(2020 - 3257) - (2019 - 3255)
Những câu hỏi liên quan
Bài 1. Tính hợp lý ( nếu có thể): 1) [ (- 2) ^ 2 +7.(-10)]:2+9 2) 2019- lceil157- (357 - 1019) rceil 3) (1173 - 3978) - (173 - 2978) 4) - 113.234 + 129.113 -113.(-105) 5) (2020 - 3257) - (2019 - 3255)
Viết lại hộ mik cái
Ko nhìn đc bn ui!@##
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Viết là bài này ko cần giải đó, hihi!
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B1 : Tính hợp lý ( nếu có thể ):a) ( -72 ) . 121 + 27 . ( -121 ) -121b) 2019 - [ 157 - ( 357 -219 ) ]c) ( 1732 - 5986 ) - ( 732 - 9286 )d) 25 . 76 - ( -29 ) . 25 -105 . 125e) ( 2020 + 3257 ) - ( 2019 + 3255 ) B2 : Tìm x, biết :a) -32 + 3 . 2x + 3 3 . 23b) -13 + 17 - 117 + 113 - | 5 - x | -52c) ( 2x - 12 ) . ( x + 1 ) 0d) ( x3 + 1 ) . ( -5 - 5x ) 0e) | x | + | x + 2 | 3x
Đọc tiếp
B1 : Tính hợp lý ( nếu có thể ):
a) ( -72 ) . 121 + 27 . ( -121 ) -121
b) 2019 - [ 157 - ( 357 -219 ) ]
c) ( 1732 - 5986 ) - ( 732 - 9286 )
d) 25 . 76 - ( -29 ) . 25 -105 . 125
e) ( 2020 + 3257 ) - ( 2019 + 3255 )
B2 : Tìm x, biết :
a) -32 + 3 . 2x + 3 = 3 . 23
b) -13 + 17 - 117 + 113 - | 5 - x | = -52
c) ( 2x - 12 ) . ( x + 1 ) = 0
d) ( x3 + 1 ) . ( -5 - 5x ) = 0
e) | x | + | x + 2 | = 3x
28 tháng 6 2021 lúc 15:55
So sánh A và B
A = \(\left(2020^{2019}+2019^{2019}\right)^{2020}\)
B = \(\left(2020^{2020}+2019^{2020}\right)^{2019}\)
Ta có: \(A=\left(2020^{2019}+2019^{2019}\right)^{2020}\)
\(=\left(2019^{2019}+2020^{2019}\right)^{2019}\cdot\left(2019^{2019}+2020^{2019}\right)\)
\(\Leftrightarrow\dfrac{A}{B}=\dfrac{\left(2019^{2019}+2020^{2019}\right)^{2019}\cdot\left(2019^{2019}+2020^{2019}\right)}{\left(2020^{2020}+2019^{2020}\right)^{2019}}\)
\(\Leftrightarrow\dfrac{A}{B}=\dfrac{2019^{2019}+2020^{2019}}{2019+2020}>1\)
\(\Leftrightarrow A>B\)
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\(\dfrac{2020^{2018}-1}{2020^{2019}+2019}\)với B=\(\dfrac{2020^{2019}+1}{2020^{2020}+2019}\)
\(A=\dfrac{2020^{2018}-1}{2020^{2019}+2019}\)
\(B=\dfrac{2020^{2019}+1}{2020^{2020}+2019}\)
Ta có :
\(A-B=\dfrac{2020^{2018}-1}{2020^{2019}+2019}-\dfrac{2020^{2019}+1}{2020^{2020}+2019}\)
\(\Rightarrow A-B=\dfrac{\left(2020^{2018}-1\right)\left(2020^{2020}+2019\right)-\left(2020^{2019}+2019\right)\left(2020^{2019}+1\right)}{\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)}\)
\(\Rightarrow A-B=\dfrac{2020^{4038}+2019.2020^{2018}-2020^{2020}-2019-2020^{4038}-2020^{2019}-2019.2020^{2018}-2029}{\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)}\)
\(\Rightarrow A-B=\dfrac{-\left(2020^{2020}+2020^{2019}+2.2019\right)}{\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)}\)
mà \(\left\{{}\begin{matrix}-\left(2020^{2020}+2020^{2019}+2.2019\right)< 0\\\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)>0\end{matrix}\right.\)
\(\Rightarrow A-B< 0\)
\(\Rightarrow A< B\)
Vậy ta được \(A< B\)
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so sanh A=2020^2018-1/2020^2019-2019 và B=2020^2019+1/2020^2020+2019
cho a = 1 - 2019 /2020 + ( 2019/2020)^2 -(2019-2020)^3 +....+(2019/2020) ^2020 chứng tỏ a ko phải là một số nguyên
Ta có:
\(a=1-\frac{2019}{2020}+\left(\frac{2019}{2020}\right)^2-\left(\frac{2019}{2020}\right)^3+...+\left(\frac{2019}{2020}\right)^{2020}\)
=> \(\frac{2019}{2020}.a=\frac{2019}{2020}-\left(\frac{2019}{2020}\right)^2+\left(\frac{2019}{2020}\right)^3-...+\left(\frac{2019}{2020}\right)^{2020}-\left(\frac{2019}{2020}\right)^{2021}\)
Lấy
\(a+\frac{2019}{2020}a=1-\left(\frac{2019}{2020}\right)^{2021}\)
<=> \(a\left(1+\frac{2019}{2020}\right)=\left[1-\left(\frac{2019}{2020}\right)^{2021}\right]\)
<=> \(a.\frac{4039}{2020}=\left[1-\left(\frac{2019}{2020}\right)^{2021}\right]\)
<=> \(a.=\left[1-\left(\frac{2019}{2020}\right)^{2021}\right].\frac{2020}{4039}\)
Vì : \(0< \left(\frac{2019}{2020}\right)^{2021}< 1\)
=> \(0< 1-\left(\frac{2019}{2020}\right)^{2021}< 1\)
và \(0< \frac{2020}{4039}< 1\)
=> \(0< \left[1-\left(\frac{2019}{2020}\right)^{2021}\right].\frac{2020}{4039}< 1\)
=> 0 < a < 1
=> a không phải là một số nguyên.
toan lop may vay ban ?
A=2020/20192+1 + 2020/20192+2 + 2020/20192+3 + ... + 2020/20192+2019. CMR 1<A<2.
Mục tiêu -500 sp mong giúp đỡ
2019/2020 x 7/8+2019/2020 x3/8-2019/2020 x1/4
So sánh x = 20192020 + 1 / 20192019 + 1 và y = 20192019 + 2020 / 20192018 + 2020
\(x=\frac{2019^{2020}+1}{2019^{2019}+1}>\frac{2019^{2020}+1+2018}{2019^{2019}+1+2018}=\frac{2019^{2020}+2019}{2019^{2019}+2019}=\frac{2019\left(2019^{2019}+1\right)}{2019\left(2019^{2018}+1\right)}=\frac{2019^{2019}+1}{2019^{2018}+1}\)(1)
\(y=\frac{2019^{2019}+2020}{2019^{2018}+2020}< \frac{2019^{2019}+2020-2019}{2019^{2018}+2020-2019}=\frac{2019^{2019}+1}{2019^{2018}+1}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow x>y\)