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Ta có: \(A=a\left(a^2-bc\right)+b\left(b^2-ac\right)+c\left(c^2-ab\right)=0\)
\(\Rightarrow A=a^3+b^3+c^3-3abc=0\) \(\Rightarrow A=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Rightarrow A=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow A=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
Vì \(a+b+c\ne0\Rightarrow a^2+b^2+c^2-ab-ac-bc=0\)
Xét \(M=a^2+b^2+c^2-ab-ac-bc=0\)
\(\Rightarrow2M=2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Rightarrow2M=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Vì \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(c-a\right)^2\ge0\forall a,b,c\)
\(\Rightarrow a-b=0;b-c=0;c-a=0\) \(\Rightarrow a=b=c\)
\(\Rightarrow P=\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}=1+1+1=3\)
Lời giải:
Ta có: \(B=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}\)
\(B=\frac{(bc)^3+(ca)^3+(ab)^3}{a^2b^2c^2}\)
Vì \(ab+bc+ac=0\Rightarrow bc+ac=-ab\). Do đó:
\((bc)^3+(ca)^3+(ab)^3=(bc+ca)^3-3bc.ca(bc+ca)+(ab)^3\)
\(=(-ab)^3-3bc.ca(-ab)+(ab)^3\)
\(=3bc.ca.ab=3a^2b^2c^2\)
Suy ra : \(B=\frac{3a^2b^2c^2}{a^2b^2c^2}=3\)
Bài 2:
a) \(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\)
\(A=\dfrac{a^3}{abc}+\dfrac{b^3}{abc}+\dfrac{c^3}{abc}\)
\(A=\dfrac{1}{abc}\left(a^3+b^3+c^3\right)\)
\(A=\dfrac{1}{abc}\left[\left(a+b\right)^3-3ab\left(a+b\right)+c^3\right]\)
Vì \(a+b+c=0\)
Nên a + b = -c (1)
Thay (1) vào A, ta được:
\(A=\dfrac{1}{abc}\left[\left(-c\right)^3-3ab\left(-c\right)+c^3\right]\)
\(A=\dfrac{1}{abc}.3abc\)
\(A=3\)
b) \(B=\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-a^2-b^2}\)
\(B=\dfrac{a^2}{a^2-\left(b^2+c^2\right)}+\dfrac{b^2}{b^2-\left(c^2+a^2\right)}+\dfrac{c^2}{c^2-\left(a^2+b^2\right)}\)
Vì \(a+b+c=0\)
Nên b + c = -a
=> ( b + c )2 = (-a)2
=> b2 + c2 + 2bc = a2
=> b2 + c2 = a2 - 2bc (1)
Tương tự ta có: c2 + a2 = b2 - 2ac (2)
a2 + b2 = c - 2ab (3)
Thay (1), (2) và (3) vào B, ta được:
\(B=\dfrac{a^2}{a^2-\left(a^2-2bc\right)}+\dfrac{b^2}{b^2-\left(b^2-2ac\right)}+\dfrac{c^2}{c^2-\left(c^2-2ab\right)}\)
\(B=\dfrac{a^2}{a^2-a^2+2bc}+\dfrac{b^2}{b^2-b^2+2ac}+\dfrac{c^2}{c^2-c^2+2ab}\)
\(B=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}\)
\(B=\dfrac{a^3}{2abc}+\dfrac{b^3}{2abc}+\dfrac{c^3}{2abc}\)
\(B=\dfrac{1}{2abc}\left(a^3+b^3+c^3\right)\)
Mà \(a^3+b^3+c^3=3abc\) ( câu a )
\(\Rightarrow B=\dfrac{1}{2abc}.3abc\)
\(\Rightarrow B=\dfrac{3}{2}\)
Bài 1:
a) GT: abc = 2
\(M=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\)
\(M=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{abc+2cb+2b}\)
\(M=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{2+2cb+2b}\)
\(M=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{2\left(1+cb+b\right)}\)
\(M=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{bc+b+1}\)
\(M=\dfrac{1+b+bc}{bc+b+1}\)
\(M=1\)
b) GT: abc = 1
\(N=\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)
\(N=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{cb}{b\left(ac+c+1\right)}\)
