Giá trị của biểu thức \(P=\frac{ab+c}{\left(a+b\right)^2}.\frac{bc+a}{\left(b+c\right)^2}.\frac{ca+b}{\left(c+a\right)^2}\) khi \(a+b+c=1\) và a ≠ -b; b ≠ -c; c ≠ -a là:
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G
10 tháng 3 2019
\(P=\frac{ab+c}{\left(a+b\right)^2}.\frac{bc+a}{\left(b+c\right)^2}.\frac{ac+b}{\left(c+a\right)^2}\)
\(P=\frac{ab+c\left(a+b+c\right)}{\left(a+b\right)^2}.\frac{bc+a\left(a+b+c\right)}{\left(b+c\right)^2}.\frac{ac+b\left(a+b+c\right)}{\left(c+a\right)^2}\)
\(P=\frac{ab+ac+bc+c^2}{\left(a+b\right)^2}.\frac{ab+bc+ac+a^2}{\left(b+c\right)^2}.\frac{ab+bc+ac+b^2}{\left(a+c\right)^2}\)
\(P=\frac{\left(a+c\right)\left(b+c\right)}{\left(a+b\right)^2}.\frac{\left(a+b\right)\left(a+c\right)}{\left(b+c\right)^2}.\frac{\left(a+b\right)\left(b+c\right)}{\left(a+c\right)^2}=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(a+c\right)^2}{\left(a+b\right)^2\left(b+c\right)^2\left(a+c\right)^2}=1\)
S
5 tháng 2 2020
\(P=\frac{ab+c.1}{\left(a+b\right)^2}.\frac{bc+a.1}{\left(b+c\right)^2}.\frac{ca+b.1}{\left(c+a\right)^2}\)
\(P=\frac{ab+c\left(a+b+c\right)}{\left(a+b\right)^2}.\frac{bc+a\left(a+b+c\right)}{\left(b+c\right)^2}.\frac{ca+b\left(a+b+c\right)}{\left(c+a\right)^2}\)
\(P=\frac{ab+ca+bc+c^2}{\left(a+b\right)^2}.\frac{bc+a^2+ab+ac}{\left(b+c\right)^2}.\frac{ca+ab+b^2+bc}{\left(c+a\right)^2}\)
\(P=\frac{\left(a+c\right)\left(b+c\right)}{\left(a+b\right)^2}.\frac{\left(a+c\right)\left(a+b\right)}{\left(b+c\right)^2}.\frac{\left(a+b\right)\left(b+c\right)}{\left(c+a\right)^2}=1\)
KN
5 tháng 2 2020
\(a+b+c=1\Rightarrow\hept{\begin{cases}ab+c=ab+c\left(a+b+c\right)\\bc+a=bc+a\left(a+b+c\right)\\ca+b=ca+b\left(a+b+c\right)\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}ab+c=ab+ca+bc+c^2\\bc+a=bc+a^2+ab+ac\\ca+b=ca+ab+b^2+bc\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}ab+c=\left(b+c\right)\left(a+c\right)\\bc+a=\left(a+c\right)\left(a+b\right)\\ca+b=\left(b+c\right)\left(a+b\right)\end{cases}}\)
\(\Rightarrow P=\frac{\left(b+c\right)\left(a+c\right)}{\left(a+b\right)^2}.\frac{\left(a+c\right)\left(a+b\right)}{\left(b+c\right)^2}.\frac{\left(b+c\right)\left(a+b\right)}{\left(c+a\right)^2}=1\)
18 tháng 7 2017
ques này nhiều ng` hỏi r` thay ab+bc+ca=1 vào rồi phân tích rút gọn
26 tháng 9 2017
Do ab + bc + ca = 1 nên ta có :
\(a\sqrt{\frac{\left(b^2+1\right)\left(c^2+1\right)}{a^2+1}}=a\sqrt{\frac{\left(b^2+ab+ac+bc\right)\left(c^2+ab+ac+bc\right)}{a^2+ab+ac+bc}}\)
\(=a\sqrt{\frac{\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)}{\left(a+b\right)\left(a+c\right)}}=a\sqrt{\left(b+c\right)^2}=a\left(b+c\right)=ab+ac\text{ }\left(1\right)\)
Tương tự : \(b\sqrt{\frac{\left(a^2+1\right)\left(c^2+1\right)}{b^2+1}}=ab+bc\) (2)và \(c\sqrt{\frac{\left(b^2+1\right)\left(a^2+1\right)}{c^2+1}}=bc+ac\) (3)
Cộng vế với vế của (1) ; (2) ; (3) lại ta được :
\(a\sqrt{\frac{\left(b^2+1\right)\left(c^2+1\right)}{a^2+1}}+b\sqrt{\frac{\left(a^2+1\right)\left(c^2+1\right)}{b^2+1}}+c\sqrt{\frac{\left(b^2+1\right)\left(a^2+1\right)}{c^2+1}}=2\left(ab+bc+ac\right)=2\)
\(P=\frac{ab+c\left(a+b+c\right)}{\left(a+b\right)^2}.\frac{bc+a\left(a+b+c\right)}{\left(b+c\right)^2}.\frac{ca+b\left(a+b+c\right)}{\left(c+a\right)^2}\)
\(=\frac{\left(a+c\right)\left(b+c\right)\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)}{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}=1\)