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Ta có :
a^2>hoặc=0(vì mang số mũ dương)
Tương tự => b^2 và c ^2 như a^2
mà a^2+b^2+c^2=1=>a=b=c=1
=> a^2016+b^2017+c^2018=1
Mình nghĩ \(a+b+c=1\) nữa chắc oke hơn :3
\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(\Rightarrow1-3abc=1-ab-bc-ca\Rightarrow3abc=ab+bc+ca\)
\(1=\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(=1+2\left(ab+bc+ca\right)\)
\(\Rightarrow ab+bc+ca=0\Rightarrow3abc=0\)
Nếu \(a=0\Rightarrow b+c=1;b^2+c^2=1;b^3+c^3=1\)
\(\Rightarrow b^2+2bc+c^2=1\Rightarrow2bc=0\Rightarrow b=0\left(h\right)c=0\)
Cứ tiếp tục thì sẽ ra nhá :))
a+b+c=1 <=> a+b=1-c
+) Nếu 1-c=0 => a+b=0 <=> a=-b
=> A = a2015+b2015+c2015
A = (-b)2015+b2015+c2015
A = c2015 => A = 1 (Vì 1-c=0) (1)
Ta có: a3+b3+c3=1
a3+b3=1-c3
(a+b)(a2-ab+b20=(1-c)(1+c+c2)
=> (1-c)(a2-ab+b2)=(1-c)(1+c+c2)
=> a2-ab+b2=1+c+c2
(a+b)2-3ab=(1-c)2+3c
=> -3ab=3c <=> -ab=c
Thay -ab = c vào a+b+c=1, ta có:
a+b+(-ab)=1 <=> a+b-ab-1=0 <=> a(1-b)-(1-b)=0 <=> (a-1)(1-b)=0
=> a-1=0 hoặc 1-b = 0 <=> a=1 hoặc b=1
+) Nếu a=1 => b+c=0 <=> b=-c
=> A=a2015+b2015+c2015
=> A=a2015+b2015-b2015
=> A=a2015 => A=1 (2)
+) Nếu b=1 => a+c=0 <=>a=-c
=> A=a2015+b2015+c2015
=> A=a2015+b2015+-a2015
=> A=b2015 => A=1 (3)
Từ (1)(2)(3) => A = 1
Vậy A = 1 với a+b+c=1 và a3+b3+c3=1
b) B = x2-3x+2016
B=x2-3x+2,25+2013,75
B=(x-1,5)2+2013,75
Vì (x-1,5)2 ≥ 0 => (x-1,5)2+2013,75 ≥ 2013,75
=> B ≥ 2013,75
=> GTNN của B bằng 2013,75
Dấu '=' xảy ra khi (x-1,5)2=0 <=> x-1,5=0 <=> x=1,5
Vậy GTNN của B bằng 2013,75 tại x = 1,5
Câu hỏi của Thị Kim Vĩnh Bùi - Toán lớp 8 - Học toán với OnlineMath
Thya các giá trị của a, b, c., d vào M . Tính đc M = 0
+) Ta có : \(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow2\left(ab+bc+ca\right)=-2016\)
\(\Rightarrow\left(ab+bc+ca\right)^2=\left(-2013\right)^2\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=2013^2\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2=2013^2\)( Do \(a+b+c=0\) )
+) Lại có : \(a^2+b^2+c^2=2016\)
\(\Rightarrow\left(a^2+b^2+c^2\right)^2=2016^2\)
\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=2016^2\)
\(\Rightarrow a^4+b^4+c^4=2016^2-2.2013^2=-4040082\)
Hay : \(A=-4040082\)
Vậy \(A=-4040082\) với a,b,c thỏa mãn đề.
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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