Lời giải:
$\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}$

$=\frac{2-1}{1.2}+\frac{4-3}{3.4}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}$

$=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{99}-\frac{1}{100}$

$=(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99})-(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{100})$
$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100})-2(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{100})$

$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100})-(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50})$

$=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}$

$\Rightarrow (\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}):(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100})=1$

Lời giải:
$\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}$

$=\frac{2-1}{1.2}+\frac{4-3}{3.4}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}$

$=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{99}-\frac{1}{100}$

$=(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99})-(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{100})$
$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100})-2(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{100})$

$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100})-(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50})$

$=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}$

$\Rightarrow (\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}):(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100})=1$

\(\dfrac{2a-1}{4}-\dfrac{3a+2}{3}+2a\)

\(=\dfrac{3\left(2a-1\right)-4\left(3a+2\right)+24a}{12}\)

\(=\dfrac{6a-3-12a-8+24a}{12}=\dfrac{18a-11}{12}\)

Bài 1:

a: \(\dfrac{6}{13}\cdot\dfrac{5}{18}+\dfrac{-7}{24}+\dfrac{5}{18}\cdot\dfrac{7}{13}+\dfrac{-5}{24}\)

\(=\dfrac{5}{18}\left(\dfrac{6}{13}+\dfrac{7}{13}\right)+\left(-\dfrac{7}{24}+\dfrac{-5}{24}\right)\)

\(=\dfrac{5}{18}+\dfrac{-12}{24}=\dfrac{5}{18}-\dfrac{1}{2}=\dfrac{-4}{18}=-\dfrac{2}{9}\)

b: \(4\dfrac{1}{20}\left(-\dfrac{2}{3}\right)^2+\left(0,8-\dfrac{8}{15}\right):\dfrac{-4}{7}\)

\(=\dfrac{81}{20}\cdot\dfrac{4}{9}+\left(\dfrac{4}{5}-\dfrac{8}{15}\right)\cdot\dfrac{-7}{4}\)

\(=\dfrac{9}{5}+\dfrac{4}{15}\cdot\dfrac{-7}{4}=\dfrac{9}{5}-\dfrac{7}{15}=\dfrac{20}{15}=\dfrac{4}{3}\)

c: \(\left(3\dfrac{3}{29}-3\dfrac{1}{5}+2\dfrac{7}{11}\right)-\left(2\dfrac{3}{29}-3\dfrac{4}{11}\right)\)

\(=3+\dfrac{3}{29}-3-\dfrac{1}{5}+2+\dfrac{7}{11}-2-\dfrac{3}{29}+3+\dfrac{4}{11}\)

\(=\left(3-3+2-2+3\right)+\left(\dfrac{3}{29}-\dfrac{3}{29}\right)+\left(\dfrac{7}{11}+\dfrac{4}{11}\right)-\dfrac{1}{5}\)

\(=3+1-\dfrac{1}{5}=4-\dfrac{1}{5}=3,8\)

d: \(\dfrac{-1}{4}\cdot\dfrac{152}{11}+\dfrac{1}{4}\cdot\dfrac{-68}{11}\)

\(=\dfrac{1}{4}\left(-\dfrac{152}{11}-\dfrac{68}{11}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{-220}{11}=\dfrac{1}{4}\cdot\left(-20\right)=-5\)

Bài 2:

a: \(\dfrac{5}{6}+\left(5x+\dfrac{3}{2}\right):\dfrac{8}{15}=2\dfrac{1}{12}\)

=>\(\left(5x+\dfrac{3}{2}\right):\dfrac{8}{15}=\dfrac{25}{12}-\dfrac{5}{6}=\dfrac{25-10}{12}=\dfrac{15}{12}=\dfrac{5}{4}\)

=>\(5x+\dfrac{3}{2}=\dfrac{8}{15}\cdot\dfrac{5}{4}=\dfrac{2}{3}\)

=>\(5x=\dfrac{2}{3}-\dfrac{3}{2}=\dfrac{-5}{6}\)

=>\(x=-\dfrac{1}{6}\)

b: \(\dfrac{2}{3}\left(x+\dfrac{9}{5}\right)-\dfrac{3}{10}\left(5x-\dfrac{1}{3}\right)=\dfrac{7}{15}\)

=>\(\dfrac{2}{3}x+\dfrac{6}{5}-\dfrac{3}{2}x+\dfrac{1}{10}=\dfrac{7}{15}\)

=>\(x\left(\dfrac{2}{3}-\dfrac{3}{2}\right)=\dfrac{7}{15}-\dfrac{6}{5}-\dfrac{1}{10}=\dfrac{14-36-3}{30}=\dfrac{-25}{30}=\dfrac{-5}{6}\)

