=> A < B

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con bái mẹ 

Lời giải:

Ta thấy, mỗi số hạng trong $b$ đều lớn hơn $1$ (do tử số lớn hơn mẫu số)

Do đó $b>1$

Ta có đpcm.

Giải:

B=2021/52+2021/52+2021/53+...+2021/100

Nhận xét: Ta thấy các số hạng ở dãy B đều > 1

2021/51 > 1

2021/52 > 1

2021/53 > 1

...

2021/100 > 1

=>B > 1

Vậy B>1

Chúc bạn học tốt!

BÀi 1:

a: (-243+345)-(257+(-55)-129)

=-243+345-257+55+129

=(-243-257)+(345+55)+129

=-500+400+129

=-100+129

=29

b: \(\left(-2\right)^3-45:\left(-3\right)^2-\left(-2019\right)^0\cdot\left(-1\right)^{2019}\)

\(=-8-\frac{45}{9}-1\cdot\left(-1\right)\)

=-8-5+1

=-13+1

=-12

c: \(\left\lbrack27\cdot\left(-67\right)+33\cdot\left(-27\right)\right\rbrack:\left(-30\right)\)

\(=\frac{-27\left(67+33\right)}{-30}\)

\(=\frac{9}{10}\cdot100=90\)

d: \(\left(1^2+2^2+\cdots+100^2\right)\left(3^4-9^2\right)\)

\(=\left(1^2+2^2+\cdots+100^2\right)\left(81-81\right)\)

=0

e: \(53\cdot47+53^2-2021^0\)

\(=53\left(53+47\right)-1\)

=5300-1

=5299

i: 38+(-21)+(-58)+(-6)

=(38-58)+(-21-6)

=-20-27

=-47

h: (2020+27)-[2020+(-73)]-129

=2020+27-2020+73-129

=100-129

=-29

`2021 . (-125) - 178 . (-2011) + (-2021).53`

`= -2021.125 + (-2021).53 - 178.(-2011)`

`= -2021 (125 + 53) - 178.(-2011)`

`= -2021 . 178 - 178 (-2011)`

`= 178(-2021  + 2011)`

` = 178. (-10)`

`= -1780`

\(=2021\left(-125-53\right)-178\cdot\left(-2011\right)\)

\(=2021\cdot\left(-178\right)-178\cdot\left(-2011\right)\)

\(=-178\left(2021-2011\right)=-178\cdot10=-1780\)

53.47=(50+3).(50-3)=50²+3²

                                     =2500+9=2509

53.47 = (50 + 3)(50 - 3) = 50² - 3² = 2500 - 9 = 2491

Ta có: \(b^2=ac\)

=>\(\frac{a}{b}=\frac{b}{c}\)

Đặt \(\frac{a}{b}=\frac{b}{c}=k\)

=>a=bk; b=ck

=>\(a=ck\cdot k=ck^2;b=ck\)

\(\frac{\left(a+b\right)^{2021}}{\left(b+c\right)^{2021}}=\frac{\left(ck^2+ck\right)^{2021}}{\left(ck+c\right)^{2021}}=\frac{\left\lbrack ck\left(k+1\right)\right\rbrack^{2021}}{\left\lbrack c\left(k+1\right)\right\rbrack^{2021}}=k^{2021}\)

\(\frac{a^{2021}+b^{2021}}{b^{2021}+c^{2021}}=\frac{\left(ck^2\right)^{2021}+\left(ck\right)^{2021}}{\left(ck\right)^{2021}+c^{2021}}\)

\(=\frac{c^{2021}\cdot k^{2021}\left(k^{2021}+1\right)}{c^{2021}\left(k^{2021}+1\right)}=k^{2021}\)

Do đó: \(\frac{\left(a+b\right)^{2021}}{\left(b+c\right)^{2021}}=\frac{a^{2021}+b^{2021}}{b^{2021}+c^{2021}}\)

53.47 = 2491

Kết bạn nhé

\(53\cdot47=2491\)