cho x^2/(x+y) + y^2/(y+z) + z^2/(z+x) = 2011 tính y^2/(x+y) + z^2/(y+z) + x^2/(z+x) 07/07/2021 Bởi Reese cho x^2/(x+y) + y^2/(y+z) + z^2/(z+x) = 2011 tính y^2/(x+y) + z^2/(y+z) + x^2/(z+x)
Đáp án: \[\frac{{{y^2}}}{{x + y}} + \frac{{{z^2}}}{{y + z}} + \frac{{{x^2}}}{{z + x}} = 2011\] Giải thích các bước giải: Ta có: \(\begin{array}{l}\frac{{{x^2}}}{{x + y}} + \frac{{{y^2}}}{{y + z}} + \frac{{{z^2}}}{{z + x}} = 2011\\ \Rightarrow \left( {x – \frac{{{x^2}}}{{x + y}}} \right) + \left( {y – \frac{{{y^2}}}{{y + z}}} \right) + \left( {z – \frac{{{z^2}}}{{z + x}}} \right) = x + y + z – 2011\\ \Leftrightarrow \frac{{xy}}{{x + y}} + \frac{{yz}}{{y + z}} + \frac{{zx}}{{z + x}} = x + y + z – 2011\\2\left( {x + y + z} \right) = \left( {x + y} \right) + \left( {y + z} \right) + \left( {z + x} \right)\\ \Leftrightarrow 2.\left( {x + y + z} \right) = \frac{{{{\left( {x + y} \right)}^2}}}{{x + y}} + \frac{{{{\left( {y + z} \right)}^2}}}{{y + z}} + \frac{{{{\left( {z + x} \right)}^2}}}{{z + x}}\\ \Leftrightarrow 2.\left( {x + y + z} \right) = \left( {\frac{{{x^2}}}{{x + y}} + \frac{{2xy}}{{x + y}} + \frac{{{y^2}}}{{x + y}}} \right) + \left( {\frac{{{y^2}}}{{y + z}} + \frac{{2yx}}{{y + z}} + \frac{{{z^2}}}{{y + z}}} \right) + \left( {\frac{{{z^2}}}{{z + x}} + \frac{{2zx}}{{z + x}} + \frac{{{x^2}}}{{z + x}}} \right)\\ \Leftrightarrow 2.\left( {x + y + z} \right) = \left( {\frac{{{x^2}}}{{x + y}} + \frac{{{y^2}}}{{y + z}} + \frac{{{z^2}}}{{z + x}}} \right) + 2.\left( {\frac{{xy}}{{x + y}} + \frac{{yz}}{{y + z}} + \frac{{zx}}{{z + x}}} \right) + \left( {\frac{{{y^2}}}{{x + y}} + \frac{{{z^2}}}{{y + z}} + \frac{{{x^2}}}{{z + x}}} \right)\\ \Leftrightarrow 2\left( {x + y + z} \right) = 2011 + 2.\left( {x + y + z – 2011} \right) + \left( {\frac{{{y^2}}}{{x + y}} + \frac{{{z^2}}}{{y + z}} + \frac{{{x^2}}}{{z + x}}} \right)\\ \Rightarrow \left( {\frac{{{y^2}}}{{x + y}} + \frac{{{z^2}}}{{y + z}} + \frac{{{x^2}}}{{z + x}}} \right) = 2011\end{array}\) Bình luận
Đáp án:
\[\frac{{{y^2}}}{{x + y}} + \frac{{{z^2}}}{{y + z}} + \frac{{{x^2}}}{{z + x}} = 2011\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\frac{{{x^2}}}{{x + y}} + \frac{{{y^2}}}{{y + z}} + \frac{{{z^2}}}{{z + x}} = 2011\\
\Rightarrow \left( {x – \frac{{{x^2}}}{{x + y}}} \right) + \left( {y – \frac{{{y^2}}}{{y + z}}} \right) + \left( {z – \frac{{{z^2}}}{{z + x}}} \right) = x + y + z – 2011\\
\Leftrightarrow \frac{{xy}}{{x + y}} + \frac{{yz}}{{y + z}} + \frac{{zx}}{{z + x}} = x + y + z – 2011\\
2\left( {x + y + z} \right) = \left( {x + y} \right) + \left( {y + z} \right) + \left( {z + x} \right)\\
\Leftrightarrow 2.\left( {x + y + z} \right) = \frac{{{{\left( {x + y} \right)}^2}}}{{x + y}} + \frac{{{{\left( {y + z} \right)}^2}}}{{y + z}} + \frac{{{{\left( {z + x} \right)}^2}}}{{z + x}}\\
\Leftrightarrow 2.\left( {x + y + z} \right) = \left( {\frac{{{x^2}}}{{x + y}} + \frac{{2xy}}{{x + y}} + \frac{{{y^2}}}{{x + y}}} \right) + \left( {\frac{{{y^2}}}{{y + z}} + \frac{{2yx}}{{y + z}} + \frac{{{z^2}}}{{y + z}}} \right) + \left( {\frac{{{z^2}}}{{z + x}} + \frac{{2zx}}{{z + x}} + \frac{{{x^2}}}{{z + x}}} \right)\\
\Leftrightarrow 2.\left( {x + y + z} \right) = \left( {\frac{{{x^2}}}{{x + y}} + \frac{{{y^2}}}{{y + z}} + \frac{{{z^2}}}{{z + x}}} \right) + 2.\left( {\frac{{xy}}{{x + y}} + \frac{{yz}}{{y + z}} + \frac{{zx}}{{z + x}}} \right) + \left( {\frac{{{y^2}}}{{x + y}} + \frac{{{z^2}}}{{y + z}} + \frac{{{x^2}}}{{z + x}}} \right)\\
\Leftrightarrow 2\left( {x + y + z} \right) = 2011 + 2.\left( {x + y + z – 2011} \right) + \left( {\frac{{{y^2}}}{{x + y}} + \frac{{{z^2}}}{{y + z}} + \frac{{{x^2}}}{{z + x}}} \right)\\
\Rightarrow \left( {\frac{{{y^2}}}{{x + y}} + \frac{{{z^2}}}{{y + z}} + \frac{{{x^2}}}{{z + x}}} \right) = 2011
\end{array}\)