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Cộng biểu thức thêm 3 vao mỗi số hạng sau đó dùng tc phân phối nha
Đáp số 3882
a / (b+c) +1+b/(a+c)+1 +c/(a+b) +1-3 =(a+b+c) /(a+b)+(a+b+c)/(a+c)+(a+b+c)/(a+b)-3
=(a+b+c).(1/(b+c)+1/(a+b)+1/(a+c))-3
=259.15-3
=3882
Ta có: \(Q=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(\Rightarrow Q=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)
\(\Rightarrow Q=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3\)
\(\Rightarrow Q=\left(a+b+c\right).\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)-3\)
\(\Rightarrow Q=259.15-3=3882\)
Vậy Q=3882
\(Q=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(\Rightarrow Q=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)
\(\Rightarrow Q=\left(\frac{a+b+c}{b+c}\right)+\left(\frac{a+b+c}{a+c}\right)+\left(\frac{a+b+c}{a+b}\right)-3\)
\(\Rightarrow Q=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)
\(\Rightarrow Q=259.15-3=3882\)
\(Q=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(Q=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+a\right)+\left(\frac{c}{a+b}+1\right)-3\)
\(Q=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3\)
\(Q=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{a}{a+c}+\frac{a}{a+b}\right)-3\)
\(Q=259.15-3\)
\(Q=3882\)
Vậy \(Q=3882\)
\(Q+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)
\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\)
\(=259.15=3885\)
\(\Rightarrow Q=3885-3=3882\)
\(Q=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(Q=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)
\(Q=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3\)
\(Q=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3=259.15-3=3882\)
Vậy Q=3882
Làm lại nè:
Ta có: Q=\(\frac{a}{b+c}+1+\frac{b}{a+c}+1+\frac{c}{a+b}+1-3\)
Q=\(\frac{a+b+c}{b+c}+\frac{b+a+c}{a+c}+\frac{c+a+b}{a+b}-3\)
Q=\(259.\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)
Q=259.15-3
Q=3882
Q=\(\frac{259}{b+c}+\frac{259}{a+c}+\frac{259}{a+b}-3\)
Ta có: Q=\(\frac{a}{b+c}+1+\frac{b}{a+c}+1+\frac{c}{a+b}+1-3\)
Q=\(\frac{a+b+c}{b+c}+\frac{b+a+c}{a+c}+\frac{c+a+b}{a+b}-3\)
Q=\(\frac{259}{b+c}+\frac{259}{a+c}+\frac{259}{a+b}-3\)
Q=\(259.\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)
Q=259.13-3
Q=3882