\(32,5x4,5+32,5x5,5+3,25\)

\(=32,5x4,5+32,5x5,5+32,5x0,1\)

\(=32,5x\left(4,5+5,5+0,1\right)\)

\(=32,5x10,1=328,25\)

328,25

\(68=2^2.17\)

\(21=3.7\)

Nên \(68\&21\) không chia hết cho số nào.

\(\left(2^2\right)^8.2^{20}=2^{16}.2^{20}=2^{36};\left(3^2\right)^{12}.\left(3^3\right)^5.\left(3^4\right)^4\)\(=3^{24}.3^{15}.3^{16}=3^{55}\)

\(64^3.4^5.16^2=\left(4^3\right)^3.4^5.\left(4^2\right)^2=4^9.4^5.4^4=4^{18}\)

Sửa đề:

\(A=\dfrac{3}{5}+\dfrac{3}{20}+\dfrac{3}{44}+\dfrac{3}{77}\)

\(A=2.\left(\dfrac{3}{5}+\dfrac{3}{20}+\dfrac{3}{44}+\dfrac{3}{77}\right)\)

\(A=\dfrac{6}{10}+\dfrac{6}{40}+\dfrac{6}{88}+\dfrac{6}{154}\)

\(A=6.\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}\right)\)

\(A=6.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}\right)\)

\(A=6.\left(\dfrac{1}{2}-\dfrac{1}{14}\right)\)

\(A=6.\dfrac{6}{14}\)

\(A=\dfrac{36}{14}=\dfrac{18}{7}\)

\(=\dfrac{1}{5}.3+\dfrac{1}{5}.\dfrac{3}{4}+\dfrac{1}{11}.\dfrac{3}{4}+\dfrac{1}{11}.\dfrac{3}{7}\)

\(=\dfrac{1}{5}.\left(3+\dfrac{3}{4}\right)+\dfrac{1}{11}.\left(\dfrac{3}{4}+\dfrac{3}{7}\right)\)

\(=\dfrac{1}{5}.\dfrac{15}{4}+\dfrac{1}{11}.\dfrac{33}{28}=\dfrac{3}{4}+\dfrac{3}{28}=\dfrac{6}{7}\)

1)

xy + x - 4y = 12

x + y(x - 4) = 12

y(x - 4) = 12 - x

\(y=\dfrac{-x+12}{x-4}\)

Vì \(x,y\inℕ\) nên

\(\left(-x+12\right)⋮\left(x-4\right)\)

\(\left(-x+12\right)-\left(x-4\right)⋮\left(x-4\right)\)

\(16⋮\left(x-4\right)\)

\(\left(x-4\right)\inƯ\left(16\right)\)

\(\left(x-4\right)\in\left\{1;-1;2;-2;4;-4;8;-8;16;-16\right\}\)

\(x\in\left\{5;3;6;2;8;0;12;-4;20;-12\right\}\)

\(y\in\left\{\dfrac{-5+12}{5-4};\dfrac{-3+12}{3-4};\dfrac{-6+12}{6-4};\dfrac{-2+12}{2-4};\dfrac{-8+12}{8-4};\dfrac{-0+12}{0-4};\dfrac{-12+12}{12-4};\dfrac{4+12}{-4-4};\dfrac{-20+12}{20-4};\dfrac{12+12}{-12-4}\right\}\)

\(y\in\left\{7;-9;3;-5;1;-3;0;-2;-\dfrac{1}{2};-\dfrac{7}{5}\right\}\)

\(\left(x;y\right)\in\left\{\left(5;7\right);\left(3;-9\right);\left(6;3\right);\left(2;-5\right);\left(8;1\right);\left(0;-3\right);\left(12;0\right);\left(-4;-2\right);\left(20;-\dfrac{1}{2}\right);\left(-12;-\dfrac{7}{5}\right)\right\}\)

Mà \(x,y\inℕ\) nên các giá trị cần tìm là \(\left(x;y\right)\in\left\{\left(5;7\right);\left(6;3\right);\left(8;1\right);\left(12;0\right)\right\}\)

2)

(2x + 3)(y - 2) = 15

\(\left(2x+3\right)\inƯ\left(15\right)\)

\(\left(2x+3\right)\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

Ta lập bảng

Mà \(x,y\inℕ\) nên các giá trị cần tìm là \(\left(x;y\right)\in\left\{\left(0;7\right);\left(1;5\right);\left(6;3\right)\right\}\)

các thầy cô ơi giúp em vs ạ mai em phải nộp r ạ!!!

Đưa các số đó về cơ số nguyên tố (2;3;5..).

Có hai trường hợp:

Trường hợp 1: Hai lũy thừa có cùng số mũ

Nhân: giữ nguyên số mũ, nhân 2 cơ số: a^m.b^m=(a.b)^m

chia: giữ nguyên số mũ, chia 2 cơ số: a^m:b^m=(a:b)^m

Trường hợp 2: Khác số mũ

Viết về dạng lũy thừa của lũy thừa để đưa 2 lũy thừa về cùng cơ số hoặc số mũ

a^m.bⁿ=a^p.q.b^p.r=(a^p)^q.(b^p)^r=c^q.c^r

a^m:bⁿ=a^p.q:b^p.r=(a^p)^q:(b^p)^r=c^q:c^r

a^m.bⁿ=a^p.q.b^p.r=(a^q)^p.(b^r)^p=c^p.d^p

a^m:bⁿ=a^p.q:b^p.r=(a^q)^p:(b^r)^p=c^p:d^p

\(\overline{abc}+\overline{ab}+a=628\)

\(100xa+10xb+c+10xa+b+a=628\)

\(111xa+11xb+c=628\)

\(\Rightarrow\left\{{}\begin{matrix}a=5\\b=6\\c=7\end{matrix}\right.\) thỏa đề bài

Mỗi gói kẹo chứa số kẹo là: 192 : 8 = 24 cái kẹo 

5 gói kẹo có số kẹo là: 24 x 5 = 120 cái kẹo 

Số kẹo bà còn lại là: 192 - 120 = 72 cái kẹo

72

\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{98.99}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{99}\)

\(=\dfrac{1}{3}-\dfrac{1}{99}\)

\(=\dfrac{33}{99}-\dfrac{1}{99}\)

\(=\dfrac{32}{99}\)

\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{98.99}\\ =\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{99}\\ =\dfrac{1}{3}-\dfrac{1}{99}\\ =\dfrac{32}{99}\)