Cho a, b, c \(\ne\)0 thỏa mãn a+b+c = 0. Tính \(\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}\).
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LD
12 tháng 3 2021
\(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=0\Leftrightarrow\frac{bc+ac-ab}{abc}=0\)
Vì \(a,b,c\ne0\Rightarrow abc\ne0\)
\(\Rightarrow bc+ac-ab=0\)
\(\Rightarrow\hept{\begin{cases}\left(bc+ac\right)^2=\left(ab\right)^2\\\left(bc-ab\right)^2=\left(-ac\right)^2\\\left(ac-ab\right)^2=\left(-bc\right)^2\end{cases}\Rightarrow\hept{\begin{cases}b^2c^2+c^2a^2-a^2b^2=-2abc^2\\b^2c^2+a^2b^2-a^2c^2=2ab^2c\\a^2c^2+a^2b^2-b^2c^2=2a^2bc\end{cases}}}\)
\(\Rightarrow E=\frac{a^2b^2c^2}{2ab^2c}+\frac{a^2b^2c^2}{-2abc^2}+\frac{a^2b^2c^2}{2a^2bc}\)
\(\Rightarrow E=\frac{ac}{2}-\frac{ab}{2}+\frac{bc}{2}=\frac{ac-ab+bc}{2}=\frac{0}{2}=0\)
CHÚC BẠN HỌC TỐT
12 tháng 3 2021
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{bc+ac-ab}{abc}=0\)
Vì \(a,b,c\ne0\Rightarrow a.b.c\ne0\)
\(\Rightarrow bc+ac-ab=0\)
\(\Rightarrow\hept{\begin{cases}\left(bc+ac\right)^2=\left(ab\right)^2\\\left(bc-ab\right)^2=\left(-ac\right)^2\\\left(ac-ab\right)^2=\left(-bc\right)^2\end{cases}\Rightarrow}\hept{\begin{cases}b^2c^2+c^2a^2-a^2b^2=-abc^2\\b^2c^2+a^2b^2-a^2c^2=2ab^2c\\a^2c^2+a^2b^2-b^2c^2=2a^2bc\end{cases}}\)
\(\Rightarrow E=\frac{a^2b^2c^2}{2ab^2c}+\frac{a^2b^2c^2}{-2abc^2}+\frac{a^2b^2c^2}{2a^2bc}\)
\(\Rightarrow E=\frac{ac}{2}-\frac{ab}{2}+\frac{bc}{2}=\frac{ac-ab+bc}{2}=\frac{0}{2}=0\)
Vậy \(E=0\)
14 tháng 10 2018
\(a+b=c\Rightarrow\left(a+b\right)^2=c^2\Rightarrow a^2+2ab+b^2=c^2\Rightarrow a^2+b^2-c^2=-2ab\)
Tượng tự: \(b^2+c^2-a^2=2bc,c^2+a^2-b^2=2ac\)
Khi đó: \(B=\frac{-1}{2ab}+\frac{1}{2bc}+\frac{1}{2ac}=\frac{-c+a+b}{2abc}=0\)
Chúc bạn học tốt.
4 tháng 8 2019
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{a+b+c}=1\Rightarrow a=b=c\)
\(\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)
LN
31 tháng 7 2019
Có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}=....+2\frac{a+b+c}{abc}=.....\)
TM
11 tháng 4 2019
a+b+c=0 <=> a+b=-c ; a+c=-b ; b+c=-a
\(\frac{1}{b^2+c^2-a^2}=\frac{1}{\left(b-a\right)\left(a+b\right)+c^2}=\frac{1}{\left(b-a\right)\left(-c\right)+c^2}=\frac{1}{c\left(a-b+c\right)}=\frac{1}{-2bc}\)
Tương tự: \(\frac{1}{c^2+a^2-b^2}=\frac{1}{-2ca};\frac{1}{a^2+b^2-c^2}=\frac{1}{-2ab}\)
=>\(G=\frac{1}{-2bc}+\frac{1}{-2ca}+\frac{1}{-2ab}=\frac{a+b+c}{-2abc}=\frac{0}{-2abc}=0\)
10 tháng 9 2020
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) <=> \(\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)
<=> \(\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)
<=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{a^2b}+\frac{3}{ab^2}=-\frac{1}{c^3}\)
<=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)\)
<=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
Khi đó, A = \(\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc\cdot\frac{3}{abc}=3\)
10 tháng 9 2020
Xét: \(A=\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
Ta có đẳng thức sau: \(x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)+3xyz\)
(Đẳng thức này chứng minh rất dễ nha, chỉ cần bung hết ra là được)
Vậy ta thế \(x=\frac{1}{a},y=\frac{1}{b},z=\frac{1}{c}\)vào đẳng thức:
\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{ab}-\frac{1}{bc}-\frac{1}{ca}\right)+\frac{3}{abc}\)
\(=\frac{3}{abc}\)Vì \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)---> Thế cái này vào A:
\(\Rightarrow A=abc.\frac{3}{abc}=3\)
Xoooooooong !!!!! :)))
\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)
\(\Leftrightarrow a^2+b^2-c^2=-2c^2-2bc-2ac-2ab\)
\(\Leftrightarrow a^2+b^2-c^2=-\left[2c.\left(c+b\right)+2a.\left(c+b\right)\right]\)
\(\Leftrightarrow a^2+b^2-c^2=-2.\left(a+c\right)\left(c+b\right)\)
Tương tự \(b^2+c^2-a^2=-2.\left(a+b\right)\left(a+c\right)\)
\(c^2+a^2-b^2=-2.\left(b+c\right)\left(b+a\right)\)
Đặt \(A=\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}\)
\(=-\frac{1}{2}.\left[\frac{1}{\left(b+c\right)\left(a+c\right)}+\frac{1}{\left(a+b\right)\left(a+c\right)}+\frac{1}{\left(b+c\right)\left(a+b\right)}\right]\)
\(=-\frac{1}{2}.\frac{a+b+b+c+a+c}{\left(b+c\right).\left(a+c\right)\left(a+b\right)}=-\frac{1}{2}.\frac{2.\left(a+b+c\right)}{\left(b+c\right).\left(a+c\right).\left(a+b\right)}=0\)