Question

3. Chứng minh rằng nếu: x/a = y/b = z/c thì (x^2 + y^2 + z^2) (a^2 + b^2 + c^2) = (ax + by + cz)^2

 

Answer 1

Với a; b ; c  khác 0

Ta có: 

\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x^2}{ax}=\frac{y^2}{by}=\frac{z^2}{cz}=\frac{ax}{a^2}=\frac{by}{b^2}=\frac{cz}{c^2}\)(1)

Áp dụng dãy tỉ số bằng nhau: 

\(\frac{x^2}{ax}=\frac{y^2}{by}=\frac{z^2}{cz}=\frac{x^2+y^2+z^2}{ax+by+cz}\)(2)

\(\frac{ax}{a^2}=\frac{by}{b^2}=\frac{cz}{c^2}=\frac{ax+by+cz}{a^2+b^2+c^2}\)(3)

Từ (1) ; (2) ; (3) 

=> \(\frac{ax+by+cz}{a^2+b^2+c^2}\)\(=\frac{x^2+y^2+z^2}{ax+by+cz}\)

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

Answer 2

Do: \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\) => \(\frac{x}{a}=\frac{y}{b};\frac{y}{b}=\frac{z}{c};\frac{z}{c}=\frac{x}{a}\)

<=> \(ay=bx;bz=cy;az=cx\)

<=> \(\left(ay-bx\right)=0;bz-cy=0;az-cx=0\)

<=> \(\left(ay-bx\right)^2+\left(yc-bz\right)^2+\left(az-cx\right)^2=0\)

<=> \(a^2y^2+b^2x^2+y^2c^2+b^2z^2+a^2z^2+c^2x^2=2abxy+2bcyz+2cazx\)

<=> \(a^2y^2+b^2x^2+y^2c^2+b^2z^2+a^2z^2+c^2x^2+a^2x^2+b^2y^2+c^2z^2=a^2x^2+b^2y^2+c^2z^2+2abxy+2bcyz+2cazx\)<=> \(\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\)

=> Ta có ĐPCM

Answer 3

cho mik cảm ơn ạ