Cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\). Tính giá trị biểu thức: \(P=\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ca}{b^2}\)
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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
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\\ \)
\(\Rightarrow bc=-ab-ac,ca=-ab-bc,ab=-bc-ca\)
\(\Rightarrow\frac{a^2+bc}{a^2+2bc}=\frac{a^2+bc}{a^2+bc+bc}=\frac{a^2+bc}{a^2+bc-ca-ab}=\frac{a^2+bc}{\left(a-b\right).\left(a-c\right)}\)
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 :\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=-\frac{1}{c^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-3\cdot\frac{1}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=3\cdot\frac{1}{abc}\)
( Do \(\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\) )
Khi đó : \(P=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc\cdot\frac{3}{abc}=3\)
Với \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\)
Chứng minh : với \(a+b+c=0\) thì \(a=-\left(b+c\right)\Leftrightarrow a^3=-\left(b+c\right)^3\)
\(\Leftrightarrow a^3=-\left(b^3+c^3+3b^2c+3bc^2\right)\Leftrightarrow a^3+b^3+c^3=-\left(b^3+c^3+3b^2c+3bc^2\right)+b^3+c^3\)
\(\Leftrightarrow a^3+b^3+c^3=-3bc\left(b+c\right)=-3bc\left(-a\right)=3abc\)vì \(b+c=-a\) =>đpcm
Vì \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\)\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
Vậy \(P=\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ca}{b^2}=abc\left(\frac{1}{c^3}+\frac{1}{a^3}+\frac{1}{b^3}\right)=abc\frac{3}{abc}=3\)
**** mình nha
cho 2014=2013+1 thay vào ta có:\(B=x^{2013}-\left(2013+1\right)x^{2012}+\left(2013+1\right)x^{2011}-...-\left(2013+1\right)x^2+\left(2013+1\right)x-1\)
\(=x^{2013}-\left(x+1\right)x^{2012}+\left(x+1\right)x^{2011}-...-\left(x+1\right)x^2+\left(x+1\right)x-1\)
\(=x^{2013}-x^{2013}-x^{2012}+x^{2012}+x^{2011}-...-x^3-x^2+x^2+x-1\)
\(=x-1=2013-1=2012\)
Từ \(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)
\(\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)
\(\Rightarrow\frac{a}{ab}+\frac{b}{ab}=\frac{b}{bc}+\frac{c}{bc}=\frac{c}{ca}+\frac{a}{ca}\)
\(\Rightarrow\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\)
\(\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\)
\(\Rightarrow M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{a\cdot a+a\cdot a+a\cdot a}{a^2+a^2+a^2}=\frac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)
theo bài ra ta có:
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)
=> \(\frac{abc}{c\left(a+b\right)}=\frac{abc}{a\left(b+c\right)}=\frac{abc}{b\left(c+a\right)}\)
=> \(\frac{abc}{ca+cb}=\frac{abc}{ab+ac}=\frac{abc}{bc+ba}\)
vì a,b,c khác 0 => ca+cb = ab+ac = bc+ba
=> a = b = c
ta có:
\(M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)
vậy M = 1
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Rightarrow\frac{1}{c}=-\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Rightarrow\frac{1}{c^2}=-\left[\frac{1}{c}\left(\frac{1}{a}+\frac{1}{b}\right)\right]=-\left(\frac{1}{bc}+\frac{1}{ca}\right)\)
\(\Rightarrow\frac{ab}{c^2}=-\left(\frac{ab}{bc}+\frac{ab}{ca}\right)=-\left(\frac{a}{c}+\frac{b}{c}\right)\)
Tương tự ,ta được : \(\hept{\begin{cases}\frac{bc}{a^2}=-\left(\frac{bc}{ab}+\frac{bc}{ca}\right)=-\left(\frac{c}{a}+\frac{b}{a}\right)\\\frac{ca}{b^2}=-\left(\frac{ca}{ab}+\frac{ca}{bc}\right)=-\left(\frac{c}{b}+\frac{a}{b}\right)\end{cases}}\)
\(\Rightarrow P=\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ca}{b^2}=-\left(\frac{a}{c}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{b}\right)\)
\(=a\left[-\left(\frac{1}{b}+\frac{1}{c}\right)\right]+b\left[-\left(\frac{1}{c}+\frac{1}{a}\right)\right]+c\left[-\left(\frac{1}{a}+\frac{1}{b}\right)\right]\)
\(=a.\frac{1}{a}+b.\frac{1}{b}+c.\frac{1}{c}\)
\(=1+1+1=3\)
Vậy \(P=3\)
Trả lời:
P=\(\frac{ab}{c^2}\)+\(\frac{bc}{a^2}\)+\(\frac{ac}{b^2}\)
P=\(\frac{abc}{c^3}\)+\(\frac{abc}{a^3}\)+\(\frac{abc}{b^2}\)
P= abc (\(\frac{1}{c^3}\)+\(\frac{1}{a^3}\)+\(\frac{1}{b^3}\)) (1)
Vì (\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\))=0
-> (\(\frac{1}{a}\)+\(\frac{1}{b}\))3=(-\(\frac{1}{c}\))3
-> \(\frac{1}{a^3}\)+\(\frac{1}{b^3}\)+\(\frac{1}{c^3}\)-\(\frac{3}{abc}\)=0
-> \(\frac{1}{a^3}\)+\(\frac{1}{b^3}\)+\(\frac{1}{c^3}\)=\(\frac{3}{abc}\)(2)
Thay vào biểu thức (1), ta có:
P= abc (\(\frac{1}{c^3}\)+\(\frac{1}{a^3}\)+\(\frac{1}{b^3}\)) = abc.\(\frac{3}{abc}\)=3
Đáp số: P=3