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\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\\ \)
\(\Rightarrow bc=-ab-ac,ca=-ab-bc,ab=-bc-ca\)
\(\Rightarrow\frac{a^2+bc}{a^2+2bc}=\frac{a^2+bc}{a^2+bc+bc}=\frac{a^2+bc}{a^2+bc-ca-ab}=\frac{a^2+bc}{\left(a-b\right).\left(a-c\right)}\)
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: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
=>\(\frac{1}{a}=-\left(\frac{1}{b}+\frac{1}{c}\right)\)
=>\(\frac{1}{a^2}=-\left(\frac{1}{ab}+\frac{1}{ca}\right)\)
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
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 \(c^2+a^2-b^2=-2ac\)
\(\Rightarrow\frac{ab}{-2ab}+\frac{bc}{-2bc}+\frac{ca}{-2ca}=\frac{1}{-2}+\frac{1}{-2}+\frac{1}{-2}\)
\(=-\frac{3}{2}\)
\(a+b+c=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}\left(a+b\right)^2=c^2\\\left(b+c\right)^2=a^2\\\left(c+a\right)^2=b^2\end{cases}}\)
\(\Leftrightarrow\)\(\hept{\begin{cases}a^2+b^2-c^2=a^2+b^2-\left(a+b\right)^2=-2ab\\b^2+c^2-a^2=b^2+c^2-\left(b+c\right)^2=-2bc\\c^2+a^2-b^2=c^2+a^2-\left(c+a\right)^2=-2ca\end{cases}}\)
Vậy \(B=\frac{ab}{-2ab}+\frac{bc}{-2bc}+\frac{ca}{-2ca}=-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}=-\frac{3}{2}\)
P/s: you tham khảo nha, mk ko biết đúng hay sai
Ta có: \(\frac{ab}{a^2+b^2-c^2}\)
\(=\frac{ab}{a^2+\left(b+c\right)\left(b-c\right)}\)
\(=\frac{ab}{a^2-a\left(b-c\right)}\)
\(=\frac{ab}{a\left(a-b+c\right)}\)
\(=\frac{ab}{-2ab}\)
\(=-\frac{1}{2}\)
Tương tự mà tính
Vì \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\)
Suy ra \(\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b}{c}=\frac{\left(b+c\right)+\left(a+c\right)+\left(a+b\right)}{a+b+c}=2\)
\(\Rightarrow b+c=2a;a+c=2b;a+b=2c\)
Bằng cách rút \(b\) từ đẳng thức thứ nhất thay vào đẳng thức thứ hai ta đễ dàng suy ra được \(a=b=c\)
\(\Rightarrow\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=2+2+2=6\)
cáh khác nè:từ
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)
\(\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}=\frac{a}{ab}+\frac{b}{ab}=\frac{b}{bc}+\frac{c}{bc}=\frac{c}{ca}+\frac{a}{ca}=\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\)\(\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\)
\(\Rightarrow P=\frac{aa+aa+aa}{a^2+a^2+a^2}=1\)
bạn dưới làm sai rồi
P=1 MỚI ĐÚNG
http://diendantoanhoc.net/topic/152549-t%C3%ADnh-fraca2a2-b2-c2-fracb2b2-c2-a2fracc2c2-b2-a2/
Ta có: \(a+b+c=0\)
\(\Rightarrow1\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a^2+b^2=-2ab+c^2\\b^2+c^2=-2bc+a^2\\c^2+a^2=-2ac+b^2\end{cases}}\)
\(\Rightarrow1A=\frac{a^2}{a^2+2bc-a^2}+\frac{b^2}{b^2+2ac-b^2}+\frac{c^2}{c^2+2ab-c^2}\)
\(=\frac{a^3+b^3+c^3}{2abc}=\frac{a^3+b^3+c^3-3abc+3abc}{2abc}\)
\(=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{2abc}\)
\(=\frac{3}{2}\)
A=3
B=3