Cho \(a+b+c=0\) và \(abc\ne0\) Tính giá trị biểu thức :
P = \(\frac{1}{b^2+c^2-a^2}+\frac{1}{a^2+c^2-b^2}+\frac{1}{a^2+b^2-c^2}\)
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27 tháng 6 2019
Do \(a+b+c=0\)
\(\Rightarrow c=-a-b\)
\(\Rightarrow c^2=a^2+2ab+b^2\)
Tương tự,ta có:
\(a^2=b^2+2bc+c^2\)
\(b^2=a^2+2ac+c^2\)
Thay vào bài toán,ta được:
\(P=\frac{c^2}{a^2+b^2-\left(a^2+2ab+b^2\right)}+\frac{a^2}{b^2+c^2-\left(b^2+2bc+c^2\right)}+\frac{b^2}{c^2+a^2-\left(a^2+2ac+c^2\right)}\)
\(P=\frac{-c^2}{2ab}+\frac{-a^2}{2bc}+\frac{-b^2}{2ac}\)
\(P=\frac{-\left(a^3+b^3+c^3\right)}{2abc}\)
Do \(a+b+c=0\Rightarrow-a=b+c\)
\(\Rightarrow-a^3=b^3+c^3+3bc\left(b+c\right)\)
\(\Rightarrow-a^3=b^3+c^3-3abc\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
Khi đó,ta có:
\(P=\frac{-\left(3abc\right)}{2abc}=-\frac{3}{2}\)
KN
16 tháng 9 2020
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow\frac{bc+ca+ab}{abc}=0\)
\(\Leftrightarrow bc+ca+ab=0\)
\(\Leftrightarrow\hept{\begin{cases}bc=-ab-ca\\ca=-ab-bc\\ab=-ca-bc\end{cases}}\)
Ta có : \(A=\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\)
\(\Leftrightarrow A=\frac{a^2}{a^2+bc-ab-ca}+\frac{b^2}{b^2+ac-ab-bc}+\frac{c^2}{c^2+ab-ca-bc}\)
\(\Leftrightarrow A=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(\Leftrightarrow A=\frac{a^2}{\left(a-b\right)\left(a-c\right)}-\frac{b^2}{\left(b-c\right)\left(a-b\right)}+\frac{c^2}{\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{\left(a^2-b^2\right)\left(b-c\right)-\left(b^2-c^2\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{\left(a+b\right)\left(a-b\right)\left(b-c\right)-\left(b+c\right)\left(b-c\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{\left(a-b\right)\left(b-c\right)\left[\left(a+b\right)-\left(b+c\right)\right]}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)
TN
30 tháng 12 2017
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow ab+bc+ac=0\)
Đặt \(F=a^2+b^2+c^2\)
Từ \(a+b+c=1\Rightarrow\left(a+b+c\right)^2=1\)
\(\Rightarrow F+2\left(ab+bc+ac\right)=1\)
\(\Rightarrow F+2\cdot0=1\Rightarrow F=1\)
13 tháng 7 2016
\(=\frac{1}{a^2}+\frac{2}{ab}+\frac{1}{b^2}+\frac{2}{bc}+\frac{1}{c^2}+\frac{2}{ac}\)
\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2a+2b+2c}{abc}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
13 tháng 7 2016
\(A=\frac{1}{a^2}+\frac{2}{ab}+\frac{1}{b^2}+\frac{2}{bc}+\frac{1}{c^2}+\frac{2}{ac}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
Linh không biết a + b + c = 0 để làm gì?
30 tháng 6 2018
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\\ \)
\(\Rightarrow bc=-ab-ac,ca=-ab-bc,ab=-bc-ca\)
\(\Rightarrow\frac{a^2+bc}{a^2+2bc}=\frac{a^2+bc}{a^2+bc+bc}=\frac{a^2+bc}{a^2+bc-ca-ab}=\frac{a^2+bc}{\left(a-b\right).\left(a-c\right)}\)
Làm tương tự. có: \(\frac{b^2+ca}{b^2+2ca}=\frac{b^2+ca}{b^2+ca-ab-bc}=\frac{b^2+ca}{\left(a-b\right).\left(c-b\right)}\)
\(\frac{c^2+ab}{c^2+2ab}=\frac{c^2+ab}{c^2+ab-ca-bc}=\frac{c^2+ab}{\left(b-c\right).\left(a-c\right)}\)
\(\Rightarrow A=\frac{a^2+bc}{\left(a-b\right).\left(a-c\right)}+\frac{b^2+ca}{\left(a-b\right).\left(c-b\right)}+\frac{c^2+ab}{\left(b-c\right).\left(a-c\right)}\)
\(=\frac{\left(a^2+bc\right).\left(b-c\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}-\frac{\left(b^2+ca\right).\left(a-c\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}+\frac{\left(c^2+ab\right).\left(a-b\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}\)
Sau đó bạn thực hiện tiếp nhé.
MV
2 tháng 8 2021
Bài 1: Cho \(a,b,c\ge0:a^2+b^2+c^2=3\). CMR: \(a^4b^4+b^4c^4+c^4a^4\le3\)
Bài 2: Cho \(a,b,c\ge0\). CMR: \(a^2+b^2+c^2+2abc+1\ge2\left(ab+bc+ca\right)\)
Bài 3: Cho \(a,b,c\ge0:a^2+b^2+c^2=a+b+c\). CMR: \(a^2b^2+b^2c^2+c^2a^2\le ab+bc+ca\)
Bài 4: Cho \(a,b,c\ge0\). CMR: \(4\left(a+b+c\right)^3\ge27\left(ab^2+bc^2+ca^2+abc\right)\)
Bài 5: Cho \(a,b,c\ge0:a+b+c=3\).CMR: \(\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}+\frac{1}{2ab^2+1}\ge1\)
5 tháng 7 2016
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
=>\(\frac{1}{a}=-\left(\frac{1}{b}+\frac{1}{c}\right)\)
=>\(\frac{1}{a^2}=-\left(\frac{1}{ab}+\frac{1}{ca}\right)\)
cm tương tự: \(\frac{1}{b^2}=-\left(\frac{1}{ab}+\frac{1}{bc}\right)\)
\(\frac{1}{c^2}=-\left(\frac{1}{ca}+\frac{1}{bc}\right)\)
=> \(N=-\left[bc\left(\frac{1}{ab}+\frac{1}{ca}\right)+ca\left(\frac{1}{ab}+\frac{1}{bc}\right)+ab\left(\frac{1}{ca}+\frac{1}{bc}\right)\right]\)
\(=-\left[\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\right]\)
\(=-\left[\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\right]\) (1)
Ta có : \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
=>\(\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}=0\)
=>\(1+\frac{b+c}{a}+1+\frac{a+c}{b}+1+\frac{a+b}{c}=0\)
=>\(\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=-3\) (2)
Từ (1) và (2) =>N=3
28 tháng 2 2020
Tham khao =))
Câu hỏi của Vu Quang Huy - Toán lớp 8 - Học toán với OnlineMath
\(a+b+c=0\Leftrightarrow a+b=-c\Leftrightarrow\left(a+b\right)^2=\left(-c\right)^2\Leftrightarrow a^2+b^2+2ab=c^2\Leftrightarrow a^2+b^2-c^2=-2ab\)
tương tự ta có: b2+c2-a2=-2bc ; a2+c2-b2=-2ac
Do đó \(P=\frac{1}{-2bc}+\frac{1}{-2ca}+\frac{1}{-2ab}=\frac{a+b+c}{-2abc}=0\)