Question

Cho \(\dfrac{\text{2bz-3cy}}{a}\)=\(\dfrac{\text{3cx-az}}{2b}\)=\(\dfrac{\text{ay-2bx}}{3c}\) CMR : \(\dfrac{x}{a}\)=\(\dfrac{y}{2b}\)=\(\dfrac{z}{3c}\)

Answer 1

Từ `(2bz - 3cy)/a = (3cx - az)/(2b) = (ay - 2bx)/(3c)`

`=> (a(2bz - 2cy))/a^2 = (2b(2cx - az))/(4b^2) = (2c(ay - 2bx))/(6c^2)`

`=> (2abz - 2acy)/(a^2) = (4bcx - 2abz)/(4b^2) = (2acy - 4bcx)/(6c^2)`

Áp dụng tính chất dãy tỉ số bằng nhau, ta được:

`(2abz - 2acy)/(a^2) = (4bcx - 2abz)/(4b^2) = (2acy - 4bcx)/(6c^2)`

`= (2abz - 2acy + 4bcx - 2abz + 2acy - 4bcx)/(a^2 + 4b^2 + 6c^2)`

`= 0/(a^2 + 4b^2 + 6c^2)`

`= 0`

Từ đó suy ra:

`(2bz - 3cy)/a = 0 => 2bz - 3cy = 0 => 2bz = 3cy => y/(2b) = z/(3c)  (1)`

`(3cx - az)/b = 0 => 3cx - az = 0 => 3cx = az => x/a = z/(3c)  (2)`

Từ `(1)` và `(2)` suy ra: `x/a = y/(2b) = z/(3c)`.