\(N=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{bc+b+1}+\dfrac{bc}{abc+bc+b}\)
\(N=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{bc+b+1}\)
\(N=\dfrac{1+b+bc}{bc+b+1}\)
\(N=1\)
a)\(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\)
\(A=\dfrac{a^3}{abc}+\dfrac{b^3}{abc}+\dfrac{c^3}{abc}\)
\(A=\dfrac{a^3+b^3+c^3}{abc}\)
\(A=\dfrac{3abc}{abc}=3\)(vì a+b+c=0)
b)Ta có: a+b+c=0
\(\Rightarrow\left\{{}\begin{matrix}a=-b-c\\b=-c-a\\c=-a-b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a^2=\left(b+c\right)^2\\b^2=\left(c+a\right)^2\\c^2=\left(a+b\right)^2\end{matrix}\right.\)
\(\Rightarrow B=\dfrac{a^2}{\left(b+c\right)^2-b^2-c^2}+\dfrac{b^2}{\left(a+c\right)^2-c^2-a^2}+\dfrac{c^2}{\left(a+b\right)^2-a^2-b^2}\)
\(\Rightarrow B=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2ab}\)
\(\Rightarrow B=\dfrac{a^3+b^3+c^3}{2abc}\)
\(\Rightarrow B=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)(vì a+b+c=0)
cm:nếu a+b+c=0 thì a^3+b^3+c^3=3abc
a^3+b^3+c^3=3abc
=>a^3+b^3+c^3-3abc=0
=>(a+b)^3-3ab(a+b)+c^3-3abc=0
=>[(a+b)^3+c^3]-3ab(a+b+c)=0
=>(a+b+c)[(a+b)^2-(a+b)c+c^2] -3ab(a+b+c)=0
=>(a+b+c)[(a+b)^2-(a+b)c+c^2-3ab]=0
vì a+b+c=0 nên a^3+b^3+c^3=3abc
thay kết quả vừa chúng minh vào đề bài ta đc
\(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}=\dfrac{a^3+b^3+c^3}{abc}=\dfrac{3abc}{abc}=3\)
chúc bạn học tốt ^ ^
a/d vào công thức a^3+b^3+b^3=3abc( khi a+b+c=0)
ta đc 1/a+1/b+1/c=0
=> (1/a)^3+(1/b)^3+(1/c)^3=3. (1/abc)
lại có S=\(\dfrac{bc}{a^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}=\dfrac{abc}{a^3}+\dfrac{abc}{b^3}+\dfrac{abc}{c^3}\)
=abc (\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\))
=3.\(\dfrac{abc}{abc}\)=1
chúc bạn học tốt ^ ^
Dễ CM : nếu x+y+z=0 thì x^3+y^3+z^3=3xyz
\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
\(S=\dfrac{bc}{a^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}=\dfrac{abc}{a^3}+\dfrac{abc}{b^3}+\dfrac{abc}{c^3}=abc\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)\\ =abc.\dfrac{1}{abc}=1\)
cho mình hỏi bạn biết làm chưa nếu rồi thì giúp mình được không ạ mình ko biết làm
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\\ \)
\(\Rightarrow bc=-ab-ac,ca=-ab-bc,ab=-bc-ca\)
\(\Rightarrow\frac{a^2+bc}{a^2+2bc}=\frac{a^2+bc}{a^2+bc+bc}=\frac{a^2+bc}{a^2+bc-ca-ab}=\frac{a^2+bc}{\left(a-b\right).\left(a-c\right)}\)
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: \(\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}+\frac{1}{2ab^2+1}\ge1\)
\(A=\dfrac{bc}{8a^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}\)
\(=\dfrac{\left(bc\right)^3+8\left(ca\right)^3+8\left(ab\right)^3}{8\left(abc\right)^2}\)
\(=\dfrac{\left(bc\right)^3+\left(2ca\right)^3+\left(2ab\right)^3}{8\left(abc\right)^2}\)
\(=\dfrac{\left(bc\right)^3+\left(2ab+2ca\right)^3-3.2ca.2ab\left(2ab+2ca\right)}{8\left(abc\right)^2}\)
\(=\dfrac{\left(bc\right)^3+\left(-bc\right)^3-3.2ca.2ab.\left(-bc\right)}{8\left(abc\right)^2}\)
\(=\dfrac{12\left(abc\right)^2}{8\left(abc\right)^2}=\dfrac{12}{8}\)
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