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\(\left\{{}\begin{matrix}x+y=12\\y+z=-16\\z+x=8\end{matrix}\right.\)
\(\Rightarrow\left(x+y\right)+\left(y+z\right)+\left(z+x\right)=12+\left(-16\right)+8\)
\(\Rightarrow x+y+y+z+z+x=4\)
\(\Rightarrow2\left(x+y+z\right)=4\)
\(\Rightarrow x+y+z=2\)
\(\Rightarrow\left\{{}\begin{matrix}z=2-12=-10\\x=2-\left(-16\right)=18\\y=2-8=-6\end{matrix}\right.\)
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
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)
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: \(\Leftrightarrow\dfrac{x}{-4}=\dfrac{21}{y}=\dfrac{z}{-80}=\dfrac{3}{4}\)
=>x=-3; y=28; z=-60
b: 5/12=x/-72
=>x=-72*5/12=-6*5=-30
c: =>x+3=-5
=>x=-8
từ đề bài
suy ra x+y+y-z+z-x=-8+4+(-6)=-10
2y=-10
y=-5
tu đ1 tìm được z và x
ta có:
(*) <=> xy + 2x =9-y
<=>x(y+2) + y+2 =11
<=> (x+1).(y+2) = 11
=> x+1 = 11\(y+2)
mà x nguyên => x+1 nguyên
=> 11\ (y+2) nguyên
=> y+2 là ước 11
=> y+2 = ±1,±11
=> y = -3,1,-13,9
=> x= -12,10,-2,0
=> ta có ác cặp nghiệm:
(-12;-3) ;(10;1) ;( -2;-13) ; (0;9)
(z+x)+(x+y)=12+8
(z+x)+(x+y)=20
2x+(y+z)=20
2x+-16=20
2x=20-(-16)
2x=36
x=36:2
x=18
x+y=12
18+y=12
y=12-18
y=-6
y+z=-16
-6+z=-16
z=-16-(-6)
z=-10