\(\Leftrightarrow5x+5=4x+12\)

hay x=7

Ta có : 2(x+y+z)=32\(\Rightarrow\) x+y+z=16

Mà x+y=5 \(\Rightarrow\) z=16-5=11

y+z=7 \(\Rightarrow\) y=7-11=-4

z+x=20\(\Rightarrow\) x=20-11=9

trên violympic vòng 9 đúng ko mình vừa thi xong

x=7 tính máy tính thử lại đúng luôn

2^x+1+44=300

2^x+1=300-44

2^x+1=256=2⁸

=>x+1=8

x=8-1

x=7

Vậy x=7

2^x+1 + 44 = 300 
2^x+1         = 300 - 44 
2^x+1         = 256 
256             = 2^8 
x + 1           = 8 
x                 = 8 - 1 
x                 = 7.

\(x=15,51\)

a) x + 9 = 2 - 17

x + 9 = - 15

x = -15 – 9

x = -24

Vậy x = -24

b) x - 17 = (-11) . (-5)

x – 17 = 55

x = 17 + 55

x = 72

c)| x – 5 | = (-4)2

| x – 5 | = 16

x – 5 = 16 hoặc x – 5 = -16

x = 21 hoặc x = -11

a) Ta có: \(\left|x-9\right|=8\)

         \(\Leftrightarrow x-9=\pm8\)

         \(\Leftrightarrow\orbr{\begin{cases}x-9=8\\x-9=-8\end{cases}}\)

         \(\Leftrightarrow\orbr{\begin{cases}x=17\left(TM\right)\\x=1\left(TM\right)\end{cases}}\)

Vậy \(x=17\)hoặc \(x=1\)

b);c) Ta có: \(17-\left|x+5\right|=14\)

            \(\Leftrightarrow\left|x+5\right|=3\)

            \(\Leftrightarrow x+5=\pm3\)

            \(\Leftrightarrow\orbr{\begin{cases}x+5=3\\x+5=-3\end{cases}}\)

            \(\Leftrightarrow\orbr{\begin{cases}x=-2\left(TM\right)\\x=-8\left(TM\right)\end{cases}}\)

Vậy \(x=-2\)hoặc \(x=-8\)

\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)

\(\Rightarrow x\left(\frac{1}{2}-\frac{2}{3}\right)=\frac{7}{12}\Rightarrow x\cdot\left(\frac{3}{6}-\frac{4}{6}\right)=\frac{7}{12}\)

\(\Rightarrow x\cdot\left(-1\right)=\frac{7}{12}\Rightarrow x=\frac{7}{12}:\left(-1\right)=\frac{7}{-12}\)

\(c,\frac{\left(x-5\right)}{12}\cdot\frac{9}{29}=\frac{-6}{29}\Rightarrow\frac{\left(x-5\right)}{12}=\frac{-6}{29}:\frac{9}{26}\)

\(\frac{\Rightarrow\left(x-5\right)}{12}=\frac{-6}{9}=\frac{-2}{3}\Rightarrow x-5=-\frac{2}{3}\cdot12\)

\(\Rightarrow x-5=\frac{-24}{3}=-8\Rightarrow x=-8+5=-3\)

\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\)

\(\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)

\(\Rightarrow-\frac{1}{6}x=\frac{7}{12}\)

\(\Rightarrow x=-\frac{7}{2}\)

\(c,\frac{x-5}{12}\cdot\frac{9}{29}=-\frac{6}{29}\)

\(\Rightarrow\frac{x-5}{12}=-\frac{2}{3}\)

\(\Rightarrow x-5=12.\left(-\frac{2}{3}\right)\)

\(\Rightarrow x-5=-8\)

\(\Rightarrow x=-3\)

b)\(\frac{-8}{17}+\frac{5}{17}< \frac{x}{17}< \frac{-6}{17}+\frac{9}{17}\)\(\left(x\in Z\right)\)

\(\Rightarrow\frac{-3}{17}< \frac{x}{17}< \frac{3}{17}\)

\(\Rightarrow-3< x< 3\)

\(\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)

Vậy \(x\in\left\{-2;-1;0;1;2\right\}\)