cho 1/x+1/y+1/z=0 . tính A=yz/x^2+xz/y^2+xy/z^2

Đáp án:

$A = 3$

Giải thích các bước giải:

Ta có:

$\dfrac1x +\dfrac1y +\dfrac1z = 0$

$\to \dfrac{xy + yz + zx}{xyz}=0$

$\to xy + yz + zx = 0$

$\to \begin{cases}xy = -z(x + y)\\yz = - x(y+z)\\zx = -y(z + x)\end{cases}$

Ta được:

$A =\dfrac{yz}{x^2} +\dfrac{zx}{y^2}+\dfrac{xy}{z^2}$

$\to A =-\dfrac{x(y+z)}{x^2}-\dfrac{y(z + x)}{y^2} -\dfrac{z(x +y)}{z^2}$

$\to A = -\dfrac yx -\dfrac zx -\dfrac zy -\dfrac xy -\dfrac xz -\dfrac yz$

$\to A = -x\left(\dfrac1y +\dfrac1z\right) -y\left(\dfrac1z +\dfrac1x\right) -z\left(\dfrac1x +\dfrac1y\right)$

$\to A = -x\cdot\left(-\dfrac1x\right) - y\cdot\left(-\dfrac1y\right)-z\cdot\left(-\dfrac1z\right)$

$\to A = 1+ 1 + 1= 3$