfrac{3}{5}\)

\(=\dfrac{3}{5}\times\left(-\dfrac{9}{14}-\dfrac{5}{14}+1\right)\)

\(=\dfrac{3}{5}\times\left(\dfrac{-9-5+14}{14}\right)\)

\(=\dfrac{3}{5}\times\dfrac{0}{14}\)

\(=0\)