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\(M=\dfrac{3^{14}\cdot5^4-3^{12}\cdot5^4}{3^{12}\cdot5^6+7\cdot3^{12}\cdot5^6}=\dfrac{3^{12}\cdot5^4\left(3^2-1\right)}{3^{12}\cdot5^6\left(1+7\right)}=\dfrac{1}{25}\)
Xét \(\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=126.16=2016\)
\(\Leftrightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}=2016\)
\(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=2013\)
Vậy A = 2013
Lời giải:
Bạn tham khảo cách làm tương tự tại đây:
https://hoc24.vn/cau-hoi/cho-dfracab-2017ccdfracbc-2017aadfracca-2017bbvoi-a-b-c-ne0-tinhp-left1dfracabrightleft1dfracb.161494910584
Kết quả $P=8$ hoặc $P=-1$
E xin lỗi, e ko nhận câu trả lời này vì có chứa link tới các web khác
Đặt \(S=\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{c}{c+a}\)
\(\Rightarrow S=\dfrac{2010-\left(a+b\right)}{a+b}+\dfrac{2010-\left(b+c\right)}{b+c}+\dfrac{2010-\left(c+a\right)}{c+a}\)\(\Rightarrow S=\dfrac{2010}{a+b}+\dfrac{2010}{b+c}+\dfrac{2010}{c+a}-3\)
\(\Rightarrow S=2010\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)-3\)
\(\Rightarrow S=2010.\dfrac{1}{3}-3\)
\(\Rightarrow S=670-3\)
\(\Rightarrow S=667\)
Ta có: \(\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}=\left(\dfrac{c}{a+b}+1\right)+\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{c+a}+1\right)-3\)
\(=\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}-3\)
\(=\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)-3\)
\(=2010.\dfrac{1}{3}-3\)
\(=670-3\)
\(=667\)
Vậy \(\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}=667\)
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(=\dfrac{a+b+c}{b+c}-1+\dfrac{a+b+c}{c+a}-1+\dfrac{a+b+c}{a+b}-1\)
\(=\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)-3=\dfrac{2010}{3}-3=\dfrac{2001}{3}\)p/s: thiếu đề
Đặt S=ca+b+ab+c+cc+aS=ca+b+ab+c+cc+a
⇒S=2010−(a+b)a+b+2010−(b+c)b+c+2010−(c+a)c+a⇒S=2010−(a+b)a+b+2010−(b+c)b+c+2010−(c+a)c+a⇒S=2010a+b+2010b+c+2010c+a−3⇒S=2010a+b+2010b+c+2010c+a−3
⇒S=2010(1a+b+1b+c+1c+a)−3⇒S=2010(1a+b+1b+c+1c+a)−3
⇒S=2010.13−3⇒S=2010.13−3
⇒S=670−3⇒S=670−3
⇒S=667