\(72^{81}=\left(...2^4\right)^{20}\cdot72=...6\cdot...2=...6\)

\(82^{83}=\left(...2^4\right)^{21}:82=\frac{...6}{...2}=...3\)

cảm ơn bạn nhưng cái đề bài không sai đâu, nó là lũy thừa tầng đấy

B = 1 + 5² +  5³ + ... + 5²⁰¹⁹

 5B = 5 + 5² + 5³ + ... + 5²⁰²⁰

Lấy 5B trừ B theo vế ta có : 

5B - B = (5 + 5² + 5³ + ... + 5²⁰²⁰) - (1 + 5² +  5³ + ... + 5²⁰¹⁹)

  4B     = 5²⁰²⁰ - 1

    B     = \(\frac{5^{2020}-1}{4}\)

Vậy B = \(\frac{5^{2020}-1}{4}\)

\(B=1+5+5^2+5^3+...+5^{2019}\)

\(\Leftrightarrow5B=5+5^2+5^3+...+5^{2020}\)

\(\Leftrightarrow5B-B=\left(5+5^2+...+5^{2020}\right)-\left(1+5+5^2+...+5^{2019}\right)\)

\(\Leftrightarrow4B=5^{2020}-1\Leftrightarrow B=\frac{5^{2020}-1}{4}\)

\(\left|x-\frac{3}{4}\right|\ge0\)

Dấu "=" xảy ra tại \(x=\frac{3}{4}\)

\(\left|x+\frac{2}{3}\right|\ge0\Rightarrow\left|x+\frac{2}{3}\right|+2\ge2\)

Dấu "=" xảy ra tại \(x=-\frac{2}{3}\)

\(a;\left|x-\frac{3}{4}\right|\ge0\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x-\frac{3}{4}=0\Leftrightarrow x=\frac{3}{4}\)

Vậy |x-3/4| min là 0 chỉ khi x=3/4

\(b;\left|x+\frac{2}{3}\right|+2\ge2\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x+\frac{2}{3}=0\Leftrightarrow x=-\frac{2}{3}\)

Vậy |x+2/3|+2 min là 2 chỉ khi x=-2/3

Ta có:\(\left(\frac{5}{2}\right)^{-3}\cdot\left(\frac{-2}{5}\right)^2=\left(\frac{2}{5}\right)^3\cdot\left(\frac{2}{5}\right)^2=\left(\frac{2}{5}\right)^5< \left(\frac{2}{5}\right)^n\Leftrightarrow5< n\)

a. 4872 - [567 + (124 - 31) x 38]

= 4872 - [567 + 93 x 38]

= 4872 - [567 + 3534]

= 4872 - 4101

= 771

b. 8³ : 2⁴ + 5¹³ : 5¹¹ x 3⁰

= 32 + 25

= 57

c. 748 x 19 -19 + 19 x 253 

= 748 x 19 - 19 x 1 + 19 x 253

= 19 x (748 - 1 + 253)

=  19 x 1000

= 19000

d. 3³ x 48 + 3⁴ x 42 + 81 x 42

= 3⁴ x 16 + 3⁴ x 42 + 3⁴ x 42

= 81 x (16 + 42 + 42)

= 81 x 100

= 8100

~Study well~

#Seok_Jin

a.4872-[567+(124-31).38]

a=4872-[567+93.38]

a=4872-[567+3534]

a=4872-4101

a=771

Câu a thôi nhé , các câu còn lại cậu tự làm nhé!

a=12

b=21

Ta có : \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}\)

= \(\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{n^2}\right)\)

< \(\frac{1}{2^2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{\left(n-\right).n}\right)\)

= \(\frac{1}{4}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)

= \(\frac{1}{4}.\left(1-\frac{1}{n}\right)\)

<  \(\frac{1}{4}.1=\frac{1}{4}\)

 \(\Rightarrow\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}< \frac{1}{4}.\left(1-\frac{1}{n}\right)< \frac{1}{4}\)

 \(\Rightarrow\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}< \frac{1}{4}\left(đpcm\right)\)

5x + 3x + 6x = 280

14x = 280 

x = 280 : 14 

x = 20

Vậy x = 20 

( 2x - 18 ) - 125 = 325 

( 2x - 18 ) = 325 + 125 

2x - 18 = 450

2x = 450 + 18 

2x = 468

x = 234

Vậy x = 234 

\(5x+3x+6x=280\)

\(\Rightarrow x\left(5+3+6\right)=280\)

\(\Rightarrow x.14=280\)

\(\Rightarrow x=20\)

b, \(\left(2x-18\right)-125=325\)

\(\Rightarrow2x-18=450\)

\(\Rightarrow2x=468\)

\(\Rightarrow x=234\)