Cho a+b+c=1. Tính E = a^3/((a-b)*(a-c)) + b^3/((b-c)*(b-a)) + c^3/((c-a)*(c-b))
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NQ
2
2 tháng 9 2015
CM a + b + c = 0
=> a + b = -c ; b + c = -a ; c+a a = -b
E = \(\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{c+a}{a}=\frac{\left(-a\right)\left(-b\right)\left(-c\right)}{abc}=1\)
2 tháng 9 2015
Như thế này :
\(a^3+b^3+c^3=3abc\)
=> (a+b)^3 - 3ab(a+b) - 3abc + c^3 = 0
=> ( a+ b +c )^3 - 3(a+b)c(a+b+c) - 3ab(a+b+c) = 0
=> \(\left(a+b+c\right)\left[\left(a+b+c\right)^2-3bc-3ac-3ab\right]=0\)
=> ( a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca ) = 0
=> 1/2 ( a + b + c )(2a^2 + 2b^2 + 2x^2 - 2ab - 2bc - 2 ca ) = 0
=> 1/2 (a+b+c) [ ( a- b)^2 + ( b - c)^2 + (c-a)^2] = 0
Bì ngoặc thứ hai luôn >= 0 => a + b + c = 0
hoặc a = b ; b =c = c=a => a = =b =c
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 7 2021
Đặt vế trái BĐT là P
Ta có:
\(\left(\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b}\right)\left(a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)\right)\ge\left(a^2+b^2+c^2\right)^2\)
\(\Rightarrow P.\left(2ab+2bc+2ca\right)\ge1\)
\(\Rightarrow P\ge\dfrac{1}{2\left(ab+bc+ca\right)}\ge\dfrac{1}{2\left(a^2+b^2+c^2\right)}=\dfrac{1}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
NC
31 tháng 8 2020
Bài làm:
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)
\(\Leftrightarrow a^2b+ab^2+c^2a+ca^2+b^2c+bc^2+2abc=0\)
\(\Leftrightarrow\left(a^2+2ab+b^2\right)c+ab\left(a+b\right)+c^2\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca+c^2\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
=> Hoặc a+b=0 hoặc b+c=0 hoặc c+a=0
=> Hoặc a=-b hoặc b=-c hoặc c=-a
Ko mất tổng quát, g/s a=-b
a) Ta có: vì a=-b thay vào ta được:
\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-\frac{1}{b^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{c^3}\)
\(\frac{1}{a^3+b^3+c^3}=\frac{1}{-b^3+b^3+c^3}=\frac{1}{c^3}\)
=> đpcm
b) Ta có: \(a+b+c=1\Leftrightarrow-b+b+c=1\Rightarrow c=1\)
=> \(P=-\frac{1}{b^{2021}}+\frac{1}{b^{2021}}+\frac{1}{c^{2021}}=\frac{1}{1^{2021}}=1\)
11 tháng 2 2021
Ta có: \(a^3+b^3+c^3=3abc\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\Rightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)\cdot c+c^2\right]-3ab\left(a+b+c\right)=0\)\(\Rightarrow\left(a+b+c\right)\left[a^2+b^2+2ab-ac-bc+c^2\right]-3ab\left(a+b+c\right)=0\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\left(1\right)\\a^2+b^2+c^2-ab-bc-ca=0\left(2\right)\end{matrix}\right.\)
Từ (1) \(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\) \(\Rightarrow P=\left(\dfrac{a+b}{b}\right)\cdot\left(\dfrac{b+c}{c}\right)\cdot\left(\dfrac{c+a}{a}\right)=\left(\dfrac{-c}{b}\right)\cdot\left(-\dfrac{a}{c}\right)\cdot\left(-\dfrac{b}{a}\right)=-1\)
Từ (2) \(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\Rightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=0\) \(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}a=b\\c=b\\a=c\end{matrix}\right.\) \(\Rightarrow a=b=c\) \(\Rightarrow P=\left(\dfrac{a+b}{b}\right)\cdot\left(\dfrac{b+c}{c}\right)\cdot\left(\dfrac{c+a}{a}\right)=\dfrac{2b}{b}\cdot\dfrac{2c}{c}\cdot\dfrac{2a}{a}=8\)
Vậy...
12 tháng 2 2021
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\cdot\left(2a^2+2b^2+2c^2-2ab-2ac-2bc\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\2a^2+2b^2+2c^2-2ab-2ac-2bc=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a^2-2ab+b^2+a^2-2ac+c^2+b^2-2bc+c^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\end{matrix}\right.\)
Ta có: \(B=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(\Leftrightarrow B=\dfrac{a+b}{b}\cdot\dfrac{b+c}{c}\cdot\dfrac{a+c}{a}\)
Trường hợp 1: a+b+c=0
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)
Thay a+b=-c; b+c=-a và c+a=-b vào biểu thức \(B=\dfrac{a+b}{b}\cdot\dfrac{b+c}{c}\cdot\dfrac{a+c}{a}\), ta được:
\(B=\dfrac{-c}{b}\cdot\dfrac{-a}{c}\cdot\dfrac{-b}{a}=\dfrac{-\left(a\cdot b\cdot c\right)}{abc}=-1\)
Trường hợp 2: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Leftrightarrow a=b=c\)
Ta có: \(B=\dfrac{a+b}{b}\cdot\dfrac{b+c}{c}\cdot\dfrac{a+c}{a}\)
mà a=b=c(cmt)
nên \(B=\dfrac{b+b}{b}\cdot\dfrac{c+c}{c}\cdot\dfrac{a+a}{a}=\dfrac{2b}{b}\cdot\dfrac{2c}{c}\cdot\dfrac{2a}{a}=2\cdot2\cdot2=8\)
QP
0
Ta có \(E=\frac{a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{\left(a+b+c\right)\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=a+b+c=1\)
Ở đây chú ý rằng \(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)=-\left(a+b+c\right)\left(a-b\right)\left(b-c\right)\left(c-a\right)\)