[toán 8] Chứng minh.

C

congchuaanhsang

Từ 3 đẳng thức dễ thấy $x+y+z=2(ax+by+cz)$

VT=$\dfrac{x}{ax+x}+\dfrac{y}{by+y}+\dfrac{z}{cz+z}$

=$\dfrac{x}{ax+by+cz}+\dfrac{y}{ax+by+cz}+\dfrac{z}{ax+by+cz}$

=$\dfrac{x+y+z}{ax+by+cz}$=$\dfrac{2(ax+by+cz)}{ax+by+cz}$=2
 
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