Question

Cho x,y,z là các số khác không và đôi một khác nhau thỏa mãn \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\) Tính giá trị biểu thức A=\(\dfrac{yz}{x^{2^{ }}+2yz}+\dfrac{xz}{y^{2^{ }}+2xz}+\dfrac{xy}{z^{2^{ }}+2xy}\)

Answer 1

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{0\Rightarrow\left(yz+xz+xy\right)}{xyz}=0\Rightarrow xy+xz+xy=0\)

ta có x²+2yz=x²+yz+yz=x²-yz-zx-xy=x.(x-z)-y.(x-z)=(x-y).(x-z)

tương tự ta có:x²+2xy=(x-z)*(y-z)

vậy\(A=\dfrac{yz}{\left(x-y\right).\left(x-z\right)}+\dfrac{xz}{\left(x-y\right)\left(z-y\right)}+\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)a

\(A=\dfrac{yz\left(y-z\right)}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}-\dfrac{xz\left(x-z\right)}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}+\dfrac{xy\left(x-y\right)}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}\)

\(=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(y-z\right)\left(x-y\right)\left(x-z\right)}\)

\(=\dfrac{\left(yz-xz\right)\left(y-z\right)+\left(xy-xz\right)\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)

\(=\dfrac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)