Cho a+b+c=0 ( abc khác 0). Rút gon:
\(A=\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-a^2-b^2}\)
K
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17 tháng 12 2021
\(\dfrac{a^2}{a^2-b^2-c^2}=\dfrac{a^2}{\left(a-b\right)\left(a+b\right)-c^2}=\dfrac{a^2}{\left(a-b\right)\left(-c\right)-c^2}=\dfrac{a^2}{c\left(b-a-c\right)}=\dfrac{a^2}{2bc}\\ \Leftrightarrow M=\sum\dfrac{a^2}{a^2-b^2-c^2}=\sum\dfrac{a^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}\\ \Leftrightarrow M=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{2abc}=0\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 12 2020
\(a+b=-c\Leftrightarrow\left(a+b\right)^3=-c^3\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)=3abc\)
\(A=\dfrac{a^3+b^3+c^3}{abc}=\dfrac{3abc}{abc}=3\)
17 tháng 11 2021
\(1,a+b+c=0\Leftrightarrow a=-b-c\Leftrightarrow a^2=b^2+2bc+c^2\Leftrightarrow b^2+c^2=a^2-2bc\)
Tương tự: \(\left\{{}\begin{matrix}a^2+b^2=c^2-2ab\\c^2+a^2=b^2-2ac\end{matrix}\right.\)
\(\Leftrightarrow N=\dfrac{a^2}{a^2-a^2+2bc}+\dfrac{b^2}{b^2-b^2+2ca}+\dfrac{c^2}{c^2-c^2+2ac}\\ \Leftrightarrow N=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{a^3+b^3+c^3-3abc+3abc}{2abc}\\ \Leftrightarrow N=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{2abc}\\ \Leftrightarrow N=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)
13 tháng 3 2022
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
TT
19 tháng 12 2020
Từ đkđb
\(\Leftrightarrow2\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow\dfrac{ab+bc+ac}{abc}=0\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}\)
\(\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=-\dfrac{1}{c^3}\)
\(\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
19 tháng 12 2020
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.
19 tháng 2 2018
Từ \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2+2ab+b^2=c^2\\a^2+2ac+c^2=b^2\\b^2+2bc+c^2=a^2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2+b^2-c^2=-2ab\\a^2+c^2-c^2=-2ac\\b^2+c^2-a^2=-2bc\\\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{1}{-2ab}+\dfrac{1}{-2ac}+\dfrac{1}{-2bc}=\dfrac{a+b+c}{-2abc}=\dfrac{0}{-2abc}=0\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 8 2021
\(a+b+c=0\Rightarrow b+c=-a\)
\(\Rightarrow\left(b+c\right)^2=a^2\) \(\Rightarrow b^2+c^2+2bc=a^2\)
\(\Rightarrow a^2-b^2-c^2=2bc\)
Tương tự: \(b^2-c^2-a^2=2ca\) ; \(c^2-a^2-b^2=2ab\)
Mặt khác ta có:
\(a+b+c=0\Rightarrow a+b=-c\Rightarrow\left(a+b\right)^3=-c^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3+c^3=-3ab\left(a+b\right)=-3ab\left(-c\right)=3abc\)
Đặt vế trái biểu thức cần chứng minh là P
\(\Rightarrow P=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2ab}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{3abc}{2abc}=\dfrac{3}{2}\) (đpcm)
17 tháng 2 2018
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
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: \(b^2-c^2-a^2=2ac;\) \(c^2-a^2-b^2=2ab\)
\(\Rightarrow A=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)