mn ơi giúp mình với thanh kiu nhìu

Để olm giúp em nhá

(99⁸⁹)⁶⁹ = 99⁶¹⁴¹ = (99²)³⁰⁷⁰.99 = (\(\overline{..01}\))³⁰⁷⁰.99 = \(\overline{..99}\)

6²⁰²¹ = (6⁵)⁴⁰⁴.6 = 7776⁴⁰⁴.6 = \(\overline{...76}.6\) = \(\overline{...56}\)

A=14²⁰²².16²⁰²²=(14.16)²⁰²²=224²⁰²²= (224²)¹⁰⁰¹= \(\overline{...76}\)¹⁰⁰¹=\(\overline{...76}\)

Lời giải:

$14.7^{2021}=35.7^{2021}-3.49^x$

$2.7^{2022}=5.7^{2022}-3.7^{2x}$

$3.7^{2x}=5.7^{2022}-2.7^{2022}=7^{2022}(5-2)=3.7^{2022}$

$\Rightarrow 7^{2x}=7^{2022}$

$\Rightarrow 2x=2022$

$\Rightarrow x=2022:2=1011$

Sr bạn mik biết kết quả nhưng mik ko biết trình bày thế nào
x=45
y=2
Sr rất nhiều

Sửa đề: \(C=\frac{2021}{\left(x-3\right)^2+\left(y+1\right)^2+1}\)

TA có: \(\left(x-3\right)^2\ge0\forall x\)

\(\left(y+1\right)^2\ge0\forall y\)

=>\(\left(x-3\right)^2+\left(y+1\right)^2\ge0\forall x,y\)

=>\(\left(x-3\right)^2+\left(y+1\right)^2+1\ge1\forall x,y\)

=>\(\frac{2021}{\left(x-3\right)^2+\left(y+1\right)^2+1}\le\frac{2021}{1}=2021\forall x,y\)

Dấu '=' xảy ra khi x-3=0 và y+1=0

=>x=3 và y=-1

\(2021S=2021+2021^2+2021^3+...+2021^{40}\\ 2021S-S=2021+2021^2+2021^3+...+2021^{40}-1-2021-2021^2-...-2021^{39}\\ 2020S=2021^{40}-1\\ S=\dfrac{2021^{40}-1}{2020}\)

Sửa đề: \(\left(x+\frac12\right)^{2021}=2020^{2021}\)

TA có: \(\left(x+\frac12\right)^{2021}=2020^{2021}\)

=>\(x+\frac12=2020\)

=>x=2020-0,5=2019,5

\(\dfrac{5^{2021}}{5^{2020}}\cdot5^2=5\cdot5^2=5^3\)

5²⁰²¹:5²⁰²⁰. 5²

= 5¹.5²

=5³

x=1
y=1