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A = 250 + 251 + 252 + .... + 22017 + 22018
=> 2A = 251 + 252 + 253 + .... + 22018 + 22019
=> 2A - A = ( 251 + 252 + 253 + ... + 22018 + 22019 ) - ( 250 + 251 + ... + 22017 + 22018 )
=> A = 22019 - 250
Ta thấy :
+) 32018 = 3 . 3 . ... . 3 = 1 số lẻ
+) 112017 = 11 . 11 . ... . 11 = 1 số lẻ
mà 1 số lẻ trừ 1 số lẻ bằng 1 số chẵn
=> 32018 - 112017 chia hết cho 2
Vậy,.......
vì mình mới hok lớp 5 ko bt làm toán lớp 6 nha thông cảm , đừng sai mik nha mình ko bt làm thôi
(2/2017 + 2/2018) / (5/2017 + 5/2018)
= 2 x (1/2017 + 1/2018) / 5 x (1/2017 + 1/2018)
= 2/5 (vì (1/2017 + 1/2018) khác 0)
\(\frac{\frac{2}{2017}+\frac{2}{2018}}{\frac{5}{2017}+\frac{5}{2018}}\)
\(=\frac{2\left(\frac{1}{2017}+\frac{1}{2018}\right)}{5\left(\frac{1}{2017}+\frac{1}{2018}\right)}\)
\(=\frac{2}{5}\)
Study well ! >_<
\(A=\frac{10^{2016}+2018}{10^{2017}+2018}\)
\(\Rightarrow10A=\frac{10^{2017}+20180}{10^{2017}+2018}\)
\(=\frac{10^{2017}+2018+18162}{10^{2017}+2018}\)
\(=\frac{10^{2017}+2018}{10^{2017}+2018}+\frac{18162}{10^{2017}+2018}\)
\(=1+\frac{18162}{10^{2017}+2018}\)
\(B=\frac{10^{2017}+2018}{10^{2018}+2018}\)
\(\Rightarrow10B=\frac{10^{2018}+20180}{10^{2018}+2018}\)
\(=\frac{10^{2018}+2018+18162}{10^{2018}+2018}\)
\(=\frac{10^{2018}+2018}{10^{2018}+2018}+\frac{18162}{10^{2018}+2018}\)
\(=1+\frac{18162}{10^{2018}+2018}\)
Ta thấy: \(1+\frac{18162}{10^{2017}+2018}>1+\frac{18162}{10^{2018}+2018}\)
=> 10A > 10B
=> A > B
Ta có : \(\left(5x-3\right)^2-\frac{1^2}{64}=0\)
\(\Leftrightarrow\left(5x-3\right)^2=\frac{1}{64}\)
\(\Leftrightarrow\left(5x-3\right)^2=\left(\frac{1}{8}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=\frac{1}{8}\\5x-3=-\frac{1}{8}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=\frac{1}{8}+3\\5x=-\frac{1}{8}+3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=\frac{25}{8}\\5x=\frac{23}{8}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{25}{8}.\frac{1}{5}\\x=\frac{23}{8}.\frac{1}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{8}\\x=\frac{23}{40}\end{cases}}\)
b) 3x - 7.(5x-1) = 6 - 2.(4-3x)
=> 3x - 35x + 7 = 6 - 8 + 6x
=> 3x - 35x - 6x = 6-8 -7
-38x = -9
x = 9/38
\(\left(1-3x\right)^3=-8\)
\( \left(1-3x\right)^3=-2^3\)
\(\Rightarrow1-3x=-2\Rightarrow-3x=-3\)
\(\Leftrightarrow x=1\)
chuc bn hok tot
\(3x_{}-2018=\left(-2\right)2\)
\(3x-2018=-4\)
\(3x=-4+2018\)
\(3x=2014\)
\(x=\frac{2014}{3}\)
Vậy \(x=\frac{2014}{3}\)