\(1\frac{2}{7}=\frac{9}{7}\)

Số đó là : 

\(-18:\frac{9}{7}=-14\)

Vậy số đó là \(-14\)

Chúc bạn học tốt !!!! 

ngu

1/2/7=9/7

9/7x(-18)

-23/1/7

con óc chó

Đặt A \(=1+5+5^2+...+5^{2000}\)

Ta có: 5A\(=5+5^2+5^3+...+5^{2001}\)

5A-A\(=5^{2001}-1\)

4A\(=5^{2001}-1\)

A\(=\frac{5^{2001}-1}{4}\)

lại gặp 1 đứa ngu

Ta có :  \(\left|x-22\right|\ge0\forall x\)

\(\left|y-12\right|\ge0\forall y\)

\(\Rightarrow\left|x-22\right|+\left|y-12\right|\ge0\forall x;y\)

Dấu \("="\)xảy ra \(\Leftrightarrow\hept{\begin{cases}\left|x-22\right|=0\\\left|y-12\right|=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-22=0\\y-12=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=22\\y=12\end{cases}}\)

Vậy  \(\hept{\begin{cases}x=22\\y=12\end{cases}}\)

Trường hợp : 

\(\left|x-22\right|+\left|y-12\right|< 0\left(l\right)\)

Để \(-\frac{1}{n+2018}\in Z\)

\(\Leftrightarrow-1⋮n+2018\)

\(\Leftrightarrow n+2018\inƯ\left(-1\right);n+2018\in Z\)

\(\Leftrightarrow n+2018\in\left\{1;-1\right\}\)

\(\Leftrightarrow n\in\left\{-2017;-2019\right\}\)

Vậy  \(n\in\left\{-2017;-2019\right\}\)

Chúc bạn học tốt !!! 

Đáp án là n thuộc -2017;-2019

a)\(=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}\)

\(=\left(\frac{-3}{7}+\frac{3}{7}\right)-\left(\frac{15}{26}+\frac{2}{13}\right)\)

\(=0-\frac{19}{26}\)

\(=-\frac{19}{26}\)

c)\(=\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)

\(=\frac{-11}{23}.2-\frac{1}{23}\)

\(=\frac{-22}{23}-\frac{1}{23}\)

\(=-1\)

\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)

\(\Rightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=-4+4\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)

\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

Mà \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}>0\)

\(\Rightarrow x+100=0\)

\(\Rightarrow x=0-100\)

\(\Rightarrow x=-100\)

Chúc bạn học tốt !!!! 

\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)

\(\Leftrightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=0\) 

\(\Leftrightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\) 

\(\Leftrightarrow x+100=0\) 

\(\Leftrightarrow x=-100\) 

Vậy...

Gọi \(d\in\left(2n+1;3n+2\right)\)

\(\Rightarrow\hept{\begin{cases}2n+1⋮d\\3n+2⋮d\end{cases}\Rightarrow\hept{\begin{cases}6n+3⋮d\\6n+4⋮d\end{cases}\Rightarrow1⋮}d\Rightarrow d=1}\)

\(\Rightarrowđpcm\)

ko hieu phan dpcm

ko hiểu đề