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\(\dfrac{a^2}{a^2-b^2-c^2}=\dfrac{a^2}{\left(a-b\right)\left(a+b\right)-c^2}=\dfrac{a^2}{\left(a-b\right)\left(-c\right)-c^2}=\dfrac{a^2}{c\left(b-a-c\right)}=\dfrac{a^2}{2bc}\\ \Leftrightarrow M=\sum\dfrac{a^2}{a^2-b^2-c^2}=\sum\dfrac{a^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}\\ \Leftrightarrow M=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{2abc}=0\)
\(a+b=-c\Leftrightarrow\left(a+b\right)^3=-c^3\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)=3abc\)
\(A=\dfrac{a^3+b^3+c^3}{abc}=\dfrac{3abc}{abc}=3\)
P= \(\dfrac{1}{b^2+c^2-a^2}+\dfrac{1}{a^2+c^2-b^2}+\dfrac{1}{a^2+b^2-c^2}\)
=
\(\dfrac{a+b+c}{\left(b^2+c^2-a^2\right)\left(a+b+c\right)}+\dfrac{a+b+c}{\left(a^2+c^2-b^2\right)\left(a+b+c\right)}+\dfrac{a+b+c}{\left(a^2+b^2-c^2\right)\left(a+b+c\right)}\)
= 0+0+0 = 0
Vậy P= 0
Ngu vãi ko bt đúng không nx
Từ đkđb
\(\Leftrightarrow2\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow\dfrac{ab+bc+ac}{abc}=0\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}\)
\(\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=-\dfrac{1}{c^3}\)
\(\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
Hớ hớ bài này mình cũng làm rồi.
Ta có: (a+b+c)2=a2+b2+c2
<=> a2+b2+c2+2(ab+bc+ca)=a2+b2+c2
<=>2(ab+bc+ca)=0
<=>ab+bc+ca=0
\(\Leftrightarrow\dfrac{ab+bc+ca}{abc}=0\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
=>\(\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3=\left(-\dfrac{1}{c}\right)^3\)
=> \(\dfrac{1}{a^3}+\dfrac{3}{a^2b}+\dfrac{3}{ab^2}+\dfrac{1}{b^3}=-\dfrac{1}{c^3}\)
=>\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=-\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=-\dfrac{3}{ab}.\left(-\dfrac{1}{c}\right)=\dfrac{3}{abc}\)
=> Đpcm.
Từ \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2+2ab+b^2=c^2\\a^2+2ac+c^2=b^2\\b^2+2bc+c^2=a^2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2+b^2-c^2=-2ab\\a^2+c^2-c^2=-2ac\\b^2+c^2-a^2=-2bc\\\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{1}{-2ab}+\dfrac{1}{-2ac}+\dfrac{1}{-2bc}=\dfrac{a+b+c}{-2abc}=\dfrac{0}{-2abc}=0\)
Vì a+b+c=0. Suy ra
* a+b=-c
=> (a+b)2=c2
=> a2+b2+2ab=c2
=>a2+b2-c2=-2ab
tương tự ta đc a2+c2-b2=-2ac và c2+b2-a2=-2bc
Ta có
A=\(\dfrac{1}{a^2+c^2-a^2}+\dfrac{1}{c^2+a^2-b^2}+\dfrac{1}{a^2+b^2-c^2}\)
=>\(A=\dfrac{-1}{2bc}-\dfrac{1}{2ac}-\dfrac{1}{2ab}\)
=>A=\(\dfrac{-a}{2abc}-\dfrac{b}{2abc}-\dfrac{c}{2abc}\)
=>A=\(\dfrac{-a-b-c}{2abc}=\dfrac{-\left(a+b+c\right)}{2abc}\)
=>\(\dfrac{0}{2abc}=0\) (vì a+b+c=0)
vậy A=0
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow ab+bc+ca=0\Rightarrow\left\{{}\begin{matrix}bc=-ab-ac\\ab=-bc-ac\\ac=-ab-bc\end{matrix}\right.\)
\(M=\dfrac{1}{a^2+bc-ab-ac}+\dfrac{1}{b^2+ac-ab-bc}+\dfrac{1}{c^2+ab-bc-ac}\)
\(=\dfrac{1}{a\left(a-b\right)-c\left(a-b\right)}+\dfrac{1}{b\left(b-c\right)-a\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)-b\left(c-a\right)}\)
\(=\dfrac{1}{\left(a-b\right)\left(a-c\right)}-\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{b-c-\left(a-c\right)+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)
\(\Leftrightarrow\dfrac{2a^2}{b^2}+\dfrac{2b^2}{c^2}+\dfrac{2c^2}{a^2}=\dfrac{2a}{c}+\dfrac{2c}{b}+\dfrac{2b}{a}\)
\(\Leftrightarrow\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}-\dfrac{2a}{c}\right)+\left(\dfrac{a^2}{b^2}+\dfrac{c^2}{a^2}-\dfrac{2c}{b}\right)+\left(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}-\dfrac{2b}{a}\right)=0\)
\(\Leftrightarrow\left(\dfrac{a}{b}-\dfrac{b}{c}\right)^2+\left(\dfrac{a}{b}-\dfrac{c}{a}\right)^2+\left(\dfrac{b}{c}-\dfrac{c}{a}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}-\dfrac{b}{c}=0\\\dfrac{a}{b}-\dfrac{c}{a}=0\\\dfrac{b}{c}-\dfrac{c}{a}=0\end{matrix}\right.\) \(\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}\Leftrightarrow a=b=c\)
Từ \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
Đầu tiên ta có hẳng đẳng thức:
\(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
\(\Rightarrow0=a^3+b^3+c^3+3\left(-c\right)\left(-b\right)\left(-a\right)\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\Rightarrow a^3+b^3+c^3=3abc\)
Biến đổi mẫu thức:
\(a^2-b^2-c^2=\left(a+b\right)\left(a-b\right)-c^2=-c\left(a-b\right)-c^2=-c\left(a-b+c\right)=2bc\)
Tương tự: \(b^2-c^2-a^2=2ac;\) \(c^2-a^2-b^2=2ab\)
\(\Rightarrow A=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)