nếu mik trả lời bn có tk mik ko z

tất nhiên bạn!

ok đề bài là j z bạn

Bài 2:

a: \(17-x=3\)

=>\(x=17-3\)

=>x=14(nhận)

b: \(2\cdot\left(x-1\right):3=6\)

=>\(2\left(x-1\right)=6\cdot3=18\)

=>x-1=18/2=9

=>x=9+1=10(nhận)

c: \(x+\left(-2\right)=\left(-11\right)+7\)

=>\(x-2=-4\)

=>\(x=-4+2=-2\left(loại\right)\)

d: \(\left(x-1\right)^2-5=20\)

=>\(\left(x-1\right)^2=25\)

=>\(\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=-4\left(loại\right)\end{matrix}\right.\)

Câu 3:

a: Đặt *=a

\(\overline{57a3}⋮9\)

=>\(5+7+a+3⋮9\)

=>\(a+15⋮3\)

mà 0<=a<=9

nên a=3

=>*=a

b: \(A=123\cdot7+8+9\)

123*7 là số lẻ

9 là số lẻ

=>123*7+9 chia hết cho 2

mà 8 chia hết cho 2

nên \(A=123\cdot7+9+8⋮2\)

\(123\cdot7⋮3;9⋮3;8⋮̸3\)

=>\(A=123\cdot7+9+8⋮̸3\)

c: \(B=3\cdot5\cdot7+10^{50}\)

\(=5\cdot3\cdot7+5\cdot5^{49}\cdot2^{49}\)

\(=5\left(3\cdot7+5^{49}\cdot2^{49}\right)⋮5\)

=>B là hợp số

a: \(61\cdot45+61\cdot23-68\cdot51\)

\(=61\left(45+23\right)-68\cdot51\)

\(=68\cdot61-68\cdot51\)

\(=68\left(61-51\right)=68\cdot10=680\)

b: \(3\cdot5^2-\left(75-4\cdot2^3\right)\)

\(=75-75+4\cdot8\)

\(=4\cdot8=32\)

c: \(36:\left\{2^2\cdot5-\left[30-\left(5-1\right)^2\right]\right\}\)

\(=\dfrac{36}{20-30+4^2}\)

\(=\dfrac{36}{-10+16}=\dfrac{36}{6}=6\)

d: \(\left(12\cdot49-3\cdot2^2\cdot7^2\right):\left(2020\cdot2021\right)\)

\(=\dfrac{\left(12\cdot49-12\cdot49\right)}{2020\cdot2021}=0\)

a) -72.17+72.31 - 36 . 228

 = -72.17 + 72. 31 - 36.2.114

 = -72 .17 + 72 . 31 -72 . 114

 = -72. ( 17-31+114)

 = -72 . 100

 = -7200

b) ..........

 = -7 . ( 16-36.9) -(-125) 

 = -7 . ( 16-324) +125

 =-7 . ( -308) +125

 = 2156+125

 = 2281

$a)$

$-72.17+72.31-36.228$

$=-72.17+72.31-36.2.114$

$=-72.17+72.31-72.114$

$=72(31-17-114)$

$=72.(-100)=-7200$

$b)$$-7.[{(-2)}^4+(-36):(-3^2)]-{(-5)}^3$

$=-7.[16-36:(-9)]+125$