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\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow abc.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\Leftrightarrow\hept{\begin{cases}bc=-\left(ab+ac\right)\\ab=-\left(bc+ac\right)\\ac=-\left(bc+ab\right)\end{cases}}\)
Ta có: \(a^2+2bc=a^2+bc+bc=a^2+bc+\left(-ab-ac\right)=\left(a-b\right)\left(a-c\right)\)
Tương tự \(b^2+2ac=\left(b-a\right)\left(b-c\right);c^2+2ab=\left(c-a\right)\left(c-b\right)\)
\(\Leftrightarrow N=\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ac}{\left(b-a\right)\left(b-c\right)}+\frac{ab}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{ab\left(a-b\right)+c^2\left(a-b\right)-c\left(a^2-b^2\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)
Nhận xét: \(\text{ *)}\) Nếu \(x+y+z=0\) thì \(x^3+y^3+z^3=3xyz\)
Thật vậy, từ \(x+y+z=0\)
Suy ra: \(x+y=-z\) \(\left(\text{*}\right)\)
\(\Leftrightarrow\) \(\left(x+y\right)^3=\left(-z\right)^3\)
\(\Leftrightarrow\) \(x^3+3x^2y+3xy^2+y^3=\left(-z\right)^3\)
\(\Leftrightarrow\) \(x^3+y^3+z^3=-3x^2y-3xy^2\)
\(\Leftrightarrow\) \(x^3+y^3+z^3=-3xy\left(x+y\right)\)
\(\Leftrightarrow\) \(x^3+y^3+z^3=3xyz\) (theo \(\left(\text{*}\right)\) )
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Theo giả thiết, ta có:
\(a+b+c=0\)
\(\Leftrightarrow\) \(b+c=-a\)
\(\Leftrightarrow\) \(\left(b+c\right)^2=\left(-a\right)^2\)
\(\Leftrightarrow\) \(b^2+2bc+c^2=a^2\)
\(\Leftrightarrow\) \(2bc=a^2-b^2-c^2\)
Tương tự, ta cũng có \(2ac=b^2-a^2-c^2\) \(;\) \(2ab=c^2-a^2-b^2\)
Mặt khác, vì \(a+b+c=0\) nên \(a^3+b^3+c^3=3abc\) (theo nhận xét trên)
Do đó, \(A=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}=\frac{a^3}{2abc}+\frac{b^3}{2abc}+\frac{c^3}{2abc}=\frac{a^3+b^3+c^3}{2abc}=\frac{3abc}{2abc}=\frac{3}{2}\) (do \(abc\ne0\)
tu a + b + c = 0 suy ra a= - (b+c) suy ra a^2 = (b+c)^2=b^2 +c^2 + 2bc suy ra a^2 - b^2 - c^2 =2bc . tuong tu ta cung co b^2-a^2-c^2=2ac ; c^2- a^2 -b^2=2ab do do A = a^2/2bc + b^2/2ac+c^2/2ab =a^3/2abc+b^3/2abc +c^3/2abc lai co a+b+c=o nen a+b=-c suyra a^3+b^3+3ab(a+b)= -c^3 do do a^3 +b^3 +c^3=3abc vay A=3abc/2abc=3/2 (abc khac 0 : a+b=c=o)
Ta có:\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow\frac{ab+bc+ac}{abc}=0\)
\(\Leftrightarrow ab+bc+ac=0\Rightarrow\hept{\begin{cases}ab=-bc-ac\\bc=-ac-ab\\ac=-ab-bc\end{cases}}\)(*)
Thay (*) vào M ta được:
\(M=\frac{1}{a^2+bc-ab-ac}+\frac{1}{b^2+ac-ab-bc}+\frac{1}{c^2+ab-bc-ac}\)
\(=\frac{1}{a\left(a-b\right)-c\left(a-b\right)}+\frac{1}{a\left(c-b\right)-b\left(c-b\right)}+\frac{1}{c\left(c-a\right)-b\left(c-a\right)}\)
\(=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(a-b\right)\left(c-b\right)}-\frac{1}{\left(c-b\right)\left(a-c\right)}\)
\(=\frac{c-b}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}+\frac{a-c}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}-\frac{a-b}{\left(a-b\right)\left(c-b\right)\left(a-c\right)}\)
\(=\frac{c-b+a-c-a+b}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}=0\)
Vậy M = 0
Có a + b + c = 0
=> a + b = - c
=> (a + b)2 = c2
=> a2 + b2 + 2ab = c2
=> a2 + b2 - c2 = - 2ab
Tương tự, b2 + c2 - a2 = - 2bc và c2 + a2 - b2 = - 2ca
Do đó \(A=\frac{ab}{-2ab}+\frac{bc}{-2bc}+\frac{ca}{-2ca}=-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}=-\frac{3}{2}\)
a+b+c=0=>a+b=-c=>a2+b2+2ab=c2=>a2+b2-c2=-2ab
Tương tự b2+c2-a2=-2bc,c2+a2-b2=-2ac
=>\(A=\frac{-ab}{2ab}+\frac{-bc}{2bc}+\frac{-ca}{2ca}=\frac{-3}{2}\)
jkghffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff giống bạn đó Nguyễn Thế An
1. a + b + c = 0 \(\Rightarrow\)a + b = -c \(\Rightarrow\)( a + b )2 = ( -c )2 \(\Rightarrow\)a2 + b2 - c2 = -2ab
Tương tự : b2 + c2 - a2 = -2bc ; c2 + a2 - b2 = -2ac
Ta có : \(\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}\)
\(=\frac{1}{-2ab}+\frac{1}{-2bc}+\frac{1}{-2ac}=\frac{-1}{2}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)
\(=\frac{-1}{2}\left(\frac{a+b+c}{abc}\right)=0\)
2. tương tự
3,4 . có ở dưới, câu hỏi của Quyết Tâm chiến thắng
ta có: a + b + c = 0 => a+b = - c => a2 + 2ab + b2 = c2 => a2 + b2 - c2 = - 2ab
tương tự như trên, ta có: b2 + c2 - a2 = -2bc; c2 + a2 - b2 = -2ac
thay vào A, có:
\(A=\frac{1}{-2bc}-\frac{1}{2ca}-\frac{1}{2ab}\)
\(A=-\frac{1}{2}.\left(\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}\right)=-\frac{1}{2}.\left(\frac{a+b+c}{abc}\right)=-\frac{1}{2}.\left(\frac{0}{abc}\right)=0\)
KL: A = 0 tại a + b + c = 0