Bài 2: 

Ta có: \(16x+40=10\cdot3^2+5\left(1+2+3\right)\)

\(\Leftrightarrow16x+40=90+30\)

\(\Leftrightarrow16x=80\)

hay x=5

Bài 1 :

[( 35 - 5 ) : 3 ]³ + 3

= [30 : 3]³ + 3

= 10³ + 3

= 1000 + 3

= 1003

Đây nha bạn!!!

Chúc bạn học tốt!!!

Câu này dễ quá

\(2\cdot\left[\left(7-3^{2021}:3^{2020}\right):2^2+99\right]-100\)

\(=2\cdot\left[\left(7-3\right):4+99\right]-100\)

\(=2\cdot\left(4:4+99\right)-100\)

\(=2\cdot\left(1+99\right)-100\)

\(=2\cdot100-100\)

\(=200-100\)

\(=100\)

#Urushi☕

M = 2019 / 2020 + 2020 / 2021 + 2021/ 2019

    = ( 1 - 1/2019 ) + ( 1 - 1/2020 ) + ( 1+ 2/2021 )

    = 3 - 1/2019 - 1/2020 + 2/2021

A - 3 = - 1/2019 - 1/2020 + 2/2021 > 0 

Vậy M > 3 

Ko bít có làm đúng ko nx nhưng bn cứ tham khảo bài mk nhé !

CẢM  ƠN X 100000000000000000

a, 23 + ( -77 ) + ( -23 ) + 77

= [ 23 + ( -23 ) ] + [ ( -77 ) + 77 ]

= 0 + 0 = 0

b, ( -2020 ) + 2021 + 21 + ( -22 )

= ( - 2020 ) + 2021 + 21 + ( -21 ) + ( -1 )

= [ ( -2020 ) + 2021 + ( -1 ) ] + [ 21 + ( -21 ) ]

= 0 + 0 = 0

a) 23 + (-77) + (-23) + 77 =
[23 + (-23)] + [(-77) + 77]
=0+0=0
b) (-2 020) + 2 021 + 21 + (-22)
=[(-2 020) + 2 021] + [21 + (-22)]
=1+(-1)
= 0. 

a, - \(\dfrac{1}{3}\).\(xy\).(3\(x^3\).y² - 6\(x^2\) + y²)

= - \(x^4\).y³ + 2\(x^3\).y - \(\dfrac{1}{3}\).\(xy^3\)

b, (2\(x\) -3).(4\(x\)² + 6\(x\) + 9)

 = (2\(x\))³ - 3³

= 8\(x^3\) - 27 

3+5+7+...+77+79+100=

(3+5+7+...+77+79)+100=

{(3+79).[(79-3)/2+1]/2} + 100=

{82.[76/2+1]/2} + 100=

{82.[38+1]/2} + 100=

{82.39/2} + 100=

{82.19,5}+ 100=

{82.19,5}+ 100=

1599+100=

1699

\(\left(5^{2021}-5^{2020}\right):5^{2020}=5^{2021}:5^{2020}-5^{2020}:5^{2020}=5-1=4\)

làm ơn đi, mình đang cần gắp.