Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(a^2+b^2+c^2=ab+bc+ac\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ac\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(c-a\right)^2+\left(b-c\right)^2=0\)
Ta có : \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(c-a\right)^2\ge0\\\left(b-c\right)^2\ge0\end{cases}}\)
\(\Rightarrow\left(a-b\right)^2+\left(c-a\right)^2+\left(b-c\right)^2=0\)
\(\Leftrightarrow a=b=c\)
a. \(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ab-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\left(đpcm\right)\)
Chọn B. Thay \(\dfrac{1}{3}\)vào x và \(\dfrac{1}{2}\)vào y
giải để ra được m
a)
\(\begin{matrix}N\left(x\right)=-4x^4+9x^3-x^2+5x+\dfrac{1}{3}\\^-M\left(x\right)=-x^4-9x^3+x^2+9x+\dfrac{4}{3}\\\overline{N\left(x\right)-M\left(x\right)=-3x^4+18x^3-2x^2-4x-1}\end{matrix}\)
b)
\(\begin{matrix}M\left(x\right)=-x^4-9x^3+x^2+9x+\dfrac{4}{3}\\^+N\left(x\right)=-4x^4+9x^3-x^2+5x+\dfrac{1}{3}\\\overline{M\left(x\right)+N\left(x\right)=-5x^4+14x+\dfrac{5}{3}}\end{matrix}\)
\(a)\) \(M_{\left(3\right)}=3+3^2+3^3+...+3^{2016}\)
\(3M_{\left(3\right)}=3^2+3^3+3^4+...+3^{2017}\)
\(3M_{\left(3\right)}-M_{\left(3\right)}=\left(3^2+3^3+3^4+...+3^{2017}\right)-\left(3+3^2+3^3+...+3^{2016}\right)\)
\(2M_{\left(3\right)}=3^{2017}-3\)
\(M_{\left(3\right)}=\frac{3^{2017}-3}{2}\)
Vậy \(M_{\left(3\right)}=\frac{3^{2017}-3}{2}\)
\(M_{\left(-3\right)}=\left(-3\right)+\left(-3\right)^2+\left(-3\right)^3+...+\left(-3\right)^{2016}\)
\(\left(-3\right)M_{\left(-3\right)}=\left(-3\right)^2+\left(-3\right)^3+\left(-3\right)^4+...+\left(-3\right)^{2017}\)
\(\left(-3\right)M_{\left(-3\right)}-M_{\left(-3\right)}=\left[\left(-3\right)^2+\left(-3\right)^3+...+\left(-3\right)^{2017}\right]-\left[\left(-3\right)+\left(-3\right)^2+...+\left(-3\right)^{2016}\right]\)\(\left(-4\right)M_{\left(-3\right)}=\left(-3\right)^{2017}+3\)
\(M_{\left(-3\right)}=\frac{\left(-3\right)^{2017}+3}{-4}\)
\(M_{\left(-3\right)}=\frac{-\left(3^{2017}-3\right)}{-4}\)
\(M_{\left(-3\right)}=\frac{3^{2017}-3}{4}\)
Vậy \(M_{\left(-3\right)}=\frac{3^{2017}-3}{4}\)
Chúc bạn học tốt ~
\(b)\) Ta có :
\(M_{\left(2\right)}=2+2^2+2^3+...+2^{2016}\)
\(M_{\left(2\right)}=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2014}+2^{2015}+2^{2016}\right)\)
\(M_{\left(2\right)}=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2014}\left(1+2+2^2\right)\)
\(M_{\left(2\right)}=2.7+2^4.7+...+2^{2014}.7\)
\(M_{\left(2\right)}=7\left(2+2^4+...+2^{2014}\right)⋮7\) \(\left(1\right)\)
Lại có :
\(M_{\left(2\right)}=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{2013}+2^{2014}+2^{2015}+2^{2016}\right)\)
\(M_{\left(2\right)}=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{2013}\left(1+2+2^2+2^3\right)\)
\(M_{\left(2\right)}=2.15+2^5.15+...+2^{2013}.15\)
\(M_{\left(2\right)}=15\left(2+2^5+...+2^{2013}\right)⋮15\) \(\left(2\right)\)
Từ (1) và (2) suy ra \(M_{\left(2\right)}\) chia hết cho \(7\) và \(15\)
\(\Rightarrow\)\(M_{\left(2\right)}⋮105\) ( vì \(7.15=105\) )
Vậy nếu \(M⋮105\)\(\Leftrightarrow\)\(x=2\)
Chúc bạn học tốt ~