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\(\dfrac{2018\times2020+2020\times2022}{2020\times4040}\\ =\dfrac{2020\times\left(2018+2022\right)}{2020\times4040}\\ =\dfrac{2020\times4040}{2020\times4040}\\ =1\)
\(\dfrac{2018\times2020+2020\times2022}{2020\times4040}\)
\(=\dfrac{2020\times\left(2018+2022\right)}{2020\times4040}\)
\(=\dfrac{2018+2022}{4040}\)
\(=\dfrac{4040}{4040}\)
\(=1\)
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
vi 2018/2019<1
2019/2020<1
2020/2021<1
nen 2018/2019 + 2019/2020 + 2020/2021<1+1+1=3
Dấu ''\(x\)'' là dấu nhân chăng ?
\(A=\frac{2019x2020}{2019x2020+1}\)và \(B=\frac{2020}{2021}\)
Bài ra ta có :
Xét \(A=\frac{2019x2020}{2019x\left(2020+1\right)}=\frac{2020}{2020+1}=\frac{2020}{2021}\)
Vì \(\frac{2020}{2021}=\frac{2020}{2021}\)
Suy ra A = B theo (ĐPCM)
Giải:
a) 2019 + 2021 - 1
= 4040 - 1
= 4039
b) 2020 x 2019 + 2018
= 4078380 + 2018
= 4080398
Học tốt!!!
\(\dfrac{2020x13+13+2007+2020x2007}{2020x\left(1+520+1500\right)}\)
\(\dfrac{2020x\left(1+13+2007\right)}{2020x\left(1+520+1500\right)}=\dfrac{2021}{2021}=1\)
A = \(\dfrac{2021\times13+2007+2020\times2007}{2020+2020\times520+1500\times2020}\)
A = \(\dfrac{2021\times13+\left(2007+2020\times2007\right)}{2020+2020\times520+1500\times2020}\)
A = \(\dfrac{2021\times13+2007\times\left(1+2020\right)}{2020\times\left(1+520+1500\right)}\)
A = \(\dfrac{2021\times13+2007\times2021}{2020\times2021}\)
A = \(\dfrac{2021\times\left(13+2007\right)}{2021\times2020}\)
A = \(\dfrac{2021\times2020}{2021\times2020}\)
A = 1
B = \(\dfrac{2021\times13+2007+2020\times2007}{2020+2020\times520+1500\times2020}\)
B = \(\dfrac{2021\times13+2007\times\left(1+2020\right)}{2020\times\left(1+520+1500\right)}\)
B = \(\dfrac{2021\times13+2007\times2021}{2020\times2021}\)
B = \(\dfrac{2021\times\left(13+2007\right)}{2021\times2020}\)
B = \(\dfrac{2021\times2020}{2021\times2020}\)
B = 1