Question

Chứng minh rằng nếu \(\frac{x}{a}\)=\(\frac{y}{b}\)=\(\frac{z}{c}\)thì :

\(\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\)

Answer 1

Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\Rightarrow x=ak,y=bk,z=ck\)

Ta có: \(\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(a^2k^2+b^2k^2+c^2k^2\right)\left(a^2+b^2+c^2\right)=k^2\left(a^2+b^2+c^2\right)^2\) (1)

\(\left(ax+by+cz\right)^2=\left(a.ak+b.bk+c.ck\right)^2=\left(a^2k+b^2k+c^2k\right)^2=\left[k\left(a^2+b^2+c^2\right)\right]^2=k^2\left(a^2+b^2+c^2\right)^2\)(2)

Từ (1),(2) => đpcm

Answer 2

Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\Rightarrow x=ka,y=kb,z=kc\)

Ta có VT=\(\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(k^2a^2+k^2b^2+k^2c^2\right)\left(a^2+b^2+c^2\right)\)=

=\(k^2\left(a^2+b^2+c^2\right)^2\)

Mà \(\left(ax+by+cz\right)^2=\left(a^2k+b^2k+c^2k\right)^2=k^2\left(a^2+b^2+c^2\right)^2\)

=> VT=VP

=> ĐPCM