=>\(x\cdot\dfrac{-5}{6}=\dfrac{-5}{6}\)

=>x=1

c: \(\dfrac{3}{5}x-\dfrac{1}{2}=\dfrac{7}{10}\)

=>\(\dfrac{3}{5}x=\dfrac{7}{10}+\dfrac{1}{2}=\dfrac{12}{10}=\dfrac{6}{5}\)

=>x=2

d: \(\dfrac{1}{5}+\dfrac{4}{5}:x=-1\)

=>\(\dfrac{4}{5}:x=-1-\dfrac{1}{5}=\dfrac{-6}{5}\)

=>\(x=-\dfrac{4}{5}:\dfrac{6}{5}=\dfrac{-2}{3}\)

Bài 1:

a. 

$=(\frac{6}{13}.\frac{5}{18}+\frac{5}{18}.\frac{7}{13})-(\frac{7}{24}+\frac{5}{24})$

$=\frac{5}{18}(\frac{6}{13}+\frac{7}{13})-\frac{12}{24}$

$=\frac{5}{18}.1-\frac{1}{2}=\frac{5}{18}-\frac{1}{2}=\frac{-2}{9}$

b.

$=\frac{81}{20}.\frac{4}{9}+\frac{4}{15}.\frac{-7}{4}$

$=\frac{9}{5}+\frac{-7}{15}=\frac{4}{3}$

c.

$=3\frac{3}{29}-3\frac{1}{5}+2\frac{7}{11}-2\frac{3}{29}+3\frac{4}{11}$

$=(3-3+2-2+3)+(\frac{3}{29}-\frac{3}{29})+(\frac{7}{11}+\frac{4}{11})$

$=3+0+\frac{11}{11}=3+1=4$

d.

$=\frac{1}{4}.\frac{-152}{11}+\frac{1}{4}.\frac{-68}{11}$

$=\frac{1}{4}(\frac{-152}{11}+\frac{-68}{11})$

$=\frac{1}{4}.(-20)=-5$

\(A=\dfrac{2n+2}{n+2}+\dfrac{5n+17}{n+2}-\dfrac{3n}{n+2}\)

\(=\dfrac{2n+2+5n+17-3n}{n+2}=\dfrac{4n+19}{n+2}\)

Để A là số nguyên thì \(4n+19⋮n+2\)

=>\(4n+8+11⋮n+2\)

=>\(11⋮n+2\)

=>\(n+2\in\left\{1;-1;11;-11\right\}\)

=>\(n\in\left\{-1;-3;9;-13\right\}\)

Lời giải:
\(A=\frac{2n+2}{n+2}+\frac{5n+17}{n+2}-\frac{3n}{n+2}=\frac{2n+2+5n+17-3n}{n+2}\\ =\frac{4n+19}{n+2}=\frac{4(n+2)+11}{n+2}=4+\frac{11}{n+2}\)

Với $n$ nguyên, để $A$ nguyên thì $11\vdots n+2$

$\Rightarrow n+2\in\left\{1; -1; 11; -11\right\}$

$\Rightarrow n\in\left\{-1; -3; 9; -13\right\}$

tạo ra nước nóng nhờ ánh sáng mặt trời; tạo hệ thống sưởi ấm, làm mát, thông gió; giúp chưng cất nước, biến nước mặn hoặc nước lợ thành nước uống được; dùng để nấu nướng, làm khô, khử trùng…

#sarin

Nhan Chào bạn mình đang sống trong một gia đình nghèo Chào bán cho các em học sinh giỏi cấp phép hoạt động trong cuộc sống này không chê vào đâu rồi em cảm ơn dành riêng cho e xin phép được chưa được biết đến như em nói với em qua nho nho chuc Chào bán cho e xin cái tên được nhắc tới nhiều 

1 How about playing badminton this weekend
2 I am more interested in basketball than my brother

😁

Lời giải:
$4\equiv 1\pmod 3$

$\Rightarrow 4^{2024}+1\equiv 1^{2024}+1\equiv 2\pmod 3$
Một scp khi chia cho 3 thì chỉ có thể có số dư là $0$ hoặc $1$

$\Rightarrow 4^{2024}+1$ không phải số chính phương.

a: Số học sinh khá là \(45\cdot40\%=18\left(bạn\right)\)

Số học sinh còn lại là 45-18=27(bạn)

Số học sinh trung bình là \(27\left(1-\dfrac{5}{9}\right)=27\cdot\dfrac{4}{9}=12\left(bạn\right)\)

b: Số học sinh nữ là \(12:\dfrac{5}{6}=12\cdot\dfrac{6}{5}=\dfrac{72}{5}=14,4\left(bạn\right)\)

=>Đề sai rồi bạn