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ta có: a+b+c=0
\(\left(a+b\right)^2=\left(-c\right)^2\Rightarrow a^2+2ab+b^2=c^2\Rightarrow a^2+b^2-c^2=-2ab\\ \)
tương tự \(\left(b+c\right)^2=\left(-a\right)^2\Rightarrow b^2+2bc+c^2=a^2\Rightarrow b^2+c^2-a^2=-2bc\\ \)
\(\left(a+c\right)^2=\left(-b\right)^2\Rightarrow a^2+2ac+c^2=b^2\Rightarrow a^2+c^2-b^2=-2ac\\ \)
thế vào biểu thức \(B=\frac{ab}{a^2+b^2-c^2}+\frac{bc}{b^2+c^2-a^2}+\frac{ac}{a^2+c^2-b^2}\)
\(B=\frac{ab}{-2ab}+\frac{bc}{-2bc}+\frac{ac}{-2ac}=\frac{-1}{2}+\frac{-1}{2}+\frac{-1}{2}=\frac{-3}{2}\)
ta có a+b+c=0 => a+b= - c => a^2+2ab+b^2=c^2 => a^2+b^2=c^2-2ab
=>b+c= - a => b^2+2bc+c^2=a^2 => b^2+c^2=a^2-2bc
=>a+c= - b => a^2+2ac+c^2=b^2 =>a^2+c^2=b^2-2ac
từ đó ta có: B = ab / (a^2+b^2-c^2) + bc/ (b^2+c^2-a^2) + ca/ (c^2+a^2-b^2)
=> ab/(c^2 -2ab+c^2)+bc/ (a^2 -2bc+a^2) + ca/ (b^2 -2ac+b^2)
=> B= -ab/2ab - bc/2bc - ca/2ca
=> B = -3/2
Ta có : \(a+b+c=0\Leftrightarrow a+b=-c\Leftrightarrow\left(a+b\right)^2=c^2\Leftrightarrow a^2+b^2+2ab=c^2\)
\(\Leftrightarrow a^2+b^2-c^2=-2ab\Rightarrow\frac{ab}{a^2+b^2-c^2}=-\frac{1}{2}\)
Tương tự : \(\frac{bc}{b^2+c^2-a^2}=-\frac{1}{2};\frac{ac}{a^2+c^2-b^2}=-\frac{1}{2}\)
Cộng các vế với nhau được \(M=-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}=-\frac{3}{2}\)
Bạn ơi hình như phân thức cuối cùng bạn bị sai bạn thử xem lại đi nha!
Ta có :a+b+c=0
a+b=-c
(a+b)2=(-c)2
a2+b2+2ab=c2
a2+b2-c2+2ab=0
\(\Rightarrow\)a2+b2-c2=-2ab (1)
Tương tự như trên , nên ta có :
b2+c2-a2=-2ab (2)
c2+b2-a2=-2cb (3)
Ta thay (1) , (2) và (3) , vào phân thức trên , ta có :
\(\frac{ab}{-2ab}+\frac{bc}{-2bc}+\frac{ca}{-2cb}\)
\(=-\frac{1}{2}+-\frac{1}{2}+-\frac{1}{2}\)
\(=-\frac{3}{2}\)
Ta xét hiệu :
\(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+a^2}-\left(\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ac+a^2}\right)\)
\(=a-b+b-c+c-a=0\)
Do đó : \(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+a^2}=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ac+a^2}=1006\)
Khi đó \(M=2\cdot1006=2012\)
Từ \(a+b+c=0\)\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
Ta có: \(B=\frac{ab}{a^2+b^2-c^2}+\frac{bc}{b^2+c^2-a^2}+\frac{ca}{c^2+a^2-b^2}\)
\(=\frac{ab}{a^2+\left(b-c\right)\left(b+c\right)}+\frac{bc}{b^2+\left(c-a\right)\left(c+a\right)}+\frac{ca}{c^2+\left(a-b\right)\left(a+b\right)}\)
\(=\frac{ab}{a^2-a\left(b-c\right)}+\frac{bc}{b^2-b\left(c-a\right)}+\frac{ca}{c^2-c\left(a-b\right)}\)
\(=\frac{ab}{a\left(a-b+c\right)}+\frac{bc}{b\left(b-c+a\right)}+\frac{ca}{c\left(c-a+b\right)}\)
\(=\frac{b}{a-b+c}+\frac{c}{b-c+a}+\frac{a}{c-a+b}\)
\(=\frac{b}{a+c-b}+\frac{c}{b+a-c}+\frac{a}{c+b-a}\)
\(=\frac{b}{-2b}+\frac{c}{-2c}+\frac{a}{-2a}=\left(-\frac{1}{2}\right)+\left(-\frac{1}{2}\right)+\left(-\frac{1}{2}\right)=-\frac{3}{2}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\\ \)
\(\Rightarrow bc=-ab-ac,ca=-ab-bc,ab=-bc-ca\)
\(\Rightarrow\frac{a^2+bc}{a^2+2bc}=\frac{a^2+bc}{a^2+bc+bc}=\frac{a^2+bc}{a^2+bc-ca-ab}=\frac{a^2+bc}{\left(a-b\right).\left(a-c\right)}\)
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.
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: \(\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}+\frac{1}{2ab^2+1}\ge1\)
Ta có:
ab + bc + ac = 0
=> \(\frac{ab+bc+ac}{abc}=0\)
=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
Em làm tiếp theo link: Câu hỏi của Conan Kudo - Toán lớp 8 - Học toán với OnlineMath
Ta có :
\(ab+bc+ca=0\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
Quay lại bài toán ta có :
\(B=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=\frac{3abc}{abc}=3\)
Chúc bạn học tốt !!!
Ta có: \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\)
\(\Rightarrow\frac{bcx+acy+abz}{abc}=0\)
\(\Rightarrow bcx+acy+abz=0\)
Lại có:\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2\)
\(\Rightarrow\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}+2.\frac{bcx+acy+abz}{xyz}=4\)(bình phương hai vế)
\(\Rightarrow\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}=4\)(Vì \(bcx+acy+abz=0\))
Từ (1) \(\Rightarrow bcx+acy+abz=0\)
Gọi \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2\left(2\right)\)
Từ (2) \(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{ab}{xy}+\frac{ac}{xz}+\frac{bc}{yz}\right)=0\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=4-\left(\frac{abz+acy+bcx}{xyz}\right)\)
\(=4\)
\(b,\frac{ab}{a^2+b^2+c^2}+\frac{bc}{b^2+c^2-a^2}+\frac{ca}{c^2+a^2-b^2}\)
Từ \(a+b+c=0\Rightarrow a+b=-c\Rightarrow a^2+b^2-c^2=-2ab\)
Tương tự \(b^2+c^2-a^2=-2bc\)và \(c^2+a^2-b^2=-2ac\)
\(\Rightarrow\frac{ab}{-2ab}+\frac{bc}{-2bc}+\frac{ca}{-2ca}=\frac{1}{-2}+\frac{1}{-2}+\frac{1}{-2}\)
\(=-\frac{3}{2}\)