Trong 1 giây rùa bò được : 72000 : 3600 = 20 (mm)

Tỉ số vận tốc của rùa và sên là : 20 : 1,5 = 40 : 3.

Vậy, vận tốc của rùa gấp hơn 13 lần vận tốc của sên !

\(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2022^2}\)

\(< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2021.2022}\)

\(=\frac{1}{4}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{2022-2021}{2021.2022}\)

\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2021}-\frac{1}{2022}\)

\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{2022}< \frac{1}{4}+\frac{1}{2}=\frac{3}{4}\)

`Answer:`

Câu 4:

a. Để cho `6a+4;a+2` đều là luỹ thừa của `2` thì \(\hept{\begin{cases}6a+2=2^m\\a+2=2^n\end{cases}}\left(m,n\inℤ\right)\)

Ta có: \(a\inℤ\Rightarrow6a+4>4,a^2>2\Rightarrow m>2,n>0\)

Mặt khác: \(6a+4>a+2\rightarrow2^m>2^n\rightarrow m>n\)

\(\Rightarrow\hept{\begin{cases}6a=2^m-4\\a=2^n-2\end{cases}}\left(m,n\inℤ\right)\)

\(\Rightarrow6.\left(2^n-2\right)=2^m-4\)

\(\Rightarrow6.2^n-12=2^m-4\)

\(\Rightarrow6.2^n-8=2^m\)

\(\Rightarrow6.2^n-2^m=8\)

\(\Rightarrow2^n.\left(6-2^{m-n}\right)=8\)

\(\Rightarrow\left(2^n,6-2^{m-n}\right)\) là cặp \(Ư\left(8\right)\)

Ta có: `2^n>2`

\(\Rightarrow\left(2^n,6-2^{m-n}\right)\in\left\{\left(2,4\right);\left(4,2\right);\left(8,1\right)\right\}\)

\(\Rightarrow\left(n,2^{m-n}\right)\in\left\{\left(1,2\right);\left(2,4\right);\left(3,7\right)\right\}\)

Mà `m,n\inZZ`

\(\Rightarrow\left(n,2^{m-n}\right)\in\left\{\left(1,2\right);\left(2,4\right)\right\}\)

\(\Rightarrow\left(n,m-n\right)\in\left\{\left(1,1\right);\left(2,2\right)\right\}\)

\(\Rightarrow\left(n,m\right)\in\left\{\left(1,2\right);\left(2,4\right)\right\}\)

Mà `n>1`

\(\Rightarrow\left(n,m\right)=\left(2,4\right)\)

`=>a=2`

`Answer:`

\(C=-\frac{1}{3}.\left(1+2+3\right)-\frac{1}{4}.\left(1+2+3+4\right)-...-\frac{1}{50}.\left(1+2+3+...+50\right)\)

\(\Rightarrow C=-[\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+...+\frac{1}{50}.\left(1+2+3+...+50\right)]\)

Ta có: 

\(\frac{3.4}{2}=1+2+3\)

\(\frac{4.5}{2}=1+2+3+4\)

...

\(\frac{50.51}{2}=1+2+3+...+50\)

\(\Rightarrow C=-[\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{50}.\frac{50.51}{2}]\)

\(\Rightarrow C=-\left(\frac{4}{2}+\frac{5}{2}+...+\frac{51}{2}\right)\)

\(\Rightarrow C=-\frac{1}{2}.\left(4+5+...+51\right)\)

Đặt \(D=4+5+...+51\)

\(=\left(51+4\right).[\left(51-4\right):1+1]:2\)

\(=55.48:2\)

\(=1320\)

\(\Rightarrow C=-\frac{1}{2}.1320\)

\(\Rightarrow C=-660\)

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\)

\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)

\(=\frac{1}{3}\left(\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{100-97}{97.100}\right)\)

\(=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(=\frac{1}{3}\left(1-\frac{1}{100}\right)=\frac{33}{100}\)

\(\frac{33}{100}=\frac{0,33.x}{2009}\Leftrightarrow x=2009\)

chịuuuuuuuuuuuuuuuuuuuuuuuuuu

`Answer:`

\(3.\left(3x.2\right)+50\%=-\frac{9}{4}\)

\(\Leftrightarrow3.\left(3x.2\right)+\frac{1}{2}=-\frac{9}{4}\)

\(\Leftrightarrow3.\left(3x.2\right)=-\frac{9}{4}-\frac{1}{2}\)

\(\Leftrightarrow3.\left(3x.2\right)=-\frac{11}{4}\)

\(\Leftrightarrow3x.2=-\frac{11}{4}:3\)

\(\Leftrightarrow3x.2=-\frac{11}{12}\)

\(\Leftrightarrow x.2=-\frac{11}{12}:3\)

\(\Leftrightarrow x.2=-\frac{11}{36}\)

\(\Leftrightarrow x=-\frac{11}{36}:2\)

\(\Leftrightarrow x=-\frac{11}{72}\)

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)

\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{10-9}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}< 1\)