Tính 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 \(a\ne-b;b\ne-c;c\ne-a\).
K
Khách
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Những câu hỏi liên quan
KN
5 tháng 2 2020
\(a+b+c=1\Rightarrow\hept{\begin{cases}ab+c=ab+c\left(a+b+c\right)\\bc+a=bc+a\left(a+b+c\right)\\ca+b=ca+b\left(a+b+c\right)\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}ab+c=ab+ca+bc+c^2\\bc+a=bc+a^2+ab+ac\\ca+b=ca+ab+b^2+bc\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}ab+c=\left(b+c\right)\left(a+c\right)\\bc+a=\left(a+c\right)\left(a+b\right)\\ca+b=\left(b+c\right)\left(a+b\right)\end{cases}}\)
\(\Rightarrow P=\frac{\left(b+c\right)\left(a+c\right)}{\left(a+b\right)^2}.\frac{\left(a+c\right)\left(a+b\right)}{\left(b+c\right)^2}.\frac{\left(b+c\right)\left(a+b\right)}{\left(c+a\right)^2}=1\)
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 2 2020
\(P=\frac{ab+c\left(a+b+c\right)}{\left(a+b\right)^2}.\frac{bc+a\left(a+b+c\right)}{\left(b+c\right)^2}.\frac{ca+b\left(a+b+c\right)}{\left(c+a\right)^2}\)
\(=\frac{\left(a+c\right)\left(b+c\right)\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)}{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}=1\)
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
\(P=\frac{ab+c.1}{\left(a+b\right)^2}.\frac{bc+a.1}{\left(b+c\right)^2}.\frac{ca+b.1}{\left(c+a\right)^2}\)
\(P=\frac{ab+c\left(a+b+c\right)}{\left(a+b\right)^2}.\frac{bc+a\left(a+b+c\right)}{\left(b+c\right)^2}.\frac{ca+b\left(a+b+c\right)}{\left(c+a\right)^2}\)
\(P=\frac{ab+ca+bc+c^2}{\left(a+b\right)^2}.\frac{bc+a^2+ab+ac}{\left(b+c\right)^2}.\frac{ca+ab+b^2+bc}{\left(c+a\right)^2}\)
\(P=\frac{\left(a+c\right)\left(b+c\right)}{\left(a+b\right)^2}.\frac{\left(a+c\right)\left(a+b\right)}{\left(b+c\right)^2}.\frac{\left(a+b\right)\left(b+c\right)}{\left(c+a\right)^2}=1\)
18 tháng 7 2017
ques này nhiều ng` hỏi r` thay ab+bc+ca=1 vào rồi phân tích rút gọn