Cho a+b+c=7, 1/a+b+1/b+c+1/c+a=7/10 Tính: c/a+b+a/b+c+b/c+a 27/08/2021 Bởi Natalia Cho a+b+c=7, 1/a+b+1/b+c+1/c+a=7/10 Tính: c/a+b+a/b+c+b/c+a
Giải thích các bước giải: Ta có: \[\begin{array}{l}\frac{1}{{a + b}} + \frac{1}{{b + c}} + \frac{1}{{c + a}} = \frac{7}{{10}}\\ \Leftrightarrow \left( {a + b + c} \right)\left( {\frac{1}{{a + b}} + \frac{1}{{b + c}} + \frac{1}{{c + a}}} \right) = \frac{7}{{10}}.\left( {a + b + c} \right)\\ \Leftrightarrow \frac{{a + b + c}}{{a + b}} + \frac{{a + b + c}}{{b + c}} + \frac{{a + b + c}}{{c + a}} = \frac{7}{{10}}.7\\ \Leftrightarrow \left( {1 + \frac{c}{{a + b}}} \right) + \left( {1 + \frac{a}{{b + c}}} \right) + \left( {1 + \frac{b}{{c + a}}} \right) = \frac{{49}}{{10}}\\ \Rightarrow \frac{c}{{a + b}} + \frac{a}{{b + c}} + \frac{b}{{c + a}} = \frac{{49}}{{10}} – 3 = \frac{{19}}{{10}}\end{array}\] Bình luận
Giải thích các bước giải:
Ta có:
\[\begin{array}{l}
\frac{1}{{a + b}} + \frac{1}{{b + c}} + \frac{1}{{c + a}} = \frac{7}{{10}}\\
\Leftrightarrow \left( {a + b + c} \right)\left( {\frac{1}{{a + b}} + \frac{1}{{b + c}} + \frac{1}{{c + a}}} \right) = \frac{7}{{10}}.\left( {a + b + c} \right)\\
\Leftrightarrow \frac{{a + b + c}}{{a + b}} + \frac{{a + b + c}}{{b + c}} + \frac{{a + b + c}}{{c + a}} = \frac{7}{{10}}.7\\
\Leftrightarrow \left( {1 + \frac{c}{{a + b}}} \right) + \left( {1 + \frac{a}{{b + c}}} \right) + \left( {1 + \frac{b}{{c + a}}} \right) = \frac{{49}}{{10}}\\
\Rightarrow \frac{c}{{a + b}} + \frac{a}{{b + c}} + \frac{b}{{c + a}} = \frac{{49}}{{10}} – 3 = \frac{{19}}{{10}}
\end{array}\]