Lời giải:
Ô tô đi vận tốc 60 km/h trên quãng đường dài:

$300.\frac{2}{5}=120$ (km) 

Ô tô đi vận tốc 45 km/h trên quãng đường dài:
$300-120=180$ (km) 

Thời gian xe đi từ A-B: 

$120:60+180:45=6$ (h)

\(\frac{x+2}{4}+\frac{x-6}{12}=\frac{3x+6+x-6}{12}=\frac{4x}{12}=\frac{x}{3}\)là số nguyên suy ra \(x\)là bội của \(3\)

Vậy \(x\in B\left(3\right)\)thì thỏa mãn ycbt. 

Có \(\frac{6}{25}< \frac{6}{24}=\frac{1}{4}\)nên vòi thứ hai chảy nhanh hơn. 

Trong một giờ, cả hai vòi chảy được số phần bể là: 

\(\frac{6}{25}+\frac{1}{4}=\frac{49}{100}\)(bể) 

Bài 5: 

a) Đặt \(\left(3n+2,5n+3\right)=d\).

\(\Rightarrow\hept{\begin{cases}3n+2⋮d\\5n+3⋮d\end{cases}}\Rightarrow5\left(3n+2\right)-3\left(5n+3\right)=1⋮d\Rightarrow d=1\).

suy ra đpcm. 

b) Đề bài sai. 

Ví dụ: với \(n=3\): \(\frac{3n+2}{7n+1}=\frac{11}{22}\)không là phân số tối giản. 

a) \(\frac{7}{12}+\frac{x}{5}=\frac{1}{20}\)

\(\Rightarrow\frac{x}{5}=\frac{1}{20}-\frac{7}{12}\)

\(\Rightarrow\frac{x}{5}=\frac{3}{60}-\frac{35}{60}\)

\(\Rightarrow\frac{x}{5}=\frac{-8}{15}\)

b) \(\frac{-7}{x}+\frac{8}{15}=\frac{-1}{20}\)

\(\Rightarrow\frac{-7}{x}=\frac{-1}{20}-\frac{8}{15}\)

\(\frac{-7}{x}=\frac{-3}{60}-\frac{32}{60}\)

\(\frac{-7}{x}=\frac{-7}{12}\)

\(\Rightarrow x=12\)

a, \(\dfrac{x}{5}=\dfrac{1}{20}-\dfrac{7}{12}=-\dfrac{8}{15}\Rightarrow3x=-8\Leftrightarrow x=-\dfrac{8}{3}\)

b, \(-\dfrac{7}{x}=-\dfrac{7}{12}\Rightarrow x=12\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}=\frac{49}{50}\)

\(B=1.2+2.3+3.4+...+49.50\)

\(3B=1.2.3+2.3.3+3.4.3+...+49.50.3\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)

\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)

\(=49.50.51\)

\(B=\frac{49.50.51}{3}=49.50.17\)

\(50^2.A-\frac{B}{17}=49.50-49.50=0\)

Ta có \(111...11-222...22=\frac{1}{9}.999...99-\frac{2}{9}.999...99=\frac{1}{9}\left(10^{2n}-1\right)-\frac{2}{9}\left(10^n-1\right)\)

              2n cs 1             n cs 2                       2n cs 9                  n cs 9   

\(=\frac{10^{2n}-1-2.10^n+2}{9}=\frac{\left(10^n\right)^2-2.10^n+1}{9}=\left(\frac{10^n-1}{3}\right)^2\)

Nhận thấy \(10^n-1⋮3\)nên \(\frac{10^n-1}{3}\inℤ\)hay \(\left(\frac{10^n-1}{3}\right)^2\)là số chính phương

Vậy \(111...11-222...22\)là số chính phương.

           2n cs 1               n cs 2

\(RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG\)

Help mik với làm ơn

Sửa đề nhé: \(S=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{19.22}\)

\(S=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{19.22}\right)\)

\(S=\frac{1}{3}\left(\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{22-19}{19.22}\right)\)

\(S=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{19}-\frac{1}{22}\right)\)

\(S=\frac{1}{3}\left(1-\frac{1}{22}\right)\)

\(S=\frac{1}{3}.\frac{21}{22}\)

\(S=\frac{7}{22}\)