\(\left\{{}\begin{matrix}x+y=12\\y+z=-16\\z+x=8\end{matrix}\right.\)

\(\Rightarrow\left(x+y\right)+\left(y+z\right)+\left(z+x\right)=12+\left(-16\right)+8\)

\(\Rightarrow x+y+y+z+z+x=4\)

\(\Rightarrow2\left(x+y+z\right)=4\)

\(\Rightarrow x+y+z=2\)

\(\Rightarrow\left\{{}\begin{matrix}z=2-12=-10\\x=2-\left(-16\right)=18\\y=2-8=-6\end{matrix}\right.\)

Từ \(\left\{{}\begin{matrix}x+y=12\\y+z=-16\\x+z=8\end{matrix}\right.\)\(\Rightarrow2\left(x+y+z\right)=4\Rightarrow x+y+z=2\)

*)Xét \(x+y=12\Rightarrow x+y+z=z+12\)

\(\Rightarrow2=z+12\Rightarrow z=-10\)

*)Xét \(y+z=-16\Rightarrow x+y+z=-16+x\)

\(\Rightarrow2=-16+x\Rightarrow x=18\)

*)Xét \(x+z=8\Rightarrow x+y+z=8+y\)

\(\Rightarrow2=8+y\Rightarrow y=6\)

x + y = 7/12  =>  x = 7/12 - y
y + z = -19/24  =>  z = -19/24 - y
Mà z + x = 1/8  =>  7/12 - y - 19/24 - y = 1/8
=>  2y = 7/12 - 19/24 - 1/8  =>  2y = -1/3
=> y = -1/6

thank

Bài 9:

Ta có: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{z}{-17}=\dfrac{-t}{-9}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{-z}{17}=\dfrac{t}{9}=-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-2\\\dfrac{-y}{3}=-2\\\dfrac{-z}{17}=-2\\\dfrac{t}{9}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\-y=-6\\-z=-34\\t=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=6\\z=34\\t=-18\end{matrix}\right.\)

Vậy: (x,y,z,t)=(-10;6;34;-18)

Bài 11:

Ta có: \(\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}\)

\(\Leftrightarrow\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}=\dfrac{-7}{6}\)

Ta có: \(\dfrac{x}{18}=\dfrac{-7}{6}\)

\(\Leftrightarrow x=\dfrac{18\cdot\left(-7\right)}{6}=-21\)

Ta có: \(\dfrac{-98}{y}=\dfrac{-7}{6}\)

\(\Leftrightarrow y=\dfrac{-98\cdot6}{-7}=84\)

Ta có: \(\dfrac{-14}{z}=\dfrac{-7}{6}\)

\(\Leftrightarrow z=\dfrac{-14\cdot6}{-7}=12\)

Ta có: \(\dfrac{u}{-78}=\dfrac{-7}{6}\)

\(\Leftrightarrow u=\dfrac{-78\cdot\left(-7\right)}{6}=\dfrac{78\cdot7}{6}=91\)

Ta có: \(\dfrac{t}{102}=\dfrac{-7}{6}\)

\(\Leftrightarrow t=\dfrac{-7\cdot102}{6}=-7\cdot17=-119\)

Vậy: (x,y,z,t,u)=(-21;84;12;-119;91)

Nguyễn Lê Phước Thịnh giải giùm mk bài 10 đc ko ạ

ta có : x-y=8

=> y=x-8

x+z=12

=> z=12-x

thay y=x-8,z=12-x vào y-z=10 ta đc:

(x-8) -( 12 -x) =10

x-8-12+x =10

2x-20=10

2x=30

x=15

thay x=15 vào x-y=8

=> 15-y=8

y=7

thay y=7 vào y-z=10

=> 7-z=10

z=-3

Vậy x=15,y=7,z=-3

p/s : mk lm ko bk có đúng ko, bn k nha !~

Ta có \(x-y=8;y-z=10;x+z=12\)

\(\Leftrightarrow x-y+y-z+x+z=30\)

\(\Leftrightarrow\left(x+x\right)+\left(-y+y\right)+\left(-z+z\right)=30\)

\(\Leftrightarrow2x=30\Rightarrow x=15\)

\(x-y=15-y=8\Rightarrow y=7\)

\(y-z=7-z=10\Rightarrow z=-3\)

Vậy \(x=15;y=7;z=-3\)

a) 2^y+1.3^x=12^y=3^y.2^2y

<=> 2^y+1.3^x=3^y.2^2y <=> 3^x-y=2^2y-y-1 <=> 3^x-y=2^y-1

Nếu x-y và y-1 khác 0 thì 2 vế 1 số là lẻ, 1 số là chẵn => ko có giá trị nào.

=> x-y=y-1=0 => x=y=1