Cho \(a^3+b^3+c^3=3abc\) với \(a+b+c\ne0;a,b,c\ne0\)
Tình giá trị biểu thức \(P=\left(2017+\frac{a}{b}\right)\left(2017+\frac{b}{c}\right)\left(2017+\frac{c}{a}\right)\)
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D
2
11 tháng 12 2023
\(a^3+b^3+c^3=3bac\)
=>\(\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
=>\(\left[\left(a+b\right)^3+c^3\right]-3ba\left(a+b+c\right)=0\)
=>\(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
=>\(a^2+b^2+c^2-ab-ac-bc=0\)
=>\(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
=>\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
=>a=b=c
T
11 tháng 12 2023
\(a^3+b^3+c^3=3abc\\\Rightarrow a^3+b^3+c^3-3abc=0\\\Rightarrow(a+b)^3+c^3-3ab(a+b)-3abc=0\\\Rightarrow (a+b+c)^3-3(a+b)c(a+b+c)-3ab(a+b+c)=0\\\Rightarrow(a+b+c)[(a+b+c)^2-3(a+b)c-3ab]=0\\\Rightarrow(a+b+c)(a^2+b^2+c^2+2ab+2bc+2ca-3ac-3bc-3ab)=0\\\Rightarrow(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0\\\Rightarrow a^2+b^2+c^2-ab-bc-ca=0(vì.a+b+c\ne0)\\\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\\\Rightarrow(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)=0\\\Rightarrow(a-b)^2+(b-c)^2+(c-a)^2=0\)
Ta thấy: \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\forall a,b\\\left(b-c\right)^2\ge0\forall b,c\\\left(c-a\right)^2\ge0\forall a,c\end{matrix}\right.\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)
Mà: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
nên: \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Rightarrow a=b=c\)
Vậy: ...
\(---\)
Các HĐT được sử dụng trong bài:
\(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca\)
$\text{#}Toru$
D
2
KB
5 tháng 12 2017
Ta có:\(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2-3abc=0\)
\(\Rightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right).c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Rightarrow\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}a+b+c=0\left(loai\right)\\a=b=c\end{cases}}\)
\(\Rightarrow P=2007.2007.2007=2007^3\)
NT
10 tháng 2 2018
\(a^3+b^3+c^3=3abc\\ \Rightarrow a^3+b^3+c^3-3abc=0\\ \Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\\ \Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{matrix}\right.\)
\(\Rightarrow a^2+b^2+c^2=ab+bc+ac\left(a+b+c\ne0\right)\\ \Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\\ \Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\\ \Rightarrow a=b=c\\ \Rightarrow B=\dfrac{2}{a}.\dfrac{2}{b}.\dfrac{2}{c}=\dfrac{8}{abc}\)
12 tháng 4 2018
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)(tự nhân lại rồi phân tích)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{matrix}\right.\)
+)Xét a+b+c=0\(\Rightarrow P=\dfrac{b+a}{b}\cdot\dfrac{c+b}{c}\cdot\dfrac{a+c}{a}=\dfrac{-c}{b}\cdot\dfrac{-a}{c}\cdot\dfrac{-b}{a}=-1\)
+Xét \(a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow\dfrac{1}{2}\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\right]=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow a=b=c\)
\(\Rightarrow P=2\cdot2\cdot2=8\)
9 tháng 4 2017
Giải:
