Câu 1.

\(y = \dfrac{{n + \sin 2n}}{{n + 5}} = \dfrac{{\dfrac{n}{n} + \dfrac{{\sin 2n}}{n}}}{{\dfrac{n}{n} + \dfrac{5}{n}}} = \dfrac{{1 + \dfrac{{2.\sin 2n}}{{2n}}}}{{1 + \dfrac{5}{n}}}\\ \Rightarrow \lim y = \dfrac{{1 + 0}}{{1 + 0}} = 1 \)

Câu 2.

\(\lim \dfrac{{3\sin n + 4\cos n}}{{n + 1}}\)

Vì \( - 1 \le \sin n \le 1; - 1 \le \cos n \le 1 \Rightarrow \) khi \(x \to \infty \) thì \(3\sin n + 4{\mathop{\rm cosn}\nolimits} = const \)

\(\Rightarrow T = \lim \dfrac{{3\sin n + 4\cos n}}{{n + 1}} = 0 \)

Chú thích: $const$ là kí hiệu hằng số, giống như dạng giới hạn L/vô cùng.

Câu 3.

\(\lim \dfrac{{1 + a + {a^2} + ... + {a^n}}}{{1 + b + {b^2} + ... + {b^n}}} = \lim \dfrac{{\left( {1 + a + {a^2} + ... + {a^n}} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}{{\left( {1 + b + {b^2} + ... + {b^n}} \right)\left( {1 - b} \right)\left( {1 - a} \right)}} = \lim \dfrac{{\left( {1 - {a^{n + 1}}} \right)\left( {1 - b} \right)}}{{\left( {1 - {b^{n + 1}}} \right)\left( {1 - a} \right)}} = \dfrac{{1 - b}}{{1 - a}}\)

\(2^{2n}\left(2^{2n+3}-1\right)-1\)

=\(4n\left(4n+2^3-1\right)-1\)

=\(\left(4n.4n+4n.2^3+4n-1\right)-1\)

= (16.2n + 32n + 3n - 1n) - 1n 

= 65n chia hết cho 5

=> đpcm

Ok bạn chờ mình latý

\(=\frac{2^7.\left(2.3\right)^5-\left(2.2\right)^2.\left(3^2.2\right)^2}{\left(2^2.3\right)^5+\left(2^3\right)^4+3^5}\)

\(=\frac{2^7.2^5.3^5-2^4.3^4.2^2}{2^{10}.3^5+2^{12}.3^5}\)

Rồi bạn rút thừa số chung ra là xong nha

Mọi ng giúp em vs ạ😢😢

d: ta có: \(C=3+3^3+3^5+...+3^{1991}\)

\(=3\left(1+3^2+3^4\right)+...+3^{1987}\left(1+3^2+3^4\right)\)

\(=91\cdot\left(3+...+3^{1987}\right)⋮13\)

Bài 2:

a) \(\dfrac{4}{5}+\dfrac{-5}{4}\)

\(=\dfrac{16}{20}+\dfrac{-25}{20}\)

\(=\dfrac{-9}{20}\)

b) \(\dfrac{-1}{3}+\dfrac{2}{5}-\dfrac{5}{6}\)

\(=\dfrac{-10}{30}+\dfrac{12}{30}-\dfrac{25}{30}\)

\(=\dfrac{-23}{30}\)

c) \(\dfrac{2}{3}-\dfrac{5}{7}.\dfrac{14}{25}\)

\(=\dfrac{2}{3}-\dfrac{5.14}{7.25}\)

\(=\dfrac{2}{3}-\dfrac{70}{175}\)

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

\(=\dfrac{10}{15}-\dfrac{6}{15}\)

\(=\dfrac{4}{15}\)

Bài 1:

\(a,\dfrac{15}{60}=\dfrac{15:15}{60:15}=\dfrac{1}{4}\)

\(b,\dfrac{42}{-28}=\dfrac{-42}{28}=\dfrac{-42:14}{28:14}=\dfrac{-3}{2}\)

\(c,\dfrac{24.39}{15.48}=\dfrac{1.39}{15.2}=\dfrac{39}{30}\)

\(d,\dfrac{49.2+49.3}{49.15}=\dfrac{49.\left(2+3\right)}{49.15}=\dfrac{49.5}{49.15}=\dfrac{1.1}{1.3}=\dfrac{1}{3}\)

\(e,\dfrac{\left(-13\right).24.\left(-20\right)}{\left(-26\right).8.15}=\dfrac{1.3.\left(-4\right)}{2.1.3}=\dfrac{1.1.\left(-2\right)}{1.1.1}=-2\)

\(f,\dfrac{53.19-53}{19-72}=\dfrac{53.\left(19-1\right)}{-53}=\dfrac{53.18}{-53}=\dfrac{1.18}{-1}=-18\)

\(g,\dfrac{-12.13+12.24}{9.17-9.5}=\dfrac{-12\left(13-24\right)}{9.\left(17-5\right)}=\dfrac{-12.\left(-11\right)}{9.12}=\dfrac{-1.\left(-11\right)}{9.1}=\dfrac{11}{9}\)

\(h,\dfrac{2^{20}.125}{2^{24}.50}=\dfrac{-8.3}{2}=\dfrac{-4.3}{1}=-12.\)

Bài 2:

\(a,\dfrac{4}{5}+\dfrac{-5}{4}=\dfrac{16}{20}+\dfrac{-25}{20}=\dfrac{-9}{20}\)

\(b,\dfrac{-1}{3}+\dfrac{2}{5}-\dfrac{5}{6}=\dfrac{-10}{30}+\dfrac{12}{30}-\dfrac{25}{30}=\dfrac{-23}{30}\)

\(c,\dfrac{2}{3}-\dfrac{5}{7}.\dfrac{14}{25}=\dfrac{2}{3}-\dfrac{2}{5}=\dfrac{10}{15}-\dfrac{6}{15}=\dfrac{4}{15}.\)

tham khảo: