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Lời giải:
$M=3^{2017}-3^{2016}+3^{2015}-....+3-1$
$3M=3^{2018}-3^{2017}+3^{2016}-...+3^2-3$
$M+3M=3^{2018}-1$
$4M=3^{2018}-1$
$16M=4(3^{2018}-1)$
Ta thấy: $3^4=81\equiv 1\pmod {10}$
$\Rightarrow 3^{2018}=(3^4)^{504}.3^2\equiv 1^{504}.3^2\equiv 9\pmod {10}$
$\Rightarrow 16M=4(3^{2018}-1)\equiv 4(9-1)\equiv 32\equiv 2\pmod {10}$
Vậy $16M$ tận cùng là $2$
Bài 1 : dễ bạn tự làm được :)
Bài 2 :
Ta có :
\(B=\frac{2015+2016+2017}{2016+2017+2018}=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Vì :
\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
Nên \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(\Leftrightarrow\)\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Leftrightarrow\)\(A>B\)
Vậy \(A>B\)
Chúc bạn học tốt ~
Ta có : B = 2016 + 2017 + 2018 2015 + 2016 + 2017 = 2016 + 2017 + 2018 2015 + 2016 + 2017 + 2018 2016 + 2016 + 2017 + 2018 2017 Vì : 2016 2015 > 2016 + 2017 + 2018 2015 2017 2016 > 2016 + 2017 + 2018 2016 2018 2017 > 2016 + 2017 + 2018 2017 Nên 2016 2015 + 2017 2016 + 2018 2017 > 2016 + 2017 + 2018 2015 + 2016 + 2017 + 2018 2016 + 2016 + 2017 + 2018 2017 ⇔ 2016 2015 + 2017 2016 + 2018 2017 > 2016 + 2017 + 2018 2015 + 2016 + 2017 ⇔A > B Vậy A > B Chúc bạn học tốt ~
b) Ta có: \(5-\left|3x-1\right|=3\)
\(\Leftrightarrow\left|3x-1\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=-2\\3x-1=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\3x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=1\end{matrix}\right.\)
Vậy: \(x\in\left\{-\frac{1}{3};1\right\}\)
c) Ta có: \(\left(1-2x\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}1-2x=3\\1-2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-2\\2x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;2\right\}\)