Tính A biết A= 1008^2(1008^3 +1)-1008(1008^4-2)-1008^2
a^2016+b^2016+c^2016=a^1008. b^1008+b^1008. c^1008+c^1008. a^1008.
Tính A=(a-b)^3+(b-c)^4+(c-a)^2015
9.2^x.(2^1005-2^1002+.....+2^3-1)=2^1008-1 *
A=2^1005-2^1002+...+2^3-1 2^3 .
A=2^2.(2^1005-2^1002+......+2^3-1) 8.
A=2^1008-2^1005+.....+2^6-2^3
A=2^1005-2^1002+.....+2^3-1 8.A+A=2^1008-1 9.
A=2^1008-1
Thay 9.a=2^1008-1 vào * ta có: 2^x.(2^1008-1)=2^1008-1
2^x=(2^1008-1):(2^1008-1)
2^x=1
2^x=2^0
x=0
vậy x=0
Lập đề bài cho bài toán này
Tính bằng cách thuận tiện:
a, 1008 + 1008 + 1008 + 1008 + 1008 - 1008 × 5
b, 66 - 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6
a, 1008 + 1008 + 1008 + 1008 + 1008 - 1008 × 5
=1008x5-1008x5
=1008x(5-5)
=1008x0
=0
b,66-6+6+6+6+6+6+6+6+6+6
=66-6x10
=66-60
=6
a. = 1008 x 5 - 1008 x 5 = 0
b.= 66 - ( 6 x 10)
= 66 - 60
= 6
cho bx^2=ay^2 va x^2+y^2=1 cm x^2016/a^1008+y^2016/b^1008=2/(a+b)^1008
Cho \(a,b,x,y\) là các số thực thỏa mãn: \(x^2+y^2=1\) và \(\dfrac{x^4}{a}+\dfrac{y^4}{b}=\dfrac{1}{a+b}\) Chứng minh rằng: \(\dfrac{x^{2016}}{a^{1008}}+\dfrac{y^{2016}}{b^{1008}}=\dfrac{2}{\left(a+b\right)^{1008}}\)
Cho:
A= 1/1*2+1/3*4+1/5*6+...+1/2013*2014
B=1/1008*2014+1/1009*2013+...+1/2014*1008
Tính A/B
\(Cho bx^2=ay^2\) và \(x^2+y^2=1.CMRa,\dfrac{x^{2016}}{a^{1008}}+\dfrac{y^{2016}}{b^{1008}}=\dfrac{2}{\left(a+b\right)^{1008}} b, bx^2=ay^2\)
Cho a,b,c thỏa mãn:
\(a^{2016}+b^{2016}+c^{2016}=a^{1008}b^{1008}+b^{1008}c^{1008}+c^{1008}a^{1008}\); a,b,c > 0
Tính biểu thức \(A=\left(a-b\right)^{15}+\left(b-c\right)^{2015}\left(a-c\right)^{2016}\)
\(a^{2016}+b^{2016}+c^{2016}=a^{1008}b^{1008}+b^{1008}c^{1008}+c^{1008}a^{1008}\)
\(\Rightarrow2a^{2016}+2b^{2016}+2c^{2016}=2a^{1008}b^{1008}+2b^{1008}c^{1008}+2c^{1008}a^{1008}\)
\(\Rightarrow\left(a^{2016}-2a^{1008}b^{1008}+b^{1008}\right)+\left(b^{2016}-2b^{1008}c^{1008}+c^{1008}\right)\)\(+\left(c^{2016}-2c^{1008}a^{1008}+a^{2016}\right)=0\)
\(\Rightarrow\left(a^{1008}-b^{1008}\right)^2+\left(b^{1008}-c^{1008}\right)^2+\left(c^{1008}-a^{1008}\right)=0\)
Vì \(\hept{\begin{cases}\left(a^{1008}-b^{1008}\right)^2\ge0\\\left(b^{1008}-c^{1008}\right)^2\ge0\\\left(c^{1008}-a^{1008}\right)^2\ge0\end{cases}}\)
\(\Rightarrow\left(a^{1008}-b^{1008}\right)^2+\left(b^{1008}-c^{1008}\right)^2+\left(c^{1008}-a^{1008}\right)^2\ge0\)
Dấu " = " xảy ra: \(\Leftrightarrow\hept{\begin{cases}a^{1008}-b^{1008}=0\\b^{1008}-c^{1008}=0\\c^{1008}-a^{1008}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}a^{1008}=b^{1008}\\b^{1008}=c^{1008}\\c^{1008}=a^{1008}\end{cases}\Leftrightarrow}a=b=c\)
Thay a=b=c vào A ta có: \(A=\left(a-a\right)^{15}+\left(a-a\right)^{2015}+\left(a-a\right)^{2016}=0\)
Cho a,b,c thoả:
a2016+b2016+c2016=a1008.b1008+b1008.c1008+a1008.c1008
Tính (a-b)2013+(b-c)2014+(c-a)2015