$\text{cho }\cfrac ax+\cfrac by+\cfrac cz=2(abc\ne0;xyz\ne0)\\\text{tính giá trị của biểu thức }\cfrac{a^2}{x^2}+\cfrac{b^2}{y^2}+\cfrac{c^2}{z^2}$

bổ sung : 

Cho `x/a + y/b + z/c=0`

Giải

`x/a + y/b + z/c=0`

`⇔(xbc+yac+zab)/(abc)=0`

`⇔xbc+yac+zab=0 (abc\ne0)`

`a/x+b/y+c/z=2`

`⇔(a/x+b/y+c/z)^2=4`

`⇔(a/x+b/y)^2+2(a/x+b/y).c/z+c^2/z^2=4`

`⇔a^2/x^2+b^2/y^2+c^2/z^2+2(a/x+b/y).c/z+2. a/x.b/y=4`

`⇔a^2/x^2+b^2/y^2+c^2/z^2+2((ac)/(xz)+(bc)/(yz)+(ab)/(xy))=4`

`⇔a^2/x^2+b^2/y^2+c^2/z^2+(xbc+yac+zab)/(xyz)=4(xyz\ne0)`

$⇔\cfrac{a^2}{x^2}+\cfrac{b^2}{y^2}+\cfrac{c^2}{z^2}=4$