Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(5^{2014}-5^{2013}+5^{2012}=5^{2011}\left(5^3-5^2+5\right)\)
\(=5^{2011}.\left(125-25+5\right)=5^{2011}.105⋮105\)
\(5^{2014}-5^{2013}+5^{2012}\)
\(=5^{2011}.\left(5^3-5^2+5\right)\)
\(=5^{2011}.105\)\(⋮105\)
\(\Rightarrow5^{2014}-5^{2013}+5^{2012}⋮105\)\(\left(đpcm\right)\)
\(VP=\dfrac{2013}{1}+\dfrac{2012}{2}+...+\dfrac{2}{2012}+\dfrac{1}{2013}\)
\(VP=2013+\dfrac{2012}{2}+...+\dfrac{2}{2012}+\dfrac{1}{2013}\)
\(VP=1+\left(\dfrac{2012}{2}+1\right)+....+\left(\dfrac{2}{2012}+1\right)+\left(\dfrac{1}{2013}+1\right)\)
\(VP=\dfrac{2014}{2014}+\dfrac{2014}{2}+...+\dfrac{2014}{2012}+\dfrac{2014}{2013}\)
\(VP=2014\left(\dfrac{1}{2}+..+\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}\right)\)
\(VP-VT=2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)-x\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)=0\)
\(\Rightarrow\left(2014-x\right)\left(\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2014}\right)=0\)
\(\Rightarrow x=2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\ne0\right)\)
Ta có :
\(H=2^{2014}-2^{2013}-2^{2012}-...-2-1\)
\(H=2^{2014}-\left(2^{2013}+2^{2012}+2^{2011}+...+2+1\right)\)
Đặt \(B=2^{2013}+2^{2012}+2^{2011}+...+2+1\)
\(2B=2^{2014}+2^{2013}+2^{2012}+...+2^2+2\)
\(2B-B=\left(2^{2014}+2^{2013}+2^{2012}+...+2^2+2\right)-\left(2^{2013}+2^{2012}+2^{2011}+...+2+1\right)\)
\(B=2^{2014}-1\)
\(\Rightarrow\)\(H=2^{2014}-B=2^{2014}-\left(2^{2014}-1\right)=2^{2014}-2^{2014}+1=1\)
Suy ra :
\(A=2014^H=2014^1=2014\)
Vậy \(A=2014\)
Chúc bạn học tốt ~
+) Ta có :
\(A\left(-1\right)=\left(-1\right)+\left(-1\right)^2+\left(-1\right)^3+...+\left(-1\right)^{99}+\left(-1\right)^{100}\)
\(A\left(-1\right)=\left(-1\right)+1+\left(-1\right)+...+\left(-1\right)+1\)
\(A\left(-1\right)=\left(-1-1-...-1\right)+\left(1+1+...+1\right)\)
Do dãy 1; 3; 5; ... ; 99 có \(\frac{99-1}{2}+1=50\) số hạng nên có 50 số \(-1\)
Do dãy 2; 4; 6; ... ; 100 có \(\frac{100-2}{2}+1=50\) số hạng nên có 50 số \(1\)
Suy ra :
\(A\left(-1\right)=50.\left(-1\right)+50.1\)
\(A\left(-1\right)=-50+50\)
\(A\left(-1\right)=0\)
Vậy \(x=-1\) là nghiệm của đa thức \(A\left(x\right)=x+x^2+x^3+...+x^{99}+x^{100}\)
Chúc bạn học tốt ~
Ta có: \(5^{2014}-5^{2013}+5^{2012}=5^{2011}\left(5^3-5^2+5\right)\)
\(=5^{2011}.105⋮105\)
\(\Rightarrow5^{2014}-5^{2013}+5^{2012}⋮105\left(đpcm\right)\)
Vậy...
ta có:
\(5^{2014}-5^{2013}+5^{2012}\)
\(=5^{2012}\left(5^2-5+1\right)\)
\(=5^{2012}\left(25-5+1\right)\)
\(=5^{2012}.21\)
ta thấy: \(5^{2012}.21⋮21\)
\(5^{2012}.21⋮5\)
\(\Rightarrow5^{2012}.21⋮21.5\)
\(\Rightarrow5^{2012}.21⋮105\)
\(\Leftrightarrow5^{2014}-5^{2013}+5^{2012}⋮105\left(đpcm\right)\)