Từ \(a^3+b^3+c^3=3abc\Leftrightarrow\)\(\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
Ta xét các trường hợp:
Trường hợp \(1\): Nếu \(a+b+c=0\) thì:
\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)
Thay vào \(P\) ta có:
\(P=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(=\left(\dfrac{a+b}{b}\right)\left(\dfrac{b+c}{c}\right)\left(\dfrac{a+c}{c}\right)\)
\(=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=\dfrac{\cdot\left(-c\right).\left(-a\right).\left(-b\right)}{b.c.a}=-1\)
Trường hợp \(2\): Nếu \(a=b=c\) thì:
\(P=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(=\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)\)
\(=\left(1+1\right)\left(1+1\right)\left(1+1\right)\)
\(=2.2.2=8\)
Vậy \(P=-1\) hoặc \(P=8\)
8 tháng 4 2017
ta có : a3+b3+c3-3abc=0
\(\Rightarrow\)(a+b)3+c3-3abc-3a2b-3ab2=0
\(\Rightarrow\)(a+b+c)(a2+b2+c2+2ab-ac-bc)-3ab(a+b+c)=0
\(\Rightarrow\)(a+b+c)(a2+b2+c2-ab-ac-bc)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{matrix}\right.\\\left(a+b+c\right)^2+a^2+b^2+c^2=0\Leftrightarrow a=b=c=0\left(bỏ\right)\end{matrix}\right.\)ta có P=(1+\(\dfrac{a}{b}\))(1+\(\dfrac{b}{c}\))(1+\(\dfrac{c}{a}\))
\(\Leftrightarrow\)p=\(\left(\dfrac{b+a}{b}\right)\left(\dfrac{c+b}{c}\right)\left(\dfrac{a+c}{a}\right)\)
\(\Leftrightarrow P=\left(\dfrac{-c}{b}\right)\left(\dfrac{-a}{c}\right)\left(\dfrac{-b}{a}\right)\)
\(\Leftrightarrow\)P=-1
31 tháng 3 2018
\(\left\{{}\begin{matrix}a^3+b^3+c^3=3abc\\a+b+c\ne0\end{matrix}\right.\) \(\Leftrightarrow a^2+b^2+c^2-\left(ab+bc+ca\right)=0\)
\(P=\dfrac{\left(a^2+b^2+c^2\right)^3}{\left(a+b+c\right)^6}=\left[\dfrac{a^2+b^2+c^2}{\left(a+b+c\right)^2}\right]^3=\left[\dfrac{\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}\right]^3=\dfrac{1}{27}\)
5 tháng 11 2016
Ta có : \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-c\right)\)
Do : \(a^3+b^3+c^3=3abc\) và \(a+b+c\ne0\) nên \(a^2+b^2+c^2-ab-bc-ac=0\)
Dễ dàng suy ra \(a=b=c\).Vậy \(N=\frac{3a^2}{\left(3a\right)^2}=\frac{1}{3}.\)
10 tháng 2 2018
Mẫu của N phải là (a+b+c)^2013 chứ bạn
Đk để phân số tồn tại là : a+b+c khác 0
a^3+b^3+c^3=abc
<=> a^3+b^3+c^3-3abc = 0
<=> (a+b+c).(a^2+b^2+c^2-ab-bc-ca) = 0
<=> a^2+b^2+c^2-ab-bc-ca = 0 ( vì a+b+c khác 0 )
<=> 2a^2+2b^2+2c^2-2ab-2bc-2ca = 0
<=> (a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2) = 0
<=> (a-b)^2+(b-c)^2+(c-a)^2 = 0
<=> a-b=0 ; b-c=0 ; c-a=0
<=> a=b=c
Khi đó : N = 3a^2013/(3a)^2013 = 3/3^2013 = 1/3^2012
Tk mk nha
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\) (vì \(a+b+c\ne0\))
<=> \(2a^2+2b^2+2c^2-2ab-2bc-2ac=0\) (nhân cả hai về với hai)
<=> \(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)
<=> \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
<=> a - b = b - c = c - a = 0 (vì 3 cái đấy đều lớn hơn hoặc bằng 0)
<=> a = b = c
Nên : P = \(\left(2017+\frac{a}{b}\right)\left(2017+\frac{b}{c}\right)\left(2017+\frac{c}{a}\right)=\left(2017+\frac{a}{a}\right)\left(2017+\frac{a}{a}\right)+\left(2017+\frac{a}{a}\right)\)
\(=\left(2017+1\right)\left(2017+1\right)\left(2017+1\right)=2018.2018.2018=2018^3\)