\(B=\dfrac{2021\times13+2007+2020\times2007}{2020+2020\times520+1500\times2020}\)
Những câu hỏi liên quan
\(\dfrac{2018\times2020+2020\times2022}{2020\times4040}\)
\(\dfrac{2018\times2020+2020\times2022}{2020\times4040}\\ =\dfrac{2020\times\left(2018+2022\right)}{2020\times4040}\\ =\dfrac{2020\times4040}{2020\times4040}\\ =1\)
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\(\dfrac{2018\times2020+2020\times2022}{2020\times4040}\)
\(=\dfrac{2020\times\left(2018+2022\right)}{2020\times4040}\)
\(=\dfrac{2018+2022}{4040}\)
\(=\dfrac{4040}{4040}\)
\(=1\)
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14 tháng 3 2022 lúc 16:21
A = \(\dfrac{2018\times2020+2020\times2022}{2020\times4040}\)
B = \(\dfrac{111\times118-33}{112\times115+185}\)
Ai nhanh mình tick cho
tìm B = 2021 x 13 + 2009 + 2020 x 2007 / 2020 + 2020 x 520 x 2020
B=2021 x 13 + 2009 + 2020 x 2007/2020 + 2020 x520x2020
B=2021 x 13 +2009+2020/1x2007/2020 +2020x520x2020
B=26273+2009+20201/1x2007/20201+2121808000
B=28282+2007+2121808000
B=2121838289
so sánh:
a)C= \(\dfrac{100^{99}+1}{100^{100}+1}\) và D= \(\dfrac{100^{100}+1}{100^{101}+1}\)
b)E=\(\dfrac{2020^{2021}+1}{2020^{2022}+1}\) và F=\(\dfrac{2020^{2020}+1}{2020^{2021}+1}\)
c: \(100C=\dfrac{100^{100}+100}{100^{100}+1}=1+\dfrac{99}{100^{100}+1}\)
\(100D=\dfrac{100^{101}+100}{100^{101}+1}=1+\dfrac{99}{100^{101}+1}\)
100^100+1<100^101+1
=>\(\dfrac{99}{100^{100}+1}>\dfrac{99}{100^{101}+1}\)
=>100C>100D
=>C>D
b: \(2020E=\dfrac{2020^{2022}+2020}{2020^{2022}+1}=1+\dfrac{2019}{2020^{2022}+1}\)
\(2020F=\dfrac{2020^{2021}+2020}{2020^{2021}+1}=1+\dfrac{2019}{2020^{2021}+1}\)
2020^2022+1>2020^2021+1(Do 2022>2021)
=>\(\dfrac{2019}{2020^{2022}+1}< \dfrac{2019}{2020^{2021}+1}\)
=>2020E<2020F
=>E<F
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17 tháng 4 2021 lúc 17:53
So sánh M = \(\dfrac{2019}{2020}+\dfrac{2020}{2021}\) và N = \(\dfrac{2019+2020}{2020+2021}\)
Giải:
Ta có: N=2019+2020/2020+2021
=>N=2019/2020+2021 + 2020/2020+2021
Vì 2019/2020 > 2019/2020+2021 ; 2020/2021 > 2020/2020+2021
=>M>N
Vậy ...
Chúc bạn học tốt!
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Ta có : \(\dfrac{2019}{2020}>\dfrac{2019}{2020+2021}\)
\(\dfrac{2020}{2021}>\dfrac{2020}{2020+2021}\)
\(\Rightarrow\dfrac{2019}{2020}+\dfrac{2020}{2021}>\dfrac{2019+2020}{2020+2021}\)
\(\Rightarrow M>N\)
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chứng minh bất đẳng thức sau: \(\dfrac{2020}{\sqrt{2021}}+\dfrac{\sqrt{2021}}{2020}>\sqrt{2020}+\sqrt{2021}\)
B=\(\dfrac{\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}}{\dfrac{3}{2020}+\dfrac{3}{2021}-\dfrac{3}{2022}}-1\)
\(B=\dfrac{\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}}{\dfrac{3}{2020}+\dfrac{3}{2021}-\dfrac{3}{2022}}-1=\dfrac{\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}}{3\left(\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}\right)}-1=\dfrac{1}{3}-1=-\dfrac{2}{3}\)
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\(B=\dfrac{\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}}{\dfrac{3}{2020}+\dfrac{3}{2021}-\dfrac{3}{2022}}-1=\dfrac{\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}}{3\left(\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}\right)}-1=\dfrac{1}{3}-1=\dfrac{1}{3}-\dfrac{3}{3}=-\dfrac{2}{3}\)
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Xem thêm câu trả lời
4 tháng 5 2021 lúc 20:36
So sánh A và B
A=\(\dfrac{4-7^{2020}}{7^{2020}}\)+\(\dfrac{5+7^{2021}}{7^{2021}}\)
B=\(\dfrac{1}{7^{2019}}\)
Ta có:
\(A=\dfrac{7\left(4-7^{2020}\right)}{7^{2021}}+\dfrac{5+7^{2021}}{7^{2021}}\)
\(A=\dfrac{28-7^{2021}+5+7^{2021}}{7^{2021}}=\dfrac{33}{7^{2021}}\)
Ta có: \(B=\dfrac{7^2}{7^{2021}}=\dfrac{49}{7^{2021}}\)
=> B>A
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12 tháng 2 2023 lúc 17:57
So sánh A=\(\dfrac{2018}{2019}\)+\(\dfrac{2019}{2020}\)+\(\dfrac{2020}{2021}\)+\(\dfrac{2021}{2018}\)với 4
Lời giải:
$A=1-\frac{1}{2019}+1-\frac{1}{2020}+1-\frac{1}{2021}+1+\frac{3}{2018}$
$=4+(\frac{1}{2018}-\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2020}+\frac{1}{2018}-\frac{1}{2021})$
$> 4+0+0+0+0=4$
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