a=2020. 2020-2022.2018
b= (1/4 -1).(1/9-1).(1/16-1)....(1/400-1)
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NH
3
H
19 tháng 5 2021
1. \(\dfrac{2019}{2020}-\left(\dfrac{2019}{2020}-\dfrac{2020}{2021}\right)\)
\(=\dfrac{2019}{2020}-\dfrac{2019}{2020}+\dfrac{2020}{2021}\)
\(=0+\dfrac{2020}{2021}=\dfrac{2020}{2021}\)
19 tháng 5 2021
Giải:
1) \(\dfrac{2019}{2020}-\left(\dfrac{2019}{2020}-\dfrac{2020}{2021}\right)\)
\(=\dfrac{2019}{2020}-\dfrac{2019}{2020}+\dfrac{2020}{2021}\)
\(=\left(\dfrac{2019}{2020}-\dfrac{2019}{2020}\right)+\dfrac{2020}{2021}\)
\(=0+\dfrac{2020}{2021}\)
\(=\dfrac{2020}{2021}\)
2) \(\dfrac{2}{9}+\dfrac{7}{9}:\left(\dfrac{42}{5}-\dfrac{7}{5}\right)\)
\(=\dfrac{2}{9}+\dfrac{7}{9}:7\)
\(=\dfrac{2}{9}+\dfrac{1}{9}\)
\(=\dfrac{1}{3}\)
3) \(\dfrac{3}{4}+\dfrac{x}{4}=\dfrac{5}{8}\)
\(\dfrac{x}{4}=\dfrac{5}{8}-\dfrac{3}{4}\)
\(\dfrac{x}{4}=\dfrac{-1}{8}\)
\(\Rightarrow x=\dfrac{4.-1}{8}=\dfrac{-1}{2}\)
4) \(\left|3x+1\right|-\dfrac{1}{4}=\dfrac{-1}{4}\)
\(\left|3x-1\right|=\dfrac{-1}{4}+\dfrac{1}{4}\)
\(\left|3x-1\right|=0\)
\(3x-1=0\)
\(3x=0+1\)
\(3x=1\)
\(x=1:3\)
\(x=\dfrac{1}{3}\)
Chúc bạn học tốt!
I
1
AH
Akai Haruma
Giáo viên
30 tháng 9 2021
Lời giải:
$(4+a-3b)^{2020}(3a-5b-1)^{2020}=[(4+a-3b).(3a-5b-1)]^{2020}$
Muốn cm biểu thức này luôn chia hết cho $16$ ta chỉ cần cm $(4+a-3b)(3a-5b-1)\vdots 2$
Thật vậy:
Xét tổng: $4+a-3b+3a-5b-1=3+4a-8b$ lẻ nên $4+a-3b, 3a-5b-1$ khác tính chẵn lẻ
Do đó tồn tại 1 trong 2 số chẵn
$\Rightarrow (4+a-3b)(3a-5b-1)\vdots 2$
Do đó ta có đpcm.
VD
0
VD
0
HN
1
19 tháng 5 2022
\(A=\dfrac{2}{1}+2+\dfrac{5}{1}+2+3+\dfrac{9}{1}+2+3+4+...+\dfrac{2041210}{1}+2+3+4+...+2020\)
\(A=\left(\dfrac{2+5+9+...+2041210}{1}\right)+2+3+4+...+2020\)
\(A=\left(\dfrac{\left(2041210-2\right)\div4+1}{1}\right)+2+3+4+...+2020\)
\(A=\dfrac{510304}{1}+\left(2+3+4+...+2020\right)\)
\(A=510304+\left(2020-2\right)+1\)
\(A=510304+2019\)
\(A=512323\)
19 tháng 6 2023
\(A=\dfrac{1}{4}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{1010^2}\right)\)
1/2^2+1/3^2+...+1/2010^2<1/1*2+1/2*3+...+1/2009*2010=1-1/2010<1
=>A<1/4
a) $2020.2020-2022.2018$
$ = 2020^2-(2020+2).(2020-2)$
$ = 2020^2 - (2020^2-2^2)$
$ = 4$
b) \(\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)...\left(\dfrac{1}{400}-1\right)\)
\(=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{20^2}-1\right)\)
\(=\dfrac{\left(-1\right)\cdot3\cdot\left(-2\right)\cdot4\cdot\left(-3\right)\cdot5\cdot\cdot\cdot\left(-19\right)\cdot21}{2^2\cdot3^2\cdot4^2\cdot\cdot\cdot20^2}\)
\(=-\dfrac{1}{20}\cdot\dfrac{21}{2}=-\dfrac{21}{40}\)
Giải:
a) 2020.2020−2022.2018
=20202−(2020+2).(2020−2)
=20202−(20202−22)
=4
b) (1/4−1)(1/9−1)(1/16−1)...(/1400−1)
=(1/22−1)(1/32−1)(1/42−1)...(1/202-1)
=(−1)⋅3⋅(−2)⋅4⋅(−3)⋅5⋅⋅⋅(−19)⋅21/22⋅32⋅42⋅⋅⋅202
=−1/20⋅21/2
=−21/40