a,\(lim\dfrac{1-2n^2}{5n+5}=lim\dfrac{\left(1-n\sqrt{2}\right)\left(1+n\sqrt{2}\right)}{5n+5}=lim\dfrac{\left(\dfrac{1}{n}-\sqrt{2}\right)\left(\dfrac{1}{n}+\sqrt{2}\right)}{5+\dfrac{5}{n}}=\dfrac{-2}{5}\)

b,\(lim\dfrac{1-2n}{5n+5n^2}=lim\dfrac{\dfrac{1}{n^2}-\dfrac{2}{n}}{\dfrac{5}{n}+5}=\dfrac{0}{5}=0\)

a,\(lim\dfrac{n^2-2n}{5n+3n^2}=lim\dfrac{1-\dfrac{2}{n}}{\dfrac{5}{n}+3}=\dfrac{1}{3}\)

b,\(lim\dfrac{n^2-2}{5n+3n^2}=lim\dfrac{1-\dfrac{2}{n^2}}{\dfrac{5}{n}+3}=\dfrac{1}{3}\)

c,\(lim\dfrac{1-2n}{5n+3n^2}=lim\dfrac{1-2n}{n\left(5+3n\right)}=lim\dfrac{\dfrac{1}{n}-2}{1\left(\dfrac{5}{n}+3\right)}=-\dfrac{2}{3}\)

d,\(lim\dfrac{1-2n^2}{5n+5}=lim\dfrac{\left(1-n\sqrt{2}\right)\left(1+n\sqrt{2}\right)}{5n+5}=lim\dfrac{\left(\dfrac{1}{n}-\sqrt{2}\right)\left(\dfrac{1}{n}+\sqrt{2}\right)}{5+\dfrac{5}{n}}=\dfrac{-2}{5}\)

Viết mũ hẩn hoi ra , viết thế này khó nhìn lắm

đây là một bài toán hay đấy

\(A=\left(1+2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8+2^9\right)+...+\left(2^{5n-3}+2^{5n-2}+2^{5n-1}+2^{5n-0}\right)\)

\(A=\left(1+2+4+8+16\right)+2^5.\left(1+2+4+8+16\right)+...+2^{5n-3}.\left(1+2+4+8+16\right)\)

\(A=31+2^5.31+....+2^{5n-3}.31\)

\(A=31.\left(1+2^5+....+2^{2n-3}\right)\) CHIA HẾT CHO 31

=> A chia hết cho 31 (đpcm)

3⁵ⁿ⁺²+3⁵ⁿ⁺¹-3⁵ⁿ

=3⁵ⁿ.3²+3⁵ⁿ.3¹-3⁵ⁿ

=3⁵ⁿ.9+3⁵ⁿ.3-3⁵ⁿ

=3⁵ⁿ.(9+3-1)

=3⁵ⁿ.11 chia hết cho 11

=> 3⁵ⁿ⁺²+3⁵ⁿ⁺¹-3⁵ⁿ chia hết cho 11

\(3^{5n+2}+3^{5n+1}-3^{5n}=3^{5n}\left(3^2+3-1\right)=11.3^{5n}⋮11\)

\(3^{5n+2}+3^{5n+1}-3^{5n}(n\in N^*)\\=3^{5n}\cdot3^2+3^{5n}\cdot3-3^{5n}\\=3^{5n}\cdot(3^2+3-1)\\=3^{5n}\cdot11\)

Vì \(3^{5n}\cdot11\vdots11\) 

nên biểu thức \(3^{5n+2}+3^{5n+1}-3^{5n}\vdots11\)

Gọi d là \(ƯCLN\left(5n+2,5n+3\right)\)

\(\Rightarrow\begin{cases}5n+2⋮d\\5n+3⋮d\end{cases}\)

\(\Rightarrow\left(5n+3\right)-\left(5n+2\right)⋮d\)

\(\Rightarrow5n+3-5n-2⋮d\)

\(\Rightarrow1⋮d\Rightarrow d=1\RightarrowƯCLN\left(5n+2,5n+3\right)=1\)

