Tim cac so nguyên x,y,z biet
x+y=2,y+z=3,x+z=-5
Những câu hỏi liên quan
tim cac so m,n,p thoa man : m+n+p+8=2canm-1 + 4cann-2 +6canp-3
tim cac so x,y,z thoa man :canx+cany-1 +canz-2 = 1/2(x+y+z)
tim cac so x,y,z thoa man :x+y+z+4=2canx-2 +4cany-3+6canz-5
cau 1,(n^2-7)\(⋮\)n+3
cau 2 tim x;y;z
bietx-y=-9
y-z=-10
z+x=11
xin loi ban voi qua chua nhap cau hoi/
tim n ban nhe
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1.Tim tat ca cac cap so nguyên sao cho x^3 -x^2y+3x-2y-5=0
2. Cho0<x,y,z =<1 . CMR : x/(1+y+xz) + y/(1+z+xy) +z/(1+x+yz) =< 3/(x+y+z)
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.
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tim cac so nguen x,y,z:x/y+y/z+z/x=y/x+z/y+x/z=x+y+z=3
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
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tim cac so huu ti x,y,z biet x+y=1/2 y+z=1/3 z+y=1/4
Tim cac so nguyen x, y, z thoa man x/y +y/z+z/x =y/x+/y+x/z=x+y+z=3
cho x,y,z la cac so thuc duong thoa man x+y+z=1 tim min A=x^3/(x^2+xy+y^2)+y^3/(y^2+yz+z^2)+z^3/(z^2+zx+x^2)