Ba số x,y,z tỉ lệ với ba số a,b,c
\(\Rightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}\)(1)
Lại có: \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{xa}{a^2}+\frac{yb}{b^2}+\frac{zc}{c^2}=\frac{xa+yb+zc}{a^2+b^2+c^2}=\frac{9\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=9\) (2)
Từ (1) và (2) ta có : \(\frac{x+y+z}{a+b+c}=9\)
\(\Rightarrow\left(x+y+z\right)=9\left(a+b+c\right)\) (đpcm)
Theo bài ra ta có:\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\)
\(\Rightarrow x=ak;y=bk;z=ck\)
Khi đó:\(ax+by+cz=a\cdot ak+b\cdot bk+c\cdot ck=a^2k+b^2k+c^2k=k\left(a^2+b^2+c^2\right)\)
Vì \(ax+by+cz=9\left(a^2+b^2+c^2\right)\Rightarrow k\left(a^2+b^2+c^2\right)=9\left(a^2+b^2+c^2\right)\)
\(\Rightarrow k=9\)
Khi đó:\(x+y+z=ak+bk+ck=k\left(a+b+c\right)=9\left(a+b+c\right)\left(đpcm\right)\)
Ta có: \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
=> \(\frac{ax}{a^2}=\frac{by}{b^2}=\frac{cz}{c^2}=\frac{ax+by+cz}{a^2+b^2+c^2}\)
Mà ax + by + cz = 9 . ( a2 + b2 + c2 )
=> \(\frac{ax+by+cz}{a^2+b^2+c^2}=\frac{9.\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=9\)
=> \(\frac{x}{a}=9;\frac{y}{b}=9;\frac{z}{c}=9\)
=> x = 9a
y = 9b
z = 9c
=> x + y + z = 9 . ( a2 + b2 + c2 )
Vậy: x + y + z = 9 . ( a2 + b2 + c2 )
