Cho a^3+b^3+c^3=3abc. Tính A=(1+a/b)*(1+b/c)*(1+c/a)?
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AA
0
BC
2
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 \(B=\dfrac{b+b}{b}\cdot\dfrac{c+c}{c}\cdot\dfrac{a+a}{a}=\dfrac{2b}{b}\cdot\dfrac{2c}{c}\cdot\dfrac{2a}{a}=2\cdot2\cdot2=8\)
VT
1
NM
Nguyễn Minh Quang
Giáo viên
14 tháng 8 2021
Áp dụng bất đẳng thức Cauchy ta có :
\(a^3+b^3+c^3=\ge3abc\)
Dấu = xảy ra khi a=b=c
vậy ta có \(A=\left(\frac{a}{a}+1\right)\left(\frac{a}{a}+1\right)\left(\frac{a}{a}+1\right)=8\)
P
3
IA
24 tháng 6 2021
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\2a^2+2b^2+2c^2-2ab-2bc-2ca=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)
bạn thay vào M giải tiếp nha
24 tháng 6 2021
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a^3+b^3\right)+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\left[\left(a+b\right)^3+c^3\right]-\left[3ab\left(a+b\right)+3abc\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
Nếu \(a^2+b^2+c^2-ab-bc-ca\)
\(=\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\left(\forall a,b,c\right)\)
Dấu "=" xảy ra khi: a = b = c
Khi đó: \(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(1+1\right)^3=8\)
Nếu \(a+b+c=0\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
\(\Rightarrow M=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{-abc}{abc}=-1\)
11 tháng 2 2018
Ta có : \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-cb\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-cb-3ab\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-cb\right)=0\)
Vi a,b,c khác 0 Nên : \(a^2+b^2+c^2-ab-bc-ac=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\)
<=> a = b = c
Vậy \(N=\frac{a^{2016}+b^{2016}+c^{2016}}{\left(a+b+c\right)^{2016}}=\frac{a^{2016}+a^{2016}+a^{2016}}{\left(a+a+a\right)^{2016}}=\frac{3.a^{2016}}{3^{2016}.a^{2016}}=\frac{1}{3^{2015}}\)
SS
1
17 tháng 11 2016
\(a^3+b^3+c^3=3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2+c^2-\left(a+b\right)c\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left[a^2+b^2+2ab+c^2-ac-bc-3ab\right]=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)=0.2\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
TH1 : \(a+b+c=0\)
\(\Rightarrow A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
\(=\frac{a+b}{b}.\frac{b+c}{c}.\frac{c+a}{a}\)
\(=\frac{\left(-c\right)}{b}.\frac{\left(-a\right)}{c}.\frac{\left(-b\right)}{a}=-1\)
TH2 : \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Mà \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\)
\(\Rightarrow a-b=b-c=c-a=0\)
\(\Rightarrow a=b=c\)
\(\Rightarrow A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
Vậy ...
LV
0
TH
1
DD
Đoàn Đức Hà
Giáo viên
17 tháng 7 2021
\(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)^3-3c\left(a+b\right)\left(a+b+c\right)-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)
Nếu \(a=b=c\): \(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=2.2.2=8\)
Nếu \(a+b+c=0\):
\(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
\(=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
\(=\frac{\left(-c\right)\left(-a\right)\left(-b\right)}{abc}=-1\)
uầy
so amazing
ai lm đc là legendary đó
viết cái đề nhìn zô đã ko mún làm r