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Những câu hỏi liên quan
Tienanh nguyễn
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Nguyễn Tất Đạt
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khánhchitt3003
17 tháng 9 2017 lúc 20:33

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

Thanh Tùng DZ
2 tháng 2 2019 lúc 10:53

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 \(\ge\)\(\Rightarrow\)a2,b2,c2 \(\le\)1

\(\Rightarrow\)\(\le\)1,b \(\le\)1, c \(\le\)\(\Rightarrow\)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

Nguyen Anh Duc
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Toán full
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Tran VAN VY
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Tran VAN VY
9 tháng 3 2016 lúc 18:41

chia hết cho n+1 nha các bạn

Nguyễn Trí Hùng
30 tháng 12 2021 lúc 18:57

? nghĩa là    sao

Khách vãng lai đã xóa
Nguyễn Minh Phương
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Phùng Thị Hồng Vân
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alibaba nguyễn
10 tháng 9 2018 lúc 14:13

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: \(\left(a,b,c\right)=\left\{0,0,1;0,1,0;1,0,0\right\}\)

\(\Rightarrow S=1\) 

Nguyễn Bảo Long
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Hoang Chau Anh
5 tháng 9 2016 lúc 21:44

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Nguyễn Mạnh Tuấn
5 tháng 9 2016 lúc 21:44

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

Lê Thư
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