idk :)

Các loại cây trong rừng phòng hộ (phi lao, đước,…) sinh trưởng nhanh, cành lá xum xuê, hệ rễ phát triển, rễ cọc đâm sâu, chịu được gió bão, cát lấp,… làm giảm bớt tác động của cát và sóng tới đê biển.

\(\dfrac{-2}{5}\cdot\dfrac{4}{7}+\dfrac{-3}{5}\cdot\dfrac{2}{7}+\dfrac{-3}{5}\)

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

\(=\dfrac{-2}{5}\cdot\dfrac{4}{7}+\dfrac{-3}{5}\cdot\dfrac{9}{7}\)

\(=\dfrac{-8-27}{35}=\dfrac{-35}{35}=-1\)

\(\dfrac{20}{1\cdot6}+\dfrac{20}{6\cdot11}+...+\dfrac{20}{101\cdot106}\)

\(=4\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{101\cdot106}\right)\)

\(=4\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{101}-\dfrac{1}{106}\right)\)

\(=4\left(1-\dfrac{1}{106}\right)=4\cdot\dfrac{105}{106}=\dfrac{105\cdot2}{53}=\dfrac{210}{53}\)

Bạn thử tham khảo nhé!  

a: \(\dfrac{4}{13}+\dfrac{-12}{39}=\dfrac{4}{13}-\dfrac{4}{13}=0\)

b: \(\dfrac{27}{23}-\dfrac{-5}{21}-\dfrac{4}{23}+\dfrac{16}{21}+\dfrac{1}{2}\)

\(=\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)

\(=1+1+\dfrac{1}{2}=\dfrac{5}{2}\)

c: \(\dfrac{-8}{9}+\dfrac{1}{9}\cdot\dfrac{2}{9}+\dfrac{1}{9}\cdot\dfrac{7}{9}\)

\(=\dfrac{-8}{9}+\dfrac{1}{9}\left(\dfrac{2}{9}+\dfrac{7}{9}\right)\)

\(=\dfrac{-8}{9}+\dfrac{1}{9}=\dfrac{-7}{9}\)

d: \(\dfrac{2}{\left(-3\right)^2}+\dfrac{5}{-12}-\dfrac{-3}{4}\)

\(=\dfrac{2}{9}-\dfrac{5}{12}+\dfrac{3}{4}\)

\(=\dfrac{8}{36}-\dfrac{15}{36}+\dfrac{27}{36}=\dfrac{19}{36}\)

- Với \(p=3\Rightarrow8p-1=8.3-1=23\) là số nguyên tố và \(8p+1=25\) là hợp số

- Với \(p\ne3\Rightarrow p\) không chia hết cho 3

\(\Rightarrow p=3k+1\) hoặc \(p=3k+2\)

Với \(p=3k+1\Rightarrow8p+1=8\left(3k+1\right)+1=24k+9=3\left(8k+3\right)\) chia hết cho 3 \(\Rightarrow\) là hợp số

Với \(p=3k+2\Rightarrow8p-1=8\left(3k+2\right)-1=24k+15=3\left(8k+5\right)\) chia hết cho 3 \(\Rightarrow\) là hợp số

Vậy số còn lại luôn là hợp số

Your city is bigger than my city

\(\dfrac{-2}{7}\cdot\dfrac{7}{4}+\dfrac{-2}{7}\cdot\dfrac{-3}{4}-2\dfrac{1}{7}\)

\(=\dfrac{-2}{7}\left(\dfrac{7}{4}-\dfrac{3}{4}\right)-\dfrac{15}{7}\)

\(=\dfrac{-2}{7}-\dfrac{15}{7}=\dfrac{-17}{7}\)

Lời giải:

Gọi chiều dài và chiều rộng ban đầu là $a$ và $b$ (m) 

Theo bài ra ta có:

$(a+36)b.0,84=ab.1,05$

$\Rightarrow b[(a+36).0,84-1,05a]=0$

$\Rightarrow (a+36).0,84-1,05a=0$

$\Rightarrow 30,24=0,21a$

$\Rightarrow a=144$ (m) 

Vậy chiều dài mới là: $a+36=144+36=180$ (m)

Bài 2:

a: \(\dfrac{7}{8}+x=\dfrac{3}{5}\)

=>\(x=\dfrac{3}{5}-\dfrac{7}{8}=\dfrac{24-35}{40}=\dfrac{-11}{40}\)

b: \(\dfrac{17}{2}:x=5\)

=>\(x=\dfrac{17}{2}:5\)

=>\(x=\dfrac{17}{2\cdot5}=\dfrac{17}{10}\)

c: \(x-\dfrac{3}{8}=2+\dfrac{1}{4}\)

=>\(x-\dfrac{3}{8}=\dfrac{9}{4}\)

=>\(x=\dfrac{9}{4}+\dfrac{3}{8}=\dfrac{18}{8}+\dfrac{3}{8}=\dfrac{21}{8}\)

d: \(\dfrac{1}{2}+\dfrac{3}{5}\left(x-2\right)=\dfrac{1}{5}\)

=>\(\dfrac{3}{5}\left(x-2\right)=\dfrac{1}{5}-\dfrac{1}{2}=\dfrac{-3}{10}\)

=>\(x-2=-\dfrac{1}{2}\)

=>\(x=2-\dfrac{1}{2}=\dfrac{3}{2}\)

Bài 1:

a: \(\dfrac{-3}{4}+\dfrac{1}{5}=\dfrac{-15}{20}+\dfrac{4}{20}=\dfrac{-15+4}{20}=\dfrac{-11}{20}\)

b: \(\dfrac{-2}{5}-\dfrac{1}{3}=\dfrac{-6}{15}-\dfrac{5}{15}=\dfrac{-6-5}{15}=\dfrac{-11}{15}\)

c: \(\dfrac{3}{7}\cdot\dfrac{2}{5}-\dfrac{2}{5}=\dfrac{2}{5}\left(\dfrac{3}{7}-1\right)=\dfrac{2}{5}\cdot\dfrac{-4}{7}=\dfrac{-8}{35}\)

d: \(\dfrac{1}{4}+\dfrac{3}{4}\left(\dfrac{2}{3}-\dfrac{1}{2}\right)\)

\(=\dfrac{1}{4}+\dfrac{3}{4}\cdot\dfrac{4-3}{6}\)

\(=\dfrac{1}{4}+\dfrac{3}{4}\cdot\dfrac{1}{6}=\dfrac{1}{4}+\dfrac{1}{8}=\dfrac{3}{8}\)

e: \(\dfrac{7}{2}\cdot\dfrac{8}{13}+\dfrac{8}{13}\cdot\dfrac{-5}{12}+\dfrac{8}{13}\)

\(=\dfrac{8}{13}\left(\dfrac{7}{2}-\dfrac{5}{2}+1\right)\)

\(=\dfrac{8}{13}\cdot2=\dfrac{16}{13}\)

f: \(1+\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)

\(=1+\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+\dfrac{1}{8\cdot10}+\dfrac{1}{10\cdot12}\)

\(=1+\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{10\cdot12}\right)\)

\(=1+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{10}-\dfrac{1}{12}\right)\)

\(=1+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{12}\right)\)

\(=1+\dfrac{1}{2}\cdot\dfrac{5}{12}=1+\dfrac{5}{24}=\dfrac{29}{24}\)