Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
BT1: 20152014 có tận cùng là 5
20142015=2014.(20142)1007=2014.40561961007=2014.(...6) => Có tận cùng là ...4
=> 20152014-20142015 có tận cùng là ...5-...4=...1
BT2: f(1)=a.1+b=1 (1)
f(2)=a.2+b=4 (2)
Trừ (2) cho (1) => a=3
Thay a=3 vào (1) => b=-2
ĐS: a=3; b=-2
\(\frac{x+5}{200}+\frac{x+4}{201}=\frac{x+3}{202}+\frac{x+2}{203}\)
=> \(\left(1+\frac{x+5}{200}\right)+\left(1+\frac{x+4}{201}\right)=\left(1+\frac{x+3}{202}\right)+\left(1+\frac{x+2}{203}\right)\)
=> \(\frac{x+205}{200}+\frac{x+205}{201}=\frac{x+205}{202}+\frac{x+205}{203}\)
=> \(\frac{x+205}{200}+\frac{x+205}{201}-\frac{x+205}{202}-\frac{x+205}{203}=0\)
=> \(\left(x+205\right).\left(\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\right)=0\)
Do \(\frac{1}{200}>\frac{1}{202};\frac{1}{201}>1-\frac{1}{203}\)
=> \(\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\ne0\)
=> \(x+205=0\)
=> \(x=-205\)
\(\frac{x+5}{200}+\frac{x+4}{201}=\frac{x+3}{202}+\frac{x+2}{203}\)
\(=>\frac{x+5+200}{200}+\frac{x+4+201}{201}-\frac{x+3+202}{202}-\frac{x+2+203}{203}=0\)
\(=>\frac{x+205}{200}+\frac{x+205}{201}-\frac{x+205}{202}-\frac{x+205}{203}=0\)
\(=>\left(x+205\right).\left(\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\right)=0\)
\(Do:\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\ne0\)
\(=>x+205=0\)
\(=>x=-205\)
a)\(VT=\left(-\dfrac{1}{8}\right)^{100}=\dfrac{1}{8^{100}}=\dfrac{1}{\left(2^3\right)^{100}}=\dfrac{1}{2^{300}}\)
\(VP=\left(-\dfrac{1}{4}\right)^{200}=\dfrac{1}{\left(2^2\right)^{200}}=\dfrac{1}{2^{400}}\)
\(\Rightarrow VT>VP\)
b) \(VT=4^{100}=\left(2^2\right)^{100}=2^{200}< 2^{202}=VP\)
c) \(VT=5^{2000}=\left(5^2\right)^{1000}=25^{1000}>10^{1000}=VP\)
d) \(VT=31^5< 32^5=\left(2^5\right)^5=2^{25}\)
\(VP=17^7>16^7=\left(2^4\right)^7=2^{28}\)
\(VP>VT\)
N = 5203 - (5202 - 5201 + 5200 - 5199 + ... + 52 - 5)
N = 5203 - 5202 + 5201 - 5200 + 5199 - ... - 52 + 5
5N = 5204 - 5203 + 5202 - 5201 + 5200 - ... - 53 + 52
5N + N = (5204 - 5203 + 5202 - 5201 + 5200 - ... - 53 + 52) + (5203 - 5202 + 5201 - 5200 + 5199 - ... - 52 + 5)
6N = 5204 + 5
N = \(\frac{5^{204}+5}{6}\)