Cho \(a^2+b^2+c^2=a^3+b^3+c^3=1\) Tính \(S=a^2+b^9+c^{2016}\)
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NT
2
17 tháng 9 2017
từ giả thiết => a;b;c<=1
\(a\le1\\ \Rightarrow a^3\le a^2\)
tt b^3<=b^2;c^3<=c^2
=>a^3+b^3+c^3\(\le\)a^2+b^2+c^2
dấu = xảy ra <=> a=0hoặc a=1 tt với b;c và a^2+b^2+c^2=a^3+b^3+c^3=1
=>S=1
2 tháng 2 2019
a2 + b2 + c2 = a3 + b3 + c3 = 1
\(\Rightarrow\)a2 ( a - 1 ) + b2 ( b - 1 ) + c2 ( c - 1 ) = 0 ( 1 )
a2 + b2 + c2 = 1 ; a2,b2,c2 \(\ge\)0 \(\Rightarrow\)a2,b2,c2 \(\le\)1
\(\Rightarrow\)a \(\le\)1,b \(\le\)1, c \(\le\)1 \(\Rightarrow\)1 - a \(\ge\)0 ; 1-b \(\ge\)0 ; 1 - c \(\ge\)0
\(\Rightarrow\)a2 ( a - 1 ) + b2 ( b - 1 ) + c2 ( c - 1 ) \(\le\)0 ( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow\)a2 ( a - 1 ) = b2 ( b - 1 ) = c2 ( c - 1 ) = 0
\(\Rightarrow\)a = b = 0 ; c = 1 hoặc b = c = 0 ; a = 1 hoặc a = c = 0 ; b = 1
\(\Rightarrow\)S = 1
NA
0
TF
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0
10 tháng 9 2018
Vì \(a^2+b^2+c^2=1\)
\(\Rightarrow-1\le a,b,c\le1\)
\(\Rightarrow a-1\le0;b-1\le0;c-1\le0\)
Lây cai xau trừ cai trươc được
\(\left(a^3+b^3+c^3\right)-\left(a^2+b^2+c^2\right)=0\)
\(\Leftrightarrow a^2\left(a-1\right)+b^2\left(b-1\right)+c^2\left(c-1\right)=0\)
Ta co \(VT\le0\)
Dâu = xảy ra khi: \(\left(a,b,c\right)=\left\{0,0,1;0,1,0;1,0,0\right\}\)
\(\Rightarrow S=1\)
5 tháng 9 2016
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5 tháng 9 2016
Ta có
(m+n+p)^q >= m^q+n^q+p^q
=>a+b+c=1
=>(a+b+c)^2016=1 >= a2016 + b2016 + c2016
Mà a2016 + b2016 + c2016 >=0
=> a2016 + b2016 + c2016=1
Lời giải:
Từ \(a^2+b^2+c^2=1\Rightarrow a^2,b^2,c^2\leq 1\Rightarrow -1\leq a,b,c\leq 1\)
Có \(a^3+b^3+c^3=a^2+b^2+c^2\)
\(\Leftrightarrow a^2(a-1)+b^2(b-1)+c^2(c-1)=0\)
Vì \(a,b,c\leq 1\) nên \(\left\{\begin{matrix} a^2(a-1)\leq 0\\ b^2(b-1)\leq 0\\ c^2(c-1)\leq 0\end{matrix}\right.\Rightarrow a^2(a-1)+b^2(b-1)+c^2(c-1)\leq 0\)
Dấu bằng xảy ra khi \(\left\{\begin{matrix} a^2(a-1)=0\\ b^2(b-1)=0\\ c^2(c-1)=0\end{matrix}\right.\)
Mà \(a^3+b^3+c^3=1\) nên trong \(a,b,c\) có hai số bằng $0$ và một số bằng $1$
Suy ra \(S=a^2+b^9+c^{2016}=1\)