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BCNN(a,b)=60
=>a.b=60
mà a=12 thì 12.b=60
=>b=60:12=5
vậy b=5
|x|+|y|+|z|=0
=> x,y,z \(\in\){0}
vậy.....
sai thì đừng trách mk
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)
\(xy+y+x=0\)
\(\Rightarrow y\left(x+1\right)+x+1=1\)
\(\Rightarrow\left(x+1\right)\left(y+1\right)=1\cdot1=\left(-1\right)\left(-1\right)\)
lập bảng
tách ra từng cặp 1: 12/16 = x/4 => x= 12*4/16=3
12/16 = 21/y => y= 16*21/12=28
12/16 = z/80 => z= 12*80/16= 60
a)tách ra từng cặp 1: 12/16 = x/4 => x= 12*4/16=3
12/16 = 21/y => y= 16*21/12=28
12/16 = z/80 => z= 12*80/16= 60
(x+y+z)+(x+z+t)+(y+z+t)+(x+y+t)=46+41+44+37=168
=>3x+3y+3z+3t=168
=>3(x+y+z+t)=168
=>x+y+z+t=56
Thay x+y+z+t=168 vào rồi tính ta được
x=x+y+z+t-(y+z+t)=56-44=12
y=x+y+z+t-(x+z+t)=56-41=15
z=x+y+z+t-(x+y+t)=56-37=19
t=x+y+z+t-(x+y+z)=56-46=10