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Zyz
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Zyz
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Zyz
22 tháng 5 2016 lúc 17:15

Help me~

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

Pham Phuong Anh
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Lei
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Pham Van Hung
30 tháng 10 2018 lúc 12:34

2. Đặt c + d = x

Ta có: \(a+b+c+d=0\Rightarrow a+b+x=0\Rightarrow a^3+b^3+c^3+d^3=3abx\)

\(\Rightarrow a^3+b^3+c^3+d^3+3cd\left(c+d\right)=3ab\left(c+d\right)\)

\(\Rightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)=3\left(ab-cd\right)\left(c+d\right)\)

Câu 4:

      \(a^{2016}+b^{2016}+c^{2016}=a^{1008}b^{1008}+b^{1008}c^{1008}+c^{1008}+a^{1008}\)

\(\Rightarrow2a^{2016}+2b^{2016}+2c^{2016}-2a^{1008}b^{1008}-2b^{1008}c^{1008}-2c^{1008}a^{1008}=0\)

\(\Rightarrow\left(a^{1008}-b^{1008}\right)^2+\left(b^{1008}-c^{1008}\right)^2+\left(c^{1008}-a^{1008}\right)^2=0\)

\(\Rightarrow a^{1008}=b^{1008},b^{1008}=c^{1008},c^{1008}=a^{1008}\)

\(\Rightarrow a=b,b=c,c=a\) (vì a,b,c > 0 nên \(a\ne-b,b\ne-c,c\ne-a\) )

\(\Rightarrow a-b=0,b-c=0,a-c=0\)

Thay vào A ta tính được A = 0

Tienanh nguyễn
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phuong mai
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Nam
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Nam
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Đỗ Thanh Tùng
5 tháng 7 2016 lúc 14:31

 a+b+c=0 => a^2+b^2+c^2+2ab+2bc+2ca = 0 => a^2+b^2+c^2=0
=> a^2+b^2+c^2 = ab+bc+ca
=> 2a^2+2b^2+2c^2 = 2ab+2bc+2ca
=> (a-b)^2 + (b-c)^2 + (c-a)^2 = 0
=> a=b=c, mà a+b+c=0 => a=b=c=0

thay vào

M=(0-2016)2016+(0-2016)2016-(0-2016)2016=(-2016)2016=20162016

Chúc bạn hoc tốt ùng hộ nha