Lời giải:

Do $p$ là số nguyên tố lớn hơn $3$ nên $p$ không chia hết cho $3$. Do đó $p$ chia $3$ dư $1$ hoặc dư $2$

Nếu $p$ chia $3$ dư $1$. Đặt $p=3k+1$ với $k$ tự nhiên.

Khi đó: $8p+1=8(3k+1)+1=24k+9=3(8k+3)\vdots 3$. Mà $8p+1>3$ với mọi $p$ nguyên tố nên $8p+1$ không là số nguyên tố (trái đề)

$\Rightarrow p$ chia $3$ dư $2$.

Đặt $p=3k+2$

Khi đó:

$4p+1=4(3k+2)+1=12k+9=3(4k+3)\vdots 3$. Mà $4p+1>3$ với mọi $p$ nguyên tố nên $4p+1$ là hợp số.

\(-\dfrac{3}{4}+x=-\dfrac{7}{12}\\ x=-\dfrac{7}{12}-\left(-\dfrac{3}{4}\right)\\ x=-\dfrac{7}{12}+\dfrac{3}{4}\\ x=\dfrac{-7+3.3}{12}\\ x=\dfrac{-7+9}{12}\\ x=\dfrac{2}{12}=\dfrac{1}{6}\)

\(\dfrac{-3}{4}+x=-\dfrac{7}{12}\)

\(x=\dfrac{-7}{12}-\dfrac{-3}{4}\)

\(x=\dfrac{-7}{12}+\dfrac{9}{12}\)

\(x=\dfrac{2}{12}=\dfrac{1}{6}\)

Bài 1:

a; Các đoạn thẳng có tên trong hình là:

AB; AD; AE; BC; BD; DE; DC

b; 

Các đoạn thẳng trên đường thẳng xy là:

MN; MP; MQ; NP; NP; PQ

Bài 2:

Vì M năm giữa A và B nên

AB = AM + MB

MB = AB - AM = 8 - 2 = 6 (cm)

Vì N nằm giữa M và B nên 

MN + NB = MB 

NB = MB - MN = 6  - 3,5 = 2,5 (cm)

Kết luận: MB = 6cm; NB = 2,5 cm

Đổi: 8 phút `=4/15(h)`

Quãng đường mà ô tô đi được là: 

`40 xx 4/15 = 16/3(km) `

Đổi: 5 phút `=1/12(h)`

Để đi hết quãng đường đó trong 5 phút thì xe đó phải đi với vận tốc là: 

`16/3 : 1/12=64`(km/h)

\(\dfrac{30}{-x}=-\dfrac{6}{5}\)

\(\dfrac{30}{-x}=\dfrac{-6\cdot5}{5\cdot5}\)

\(\dfrac{30}{-x}=\dfrac{-30}{25}\)

\(\Rightarrow\dfrac{30}{-x}=\dfrac{30}{-25}\)

\(\Rightarrow x=25\)

30/(-x) = -6/5

-x.(-6) = 30.5

6x = 150

x = 150 : 6

x = 25

-12/18 - (-21/35)

= -2/3 + 3/5

= -10/15 + 9/15

= -1/15

\(\dfrac{-12}{18}\) - \(\dfrac{-21}{35}\) 

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

= \(\dfrac{-10}{15}\) + \(\dfrac{9}{15}\)

= \(-\dfrac{1}{15}\)

S =  \(\dfrac{1}{1.3}\)+\(\dfrac{1}{2.4}\)+...+\(\dfrac{1}{97.99}\)+\(\dfrac{1}{98.100}\) - \(\dfrac{49}{99}\)

S = (\(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{97.99}\))+(\(\dfrac{1}{2.4}\)+\(\dfrac{1}{4.6}\)+\(\dfrac{1}{6.8}\)+...+\(\dfrac{1}{98.100}\))- \(\dfrac{49}{99}\)

S = \(\dfrac{1}{2}\).(\(\dfrac{2}{1.3}\)+\(\dfrac{2}{3.5}\)+...+\(\dfrac{2}{97.99}\))+\(\dfrac{1}{2}\)(\(\dfrac{2}{2.4}\)+\(\dfrac{2}{4.6}\)+\(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{98.100}\))-\(\dfrac{49}{99}\)

S =\(\dfrac{1}{2}\).(\(\dfrac{1}{1}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+...+\(\dfrac{1}{97}\)-\(\dfrac{1}{99}\))+\(\dfrac{1}{2}\).(\(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{98}\)-\(\dfrac{1}{100}\))-\(\dfrac{49}{99}\)

S = \(\dfrac{1}{2}\).(\(\dfrac{1}{1}\)-\(\dfrac{1}{99}\))+\(\dfrac{1}{2}\).(\(\dfrac{1}{2}\)-\(\dfrac{1}{100}\)) - \(\dfrac{49}{99}\)

S = \(\dfrac{1}{2}\).\(\dfrac{98}{99}\) + \(\dfrac{1}{2}\).\(\dfrac{49}{100}\) - \(\dfrac{49}{99}\)

S = \(\dfrac{49}{99}\) + \(\dfrac{49}{200}\) - \(\dfrac{49}{99}\)

S = (\(\dfrac{49}{99}\)- \(\dfrac{49}{99}\)) + \(\dfrac{99}{200}\)

S = 0 + \(\dfrac{49}{200}\)

S = \(\dfrac{49}{200}\)

Bài \(13\):
\(C=\dfrac{3}{\left(1\cdot2\right)^2}+\dfrac{5}{\left(2\cdot3\right)^2}+\dfrac{7}{\left(3\cdot4\right)^2}+...+\dfrac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(=\dfrac{3}{1\cdot4}+\dfrac{5}{4\cdot9}+\dfrac{7}{9\cdot16}+...+\dfrac{n^2+2n+1-n^2}{n^2\left(n+1\right)^2}\)
\(=\dfrac{4-1}{1\cdot4}+\dfrac{9-4}{4\cdot9}+\dfrac{16-9}{9\cdot16}+...+\dfrac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}\)
\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{n^2}-\dfrac{1}{\left(n+1\right)^2}\)
\(=1-\dfrac{1}{\left(n+1\right)^2}=\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\)

Bài \(10\):
\(B=\dfrac{5}{2\cdot1}+\dfrac{4}{1\cdot11}+\dfrac{3}{11\cdot2}+\dfrac{1}{2\cdot15}+\dfrac{13}{15\cdot4}\)
\(=7\left(\dfrac{5}{2\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{1}{14\cdot15}+\dfrac{13}{15\cdot28}\right)\)
\(=7\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{28}\right)\)
\(=7\left(\dfrac{1}{2}-\dfrac{1}{28}\right)=7\cdot\dfrac{13}{28}=\dfrac{13}{4}\)