Cho x/(y+z)+y/(z+x)+z/(x+y)=1. Tính giá trị bt N=x^2/(y+z)+y^2/(z+x)+z^2/(x+y)

Đáp án:

\[N = 0\]

Giải thích các bước giải:

\(\begin{array}{l}
\frac{x}{{y + z}} + \frac{y}{{z + x}} + \frac{z}{{x + y}} = 1\\
 \Leftrightarrow \left( {x + y + z} \right)\left( {\frac{x}{{y + z}} + \frac{y}{{z + x}} + \frac{z}{{x + y}}} \right) = x + y + z\\
 \Leftrightarrow \left( {\frac{{{x^2}}}{{y + z}} + \frac{{{y^2}}}{{z + x}} + \frac{{{x^2}}}{{x + y}}} \right) + \left( {\frac{{xy}}{{z + x}} + \frac{{zx}}{{x + y}} + \frac{{xy}}{{y + z}} + \frac{{yz}}{{x + y}} + \frac{{zx}}{{y + z}} + \frac{{zy}}{{z + x}}} \right) = x + y + z\\
 \Leftrightarrow \left( {\frac{{{x^2}}}{{y + z}} + \frac{{{y^2}}}{{z + x}} + \frac{{{x^2}}}{{x + y}}} \right) + \left( {\frac{{xy}}{{x + z}} + \frac{{yz}}{{z + x}}} \right) + \left( {\frac{{zx}}{{x + y}} + \frac{{yz}}{{x + y}}} \right) + \left( {\frac{{xy}}{{y + z}} + \frac{{zx}}{{y + z}}} \right) = x + y + z\\
 \Leftrightarrow \left( {\frac{{{x^2}}}{{y + z}} + \frac{{{y^2}}}{{z + x}} + \frac{{{x^2}}}{{x + y}}} \right) + y.\left( {\frac{x}{{x + z}} + \frac{z}{{z + x}}} \right) + z\left( {\frac{x}{{x + y}} + \frac{y}{{x + y}}} \right) + x\left( {\frac{y}{{z + y}} + \frac{z}{{z + y}}} \right) = x + y + z\\
 \Leftrightarrow \left( {\frac{{{x^2}}}{{y + z}} + \frac{{{y^2}}}{{z + x}} + \frac{{{x^2}}}{{x + y}}} \right) + x + y + z = x + y + z\\
 \Leftrightarrow \left( {\frac{{{x^2}}}{{y + z}} + \frac{{{y^2}}}{{z + x}} + \frac{{{x^2}}}{{x + y}}} \right) = 0
\end{array}\)