+TH1:\(x\le-7\)

Ta có: \(-x-1-x+2-x-7=5x-10\Rightarrow x=\dfrac{1}{2}\) (loại)

+TH2: \(-7\le x\le-1\)

Ta có: \(-x-1-x+2+x+7=5x-10\Rightarrow x=3\)(loại)

+TH3: \(-1\le x\le2\)

Ta có: \(x+1-x+2+x+7=5x-10\Rightarrow x=5\)(loại)

+TH4: \(x>2\)

Ta có: \(x+1+x-2+x+7=5x-10\Rightarrow x=8\)(tm)

Vậy x=8

\(a,\frac{7}{2}x+\frac{3}{4}=\frac{-3}{10}\)

\(\frac{7}{2}x=\frac{-21}{20}\)

\(x=\frac{-3}{10}\)

\(b,\left(8\frac{4}{5}x-50\right):40\%=51\)

\(8\frac{4}{5}x-50=\frac{102}{5}\)

\(\frac{44}{5}x=\frac{352}{5}\)

\(x=8\)

a,\(\frac{7}{2}x+\frac{3}{4}=-\frac{3}{10}\)

\(\frac{7}{2}x=-\frac{21}{20}\)

\(x=-\frac{3}{10}\)

b,\(\left(8\frac{4}{5}x-50\right)\div40\%=51\)

\(\left(8\frac{4}{5}x-50\right)\div\frac{2}{5}=51\)

\(8\frac{4}{5}x-50=\frac{102}{5}\)

\(8\frac{4}{5}x=\frac{352}{5}\)

\(\frac{44}{5}x=\frac{352}{5}\)

\(x=8\)

Ta co : x/10=y/6=z/21=>5x/50=y/6=2z/42 va 5x+y-2z=-56

Ap dung tinh chat day ti so bang nhau ta co : 

5x/50=y/6=2z/42=5x+y-2z/50+6-42=-56/14

Suy ra :5x/50=-4=>x=-4.50:5=-40

y/6=-4=>y=-4.6=-24

2z/42=-4=>z=-4.42:2=-84

**** nhe

a)\(A=\frac{x^2}{5x+25}+\frac{2x-10}{x}+\frac{50+5x}{x^2+5x}\left(ĐK:x\ne0;-5\right)\)

\(\Leftrightarrow A=\frac{x^2}{5\left(x+5\right)}+\frac{2\left(x-5\right)}{x}+\frac{5\left(x+10\right)}{x\left(x+5\right)}\)

\(\Leftrightarrow A=\frac{x^3+10\left(x^2-25\right)+25x+250}{5x\left(x+5\right)}\)

\(\Leftrightarrow A=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}\)

\(\Leftrightarrow A=\frac{x\left(x+5\right)^2}{5x\left(x+5\right)}\)

\(\Leftrightarrow A=\frac{x+5}{5}\)

b)Để A=-4 \(\Leftrightarrow\frac{x+5}{5}=-4\)

                  \(\Leftrightarrow x+5=-20\)

                   \(\Leftrightarrow x=-25\)

a).....

\(=\frac{x^2}{5\left(x+5\right)}+\frac{2x-10}{x}+\frac{50+5x}{x\left(x+5\right)}\)                                MTC= 5x (x+5)                 ĐK\(\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)

\(=\frac{x^2.x}{5x\left(x+5\right)}+\frac{5.\left(2x-10\right).\left(x+5\right)}{5x\left(x+5\right)}+\frac{5.\left(50+5x\right)}{5x\left(x+5\right)}\)

\(=\frac{x^3+\left(10x-50\right).\left(x+5\right)+250+25x}{5x\left(x+5\right)}\)

\(=\frac{x^3+10x^2+50x-50x-250+250+25x}{5x\left(x+5\right)}\)

\(=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}\)

\(=\frac{x\left(x^2+10x+25\right)}{5x\left(x+5\right)}\)

\(=\frac{x\left(x+5\right)^2}{5x\left(x+5\right)}=\frac{x+5}{5}\)

b) A=-4

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

=> x = -25

c)

d) Để A đạt gt nguyên thì 5\(⋮\)x+5

=> \(\left(x+5\right)\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)

*x+5=1 => x=-4 \(\in Z\)

*x+5=-1 => x=-6\(\in Z\)

*x+5=5  => x=0\(\in Z\)

*x+5=-5  => x=-10\(\in Z\)

Vậy...........

a/          5x +y -2x = 28 => 3x +y = 28

       x/10 = y/6 = z/21 = 3x /30= y/6 = 3x +y  /  36 = 28 /36 = 7/9

=> x= 70/9 ; y = 14/3 ; z= 49/3

b/

          x/3 = y/4 => x/15 = y/20 [1]

        y/5 = z/7 => y/20 = z/28  [2]

Từ [1] và [2] => x/15 = y/20 = z/28 = 2x /30 = 3y/60 = z/28 = [2x +3y - z] / [30+60-28]= 124 /62 = 2

=> x= 2 .15 = 30 ; y = 2x20 = 40 ; z= 2 . 28= 56