\(\frac{-\left(-x\right)}{5}-\frac{2}{10}=\frac{1}{-5}-\frac{7}{50}\)

\(\frac{x}{5}-\frac{2}{10}=\frac{-1}{5}-\frac{7}{50}\)

\(\frac{x}{5}-\frac{2}{10}=-\frac{1}{5}+-\frac{7}{50}\)

\(\frac{x}{5}-\frac{2}{10}=-\frac{17}{50}\)

           \(\frac{x}{5}=-\frac{17}{50}+\frac{2}{10}\)

           \(\frac{x}{5}=-\frac{7}{50}\)

     \(\Rightarrow x=-\frac{7}{50}.5\)

Vậy     \(x=-\frac{7}{10}\)

\(\frac{x-4}{2015}-\frac{1}{2015}=\frac{10-2x}{2015}\)

\(\Rightarrow\frac{x-4}{2015}-\frac{10-2x}{2015}=\frac{1}{2015}\)

\(\Rightarrow\frac{x-4-\left(10-2x\right)}{2015}=\frac{1}{2015}\)

\(\Rightarrow\frac{\left(x+2x\right)-\left(4+10\right)}{2015}=\frac{1}{2015}\)

\(\Rightarrow\frac{3x-14}{2015}=\frac{1}{2015}\)

\(\Rightarrow\left(3x-14\right).2015=2015\)

\(\Rightarrow3x-14=1\) ( bớt cả 2 vế đi 2015 lần )

\(\Rightarrow3x=15\)

\(\Rightarrow x=5\)

Vậy \(x=5\)

X=5

Ta có: \(\frac{1}{x}-\frac{y}{8}=\frac{1}{16}\)

=> \(\frac{1}{x}=\frac{1}{16}+\frac{y}{8}\)

=> \(\frac{1}{x}=\frac{1+2y}{16}\)

=> 1.16 = x(1 + 2y)

=> x(1 + 2y) = 16 = 1 . 16 = 2 . 8 = 4.4

Vì 1 + 2y là số lẽ nên 1 + 2y \(\in\){1; -1} => x \(\in\){16; -16}

Lập bảng :

Vậy ...

 :

1x =116 

=>                        => 

        X = 1.16:1 =16

                                      Y=1.8:16= 0.5

y8 =116 

Vậy X = 16 ; Y=0.5       

 :                            

x-4-1 = 10-2x

x = 5

nhanh chưa bn

x=5

k mik nha

mik k lại cho bạn heng

\(\frac{x-1}{2015}-\frac{1}{2015}=\frac{10-2x}{2015}\)

\(\Rightarrow x-1-1=10-2x\)

\(\Rightarrow x-2=10-2x\)

\(\Rightarrow-2x+x=10+2\)

\(\Rightarrow-x=12\Rightarrow x=-12\)

mẫu số = nhau nên tử số = nhau

x-4 -1 = 10-2x

3x = 10+4+1

3x = 15

x = 5

Giá trị x thỏa mãn : \(\frac{x-4}{2015}\)-\(\frac{1}{2015}\)=\(\frac{10-2x}{2015}\)

mẫu số bằng nhau nên tử số cũng bằng  nhau

x-4 -1 = 10-2x

3x = 10+4+1

3x = 15

x = 5

= x-4-1 =10-2x

x= 5

tôi yêu violympic

<=> x - 4 - 1 = 10 - 2x

x = 5

(đúg chứ?)

$\text{Ta coˊ :}\genfrac{}{}{0ex}{}{x}{xy+x+1}+\genfrac{}{}{0ex}{}{y}{yz+y+1}+\genfrac{}{}{0ex}{}{z}{xz+z+1}$

$=\genfrac{}{}{0ex}{}{x}{xy+x+1}+\genfrac{}{}{0ex}{}{xy}{xyz+xy+x}+\genfrac{}{}{0ex}{}{xyz}{x^{2}yz+xyz+xy}$

$=\genfrac{}{}{0ex}{}{x}{xy+x+1}+\genfrac{}{}{0ex}{}{xy}{xy+x+1}+\genfrac{}{}{0ex}{}{1}{xy+x+1}\left(\text{Vıˋ }xyz=1\right)$

$=\genfrac{}{}{0ex}{}{x+xy+1}{xy+x+1}$

$=1$