\(\left(x-2\right)^{2016}>=0\forall x\)
\(\left|y^2-9\right|^{2017}>=0\forall y\)
Do đó: \(\left(x-2\right)^{2016}+\left|y^2-9\right|^{2017}>=0\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-2=0\\y^2-9=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\y\in\left\{3;-3\right\}\end{matrix}\right.\)
so sánh x và y biết : \(x=\dfrac{2016^{2017}+1}{2016^{2016}+1}\) và \(y=\dfrac{2016^{2016}+1}{2016^{2015}+1}\)
Tìm x, y, z thỏa mãn
\(\dfrac{x}{y}=\dfrac{2017}{2}\\ \dfrac{y}{z}=\dfrac{2}{2017}\\ x-2z=2017\)
\(2^{2017}-\left(2^{2016}+2^{2015}+2^{2014}+...+12+1\right)\)
Tính nhanh:
a) \(\left(2017^1\right)^{2.4.6.8...2016}\)
b)\(2017^{\left(225-1^2\right).\left(225-2^2\right).\left(225-3^2\right)....\left(225-56^2\right)}\)
Tìm x, biết:\(\frac{x+2}{2019}+\frac{x+3}{2018}+\frac{x+4}{2017}+\frac{x+5}{2008}+\frac{x+2076}{11}=0\)
Tìm x, y biết:
a) \(3^x=9^{y-1}\) với \(8^y=2^{x+8}\)
b) \(\left(2x-1\right)^2+\left(4y-1\right)^4=0\)
tìm x,y bt
a, (x+2)2=0
b, (x-1)2=1
c, (2x-3)2=9
d, ( x+\(\dfrac{1}{2}\) )2=1
(x-1,2)^2=4
(x+1)^3=-125
(x+1,5)^8+(2,7-y)^10=0
3^-1.4^x+ 3.$^x+5/3. 27
9^-x. 27^x= 243
b1 :tính
a) N = (1/2 - 1). (1/3 - 1). (1/4 - 1) .... (1/2016 - 1). (1/2017 - 1)
b) P = (-1/1/2). (-1/1/3). (-1/1/4) ... (-1/1/2015). (2016/01/01)
c) Q = 1/13 + 3 / 13,23 + 3 / 23,33 + ... + 3 / 2013,2023
d) R = 1 / 2017.2016-1 / 2016.2015-1 / 2015.2014 - ... -. 1 / 3.2 - 1/ 2.1
( chú ý : / là phần nha )
