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B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + \(\dfrac{2022}{1}\)
B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + 2022
B = 1 + ( 1 + \(\dfrac{1}{2022}\)) + ( 1 + \(\dfrac{2}{2021}\)) + \(\left(1+\dfrac{3}{2020}\right)\)+ ... + \(\left(1+\dfrac{2021}{2}\right)\)
B = \(\dfrac{2023}{2023}\) + \(\dfrac{2023}{2022}\) + \(\dfrac{2023}{2021}\) + \(\dfrac{2023}{2020}\) + ...+ \(\dfrac{2023}{2}\)
B = 2023 \(\times\) ( \(\dfrac{1}{2023}\) + \(\dfrac{1}{2022}\) + \(\dfrac{1}{2021}\) + \(\dfrac{1}{2020}\)+ ... + \(\dfrac{1}{2}\))
Vậy B > C
\(a,2^x+2^{x+3}=144\\ 2^x.\left(1+2^3\right)=144\\ 2^x.9=144\\ 2^x=144:9\\ 2^x=16=2^4\\ vậy:x=4\)
\(b,\left(x-5\right)^{2022}=\left(x-5\right)^{2021}\\ Vì:\left[{}\begin{matrix}0^{2022}=0^{2021}\\1^{2022}=1^{2021}\end{matrix}\right.\\ Vậy:\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
=(1-2)-(3-4)+(5-6)-(7-8)+...+(2021-2022)-2023
=(-1)-(-1)+(-1)-...+(-1)-2023
=0-2023
=-2023
a) 12 : { 400 : [500 – (125 + 25 . 7)]}
=12 : { 400 : [500 – (125 + 175)]}
=12 : [ 400 : [500 – 300)]
=12 : (400:200)
=12:2=6
b) 5 . 22 – 18 : 3
=5.4-6
=20-6=14
c) 18 : 3 + 182 + 3.(51 : 17)
=5+182+3.3
=187+9=196
d) 25 . 8 – 12.5 + 170 : 17 – 8
=200-60+10-8
=142
e) 2.52+ 3: 710 – 54: 33
=2.25+3:1-54:27
=50+3-2
=51
f) 189 + 73 + 211 + 127
=(189+211)+(73+127)
=400+200=600
g) 375 : {32 – [ 4 + (5. 32– 42)]} – 14 )
=375 : {32 – [ 4 + (160– 42)]} – 14 )
=375 : [32 – ( 4 + 118) – 14 ]
=375:(32-122-14)
=375:-104
=-375/104
h) (52022 + 52021) : 52021
=(52022:52021)+(52021:52021)
=5+1=6
B = 1 + 3 + 32 + ... + 32021
3B = 3 + 32 + 33 + ... + 32021 + 32022
3B - B = (3 + 32 + 33 + ... + 32021 + 32022) - (1 + 3 + 32 + ... + 32021)
2B = 32022 - 1
B = (32022 - 1) : 2
C - B = 32022 : 2 - (32022 - 1) : 2
C - B = [32022 - (32022 - 1)] : 2
C - B = [32022 - 32022 + 1] : 2
C - B = [0 + 1] : 2
C - B = 1 : 2
C - B = 1/2
Vậy C - B = 1/2
Học tốt nhe bạn :)