Tim cac so x,y,z biet x/y=9/7; y/2=7/3 va x-y+z=-15
bai 1 tim cac so nguyen x,y biet
a,x/3=7/y b,x/y=-3/11 c,x/y-1=5/-19
bai 2 tim cac so nguyen x,y,z,t biet
12/-6=x/5=-y/3=z/-17=-t/-9
\(a,\frac{x}{3}=\frac{7}{y}\)
\(\Rightarrow x\cdot y=3\cdot7\)
\(\Rightarrow x\cdot y=21\)
\(\Rightarrow x;y\inƯ\left(21\right)=\left\{\pm1;\pm21;\pm3;\pm7\right\}\)
tim cac so x,y,z thuoc Q biet rang (x+y):(5-z):(y+z):(y+9)=3:1:2:5
Ta có:(x+y):(5-z):(y+z):(y+9)=3:1:2:5
=> 5-z=1=>z=4.
y+9=5=>y=-4.
x+y=3=>x-4=3(do y=-4)=>x=7.
Vậy x=7,y=-4,z=4.
Tim cac so nguyen x,y,z biet
x(x+y+z)=-5
y(x+y+z)=9
z(x+y+z)=5
Giải
Ta có : \(\hept{\begin{cases}x\left(x+y+z\right)=-5\\y\left(x+y+z\right)=9\\z\left(x+y+z\right)=5\end{cases}}\Rightarrow x\left(x+y+z\right)+y\left(x+y+z\right)+z\left(x+y+z\right)=-5+9+5\)
\(\Rightarrow\left(x+y+z\right)\left(x+y+z\right)=9\)
\(\Rightarrow\left(x+y+z\right)^2=3^2\)
\(\Rightarrow x+y+z=3\)
\(\Rightarrow\hept{\begin{cases}3x=-5\\3y=9\\3z=5\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-5}{3}\\y=3\\z=\frac{5}{3}\end{cases}}\)
Mà x , y , z là các số nguyên nên không có nghiệm x , y , z cần tìm
tim cac so nguyen x,y,z biet :-6/12=x/8=-7/y=z/-18
Ta có :
\(\frac{-6}{12}=\frac{x}{8}=\frac{-7}{y}=\frac{z}{-18}=\frac{-1}{2}\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{8}=\frac{-1}{2}\Rightarrow x=\left(-4\right)\\\frac{-7}{y}=\frac{-1}{2}\Rightarrow y=14\\\frac{z}{-18}=\frac{-1}{2}\Rightarrow z=9\end{cases}}\)
Vậy ...
tim so tu nhien biet tong cac chu so cua x=y,tong cac chu so cua y=z va x+y+z=60
tim cac so x,y,z ,biet x/2=y/4=z/6 va x-y+z
Tim so tu nhien x, biet rang tong cac chu so chua x bang y, tong cac chu so cua y bang z va x+y+z =60
Tim cac so nguyên x,y,z biet
x+y=2,y+z=3,x+z=-5
x= -3
y= 5
z= -2....đúng 100000%
tích cho mk ik mọi người....>.<!!!!
tim cac so huu ti x,y,z biet x+y=1/2 y+z=1/3 z+y=1/4