a=2020. 2020-2022.2018
b= (1/4 -1).(1/9-1).(1/16-1)....(1/400-1)
1. 2019/2020-(2019/2020-2020/2021)
2.2/9+7/9 :(42/5-7/5
3.a)3/4+x/4=5/8
4./3x+1/-1/4=-1/4
1. \(\dfrac{2019}{2020}-\left(\dfrac{2019}{2020}-\dfrac{2020}{2021}\right)\)
\(=\dfrac{2019}{2020}-\dfrac{2019}{2020}+\dfrac{2020}{2021}\)
\(=0+\dfrac{2020}{2021}=\dfrac{2020}{2021}\)
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!
4) \(\left|3x+1\right|-\dfrac{1}{4}=\dfrac{-1}{4}\)
\(\left|3x+1\right|=\dfrac{-1}{4}+\dfrac{1}{4}\)
\(\left|3x+1\right|=0\)
\(3x+1=0\)
\(3x=0-1\)
\(3x=-1\)
\(x=-1:3\)
\(x=\dfrac{-1}{3}\)
cho ba số a, b, c thỏa mãn abc = 27 và 1/a+1/b+1/c = (a+b+c)/9 Chứng minh (a*2020-9*1010)(b*2020-9*1010)(c*2020-9*1010)=0
CMR: \(\left(4+a-3b\right)^{2020}.\left(3a-5b-1\right)^{2020}⋮16\) với mọi a,b nguyên
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.
Câu 30. Giá trị của tổng
S =1+ 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 +10 -... + 2018 - 2019 - 2020 + 2021 là
A. 2020 . B. 2021. C. 1. D. -1.
1/2+1/3+2/3+1/4+2/4+3/4+......+1/2020+2=2020+3/2020+.....+2019/2020
1/2+1/3+2/3+1/4+2/4+3/4+......+1/2020+2/2020+3/2020+.....+2019/2020
tìm A = 2/1+2 + 5/1+2+3 + 9/1+2+3+4 + ..... + 2041210/1+2+3+4 + ...+ 2020
\(A=\dfrac{2}{1}+2+\dfrac{5}{1}+2+3+\dfrac{9}{1}+2+3+4+...+\dfrac{2041210}{1}+2+3+4+...+2020\)
\(A=\left(\dfrac{2+5+9+...+2041210}{1}\right)+2+3+4+...+2020\)
\(A=\left(\dfrac{\left(2041210-2\right)\div4+1}{1}\right)+2+3+4+...+2020\)
\(A=\dfrac{510304}{1}+\left(2+3+4+...+2020\right)\)
\(A=510304+\left(2020-2\right)+1\)
\(A=510304+2019\)
\(A=512323\)
Chứng minh rằng : \(\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+........+\dfrac{1}{2020^2}< \dfrac{1}{4}\)
\(A=\dfrac{1}{4}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{1010^2}\right)\)
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