\(a,\left(x-36\right):\left(2\cdot3^2\right)=2^3\cdot3\\ \Leftrightarrow x-36=432\\ x=468\\ b,2^x=32\\ \Leftrightarrow x=5\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x^3=3^3\\ \Leftrightarrow x=3\\ d,1579+\left(625-x\right)=2023\\ \Leftrightarrow x=1579+625-2023\\ \Leftrightarrow x=181\)

A. \(\left(x-36\right):\left(2.3^2\right)=2^3.3\)

\(\left(x-36\right):\left(2.9\right)=8.3\)

\(\left(x-36\right):18=24\)

\(x-36=24.18\)

\(x-36=432\)

\(x=432+36\)

\(x=468\)

B. \(2^x=32\)

\(2^x=2^5\)

\(x=5\)

C. \(x^3=27\)

\(x^3=3^3\)

\(x=3\)

D. \(1579+\left(625-x\right)=2023\)

\(625-x=2023-1579\)

\(625-x=444\)

\(x=625-444\)

\(x=181\)

Tích của số chia và thương là: 111

Số chia là: 37

Thương là: 3

Do \(\left(10a+b\right)⋮13\Rightarrow10a+b=13k\left(k\in N\right)\)

\(\Rightarrow b=13k-10a\)

\(\Rightarrow a+4b=a+4.\left(13k-10a\right)\)

\(=a+52k-40a\)

\(=52k-39a\)

\(=13\left(4k-3a\right)⋮13\)

Vậy \(\left(10a+b\right)⋮13\Rightarrow\left(a+4b\right)⋮13\)

Do $\left(10a+b\right)⋮13\Rightarrow 10a+b=13k\left(k\in N\right)$

$\Rightarrow b=13k-10a$

$\Rightarrow a+4b=a+4.\left(13k-10a\right)$

$=a+52k-40a$

$=52k-39a$

$=13\left(4k-3a\right)⋮13$

Vậy $\left(10a+b\right)⋮13\Rightarrow \left(a+4b\right)⋮13$

A.35 x 34 + 35 x 38 + 65 x 75 + 65 x 45

Câu này không tính hợp lí được

B. 12x 53 + 53 x 172 - 53 x 84

= 53 x (12 + 172 - 84)

= 53 x 100

= 5300

C. 36 x 28 + 36 x 82 + 64 x 69 + 64 x 41

= 36 x (28+82) + 64 x (69+41)

= 36 x 110 + 64 x 110

= (36+64) x 110

= 100 x 110

= 11 000

D. A = 1+2+3+...+100

= (1+ 100) x 50

= 101 x 50

= 5050

E. B =17+19+...+101+103

= (103 +17) x [(103 - 17):2 + 1]

= 120 x 44

=5280

G. D= 3x (12+13+14+15) + 3x(8+7+6+5)

= 3 x (5+6+7+8+12+13+14+15)

= 3 x (20 x 4)

= 3 x 80

=240

Xét tổng 1 + 3 +5 +... + 99:

- Số số hạng là: (99-1):2+1= 50 số

- Tổng là: (1 + 99) x 50 : 2 = 2500

Vậy ta có biểu thức:

 \(2500=\left(x-2\right)^2\\ \Leftrightarrow50^2=\left(x-2\right)^2\\\Leftrightarrow \left[{}\begin{matrix}x-2=50\\x-2=-50\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=52\\x=-48\end{matrix}\right.\)

Vậy...

\(1+3+5+...+99=\left(x-2\right)^2\)

\(\Rightarrow\left[\left(99-1\right):2+1\right]\left(1+99\right):2=\left(x-2\right)^2\)

\(\Rightarrow50.100:2=\left(x-2\right)^2\)

\(\Rightarrow\left(x-2\right)^2=2500=50^2\)

\(\Rightarrow\left[{}\begin{matrix}x-2=50\\x-2=-50\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=52\\x=-48\end{matrix}\right.\)

\(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\)

\(\Rightarrow\left(3x-1\right)^{20}-\left(3x-1\right)^{10}=0\)

\(\Rightarrow\left(3x-1\right)^{10}\left[\left(3x-1\right)^2-1\right]=0\)

\(\Rightarrow\left(3x-1\right)^{10}\left(3x-1+1\right)\left(3x-1-1\right)=0\)

\(\Rightarrow\left(3x-1\right)^{10}.3x.\left(3x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3x=0\\3x-1=0\\3x-2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\3x=1\\3x=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)

x = 0 hoặc 1/3 hoặc 2/3

Diện tích được chồng lên nhau :

\(19.19=361\left(cm^2\right)\)

Hiệu diện tích các phần không bị chồng lên nhau :

\(\left(20-19\right).20=20\left(cm^2\right)\)

20cm2

\(...=57.\left(12+15+63\right)=57.90=5130\)

\(12.57+57.15+63.57\\ \Rightarrow57.\left(12+15+63\right).\\ \Rightarrow57.90=5130.\)

Ta có:

\(8=2^3\\ 31=31\)

\(\Rightarrow BCNN\left(8,31\right)=2^3.31=248.\)

\(\Rightarrow BCNN\left(8,31\right)=B\left(248\right)=\left\{0;248;496;744;...\right\}\)

Vì 31 là số nguyên tố

=>BC(8,31)=1