Vậy 5n + 2 và 5n + 3 là hai số nguyên tố cùng nhau .

b, Gọi d là \(ƯCLN\left(7n+1,6n+1\right)\)

\(\Rightarrow\begin{cases}7n+1⋮d\\6n+1⋮d\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}42n+6⋮d\\42n+7⋮d\end{cases}\)

\(\Rightarrow\left(42n+7\right)-\left(42n+6\right)⋮d\)

\(\Rightarrow42n+7-42n-6⋮d\)

\(\Rightarrow1⋮d\Rightarrow d=1\RightarrowƯCLN\left(7n+1,6n+1\right)=1\)

Vậy 7n + 1 và 6n + 1 là hai số nguyên tố cùng nhau .

c, Gọi d là \(ƯCLN\left(5n+1,4n+1\right)\)

\(\Rightarrow\begin{cases}5n+1⋮d\\4n+1⋮d\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}20n+4⋮d\\20n+5⋮d\end{cases}\)

\(\Rightarrow\left(20n+5\right)-\left(20n+4\right)⋮d\)

\(\Rightarrow20n+5-20n-4⋮d\)

\(\Rightarrow1⋮d\Rightarrow d=1\RightarrowƯCLN\left(5n+1,4n+1\right)=1\)

Vậy 5n + 1 và 4n + 1 là hai số nguyên tố cùng nhau

98

 CM:  \(\dfrac{1}{1.6}\)+ \(\dfrac{1}{11.16}\)+...+ \(\dfrac{1}{\left(5n+1\right)\left(5n+6\right)}\) = \(\dfrac{n+1}{5n+6}\)

A = \(\dfrac{1}{5}\)(\(\dfrac{5}{1.6}\) + \(\dfrac{5}{6.11}\)+...+ \(\dfrac{5}{\left(5n+1\right).\left(5n+6\right)}\)) 

A = \(\dfrac{1}{5}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{6}\)+ \(\dfrac{1}{6}\) - \(\dfrac{1}{11}\)+...+ \(\dfrac{1}{5n+1}\) - \(\dfrac{1}{5n+6}\))

A = \(\dfrac{1}{5}\) .( \(\dfrac{1}{1}\) - \(\dfrac{1}{5n+6}\))

A = \(\dfrac{1}{5}\). \(\dfrac{5n+6-1}{5n+6}\)

A = \(\dfrac{1}{5}\). \(\dfrac{5n+5}{5n+6}\)

A = \(\dfrac{1}{5}\) . \(\dfrac{5.\left(n+1\right)}{5n+6}\)

A = \(\dfrac{n+1}{5n+6}\)

⇒\(\dfrac{1}{1.6}\) + \(\dfrac{1}{6.11}\)+ \(\dfrac{1}{11.16}\)+...+ \(\dfrac{1}{\left(5n+1\right)\left(5n+6\right)}\) = \(\dfrac{n+1}{5n+1}\) (đpcm)

\(A=\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+...+\dfrac{1}{\left(5n+1\right)\left(5n+6\right)}\)

\(A=\dfrac{1}{5}\left[1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{5n+1}-\dfrac{1}{5n+6}\right]\)

\(A=\dfrac{1}{5}\left(1-\dfrac{1}{5n+6}\right)\)

\(A=\dfrac{1}{5}\left(\dfrac{5n+6-1}{5n+6}\right)=\dfrac{1}{5}\left(\dfrac{5n+5}{5n+6}\right)=\dfrac{1}{5}.5\left(\dfrac{n+1}{5n+6}\right)=\dfrac{n+1}{5n+6}\)

\(\Rightarrow dpcm\)

Gọi A = 1/1.6 + 1/6.11 +...+ 1/(5n+1)(5n+6) 

5A = 5/1.6 + 5/6.11 + ... + 5/(5n+1)(5n+6)

     =1 - 1/6 + 1/6 - 1/11 + ... + 1/5n+1 - 1/5n+6 

    =1 - 1/5n+6 =5n+6/5n+6 - 1/5n+6=5n+5 /5n+6

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