Question

2) Tìm 380 $x,y,z$ biết $\frac{x}{3}=\frac{y}{4}=\frac{z}{5}$ và $2x^2+2y^2-3z^2=-100$
 

 

Answer 1

Đặt \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\)

=>x=3k; y=4k; z=5k

Ta có: \(2x^2+2y^2-3z^2=-100\)

=>\(2\cdot\left(3k\right)^2+2\cdot\left(4k\right)^2-3\cdot\left(5k\right)^2=-100\)

=>\(2\cdot9k^2+2\cdot16k^2-3\cdot25k^2=-100\)

=>\(-25k^2=-100\)

=>\(k^2=4\)

=>k=2 hoặc k=-2

TH1: k=2

=>\(x=3\cdot2=6;y=4\cdot2=8;z=5\cdot2=10\)

TH2: k=-2

=>\(x=3\cdot\left(-2\right)=-6;y=4\cdot\left(-2\right)=-8;z=5\cdot\left(-2\right)=-10\)