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b,\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
=>\(\dfrac{bc}{abc}+\dfrac{ac}{bac}+\dfrac{ab}{abc}=0\)
=>\(\dfrac{ab+ac+bc}{abc}=0\)
=>ab+ac+bc=0
=>ab=-ac-bc
ac=-ab-bc
bc=-ab-ac
N=\(\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ca}+\dfrac{1}{c^2+2ab}\)
N=\(\dfrac{1}{a^2+bc+bc}+\dfrac{1}{b^2+ca+ca}+\dfrac{1}{c^2+ab+ab}\)
N=\(\dfrac{1}{a^2-ab-ac+bc}+\dfrac{1}{b^2-ab-bc+ca}+\dfrac{1}{c^2-ac-bc+ab}\)
N=\(\dfrac{1}{a\left(a-b\right)-c\left(a-b\right)}+\dfrac{1}{b\left(b-a\right)-c\left(b-a\right)}+\dfrac{1}{c\left(c-a\right)-b\left(c-a\right)}\)
N=\(\dfrac{1}{\left(a-c\right)\left(a-b\right)}+\dfrac{1}{\left(b-c\right)\left(b-a\right)}+\dfrac{1}{\left(c-b\right)\left(c-a\right)}\)
N=\(\dfrac{b-c}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}-\dfrac{a-c}{\left(b-c\right)\left(a-b\right)\left(a-c\right)}+\dfrac{a-b}{\left(b-c\right)\left(a-c\right)\left(a-b\right)}\)
N=\(\dfrac{b-c-a+c+a-b}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)=0
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-c\right)\left(b-a\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
\(A=\frac{-b+c}{-\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{-c+a}{-\left(a-b\right)\left(a-c\right)\left(b-a\right)}+\frac{-a+b}{-\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(A=\frac{-b+c-c+a-a+b}{-\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(A=\frac{0}{-\left(a-b\right)\left(a-c\right)\left(b-a\right)}\)
A = 0
Ta có: \(a^2+b^2+c^2=\left(a+b+c\right)^2\)
\(\Leftrightarrow ab+bc+ca=0\)
Ta có: \(A=\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\)
\(=\frac{1}{a^2+2bc-ab-bc-ca}+\frac{1}{b^2+2ca-ab-bc-ca}+\frac{1}{c^2+2ab-ab-bc-ca}\)
\(=\frac{1}{a^2+bc-ca-ab}+\frac{1}{b^2+ca-ab-bc}+\frac{1}{c^2+ab-bc-ca}\)
\(=-\left(\frac{1}{\left(a-b\right)\left(c-a\right)}+\frac{1}{\left(b-c\right)\left(a-b\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)}\right)\)
\(=-\frac{b-c+c-a+a-b+}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
PS: Hồi tối lười để người khác làm mà không ai làm thôi t làm vậy
( a+b+c)^2 = a^2 + b^2 + c^2
=> a^2 + b^2 + c^2 + 2ab + 2bc + 2ac = a^2 + b^2 + c^2
=> a^2 + b^2 + c^2 + 2ab + 2bc + 2ac - a^2 - b^2 - c^2 = 0
=> 2ab + 2bc + 2ac = 0
ta có
A = \(\frac{1}{a^2+2bc}\)+ \(\frac{1}{b^2+2ac}\)+ \(\frac{1}{c^2+2ab}\)
= \(\frac{1}{a^2+2bc}\)+ \(\frac{1}{b^2+2ac}\)+ \(\frac{1}{c^2+2ab}\) + 2ab + 2bc + 2ac
đến đây bạn nhóm lại nhé mk giải ra thì dài lắm nên chỉ gợi ý cho bn đấy đây thôi
Ta có: \(a+b+c=\frac{1}{2}\) \(\Rightarrow\left\{\begin{matrix}a=\frac{1}{2}-b-c\\b=\frac{1}{2}-a-c\\c=\frac{1}{2}-a-b\end{matrix}\right.\) hay \(\left\{\begin{matrix}a+b=\frac{1}{2}-c\\b+c=\frac{1}{2}\\a+c=\frac{1}{2}-b\end{matrix}\right.\)
\(P=\frac{\left(2ab+c\right)\left(2bc+a\right)\left(2ac+b\right)}{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}\)
\(=\frac{\left(2ab+\frac{1}{2}-a-b\right)\left(2bc+\frac{1}{2}-b-c\right)\left(2ca+\frac{1}{2}-a-c\right)}{\left(\frac{1}{2}-c\right)\left(\frac{1}{2}-a\right)\left(\frac{1}{2}-c\right)\left(\frac{1}{2}-b\right)\left(\frac{1}{2}-a\right)\left(\frac{1}{2}-b\right)}\)
\(=\frac{2\left(ab+\frac{1}{4}-\frac{1}{2}a-\frac{1}{2}b\right).2\left(bc+\frac{1}{4}-\frac{1}{2}b-\frac{1}{2}c\right).2\left(ca+\frac{1}{4}-\frac{1}{2}a-\frac{1}{2}c\right)}{\left(ac+\frac{1}{4}-\frac{1}{2}a-\frac{1}{2}c\right)\left(bc+\frac{1}{4}-\frac{1}{2}b-\frac{1}{2}c\right)\left(ab+\frac{1}{4}-\frac{1}{2}a-\frac{1}{2}b\right)}\)
\(=2.2.2=8\)
Vậy với \(a+b+c=\frac{1}{2}\) và \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ne0\) thì \(P=8\)
đơn giản, cứ áp dụng theo công thức là ra!!!!