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

\(b=\lim\dfrac{\sqrt{\dfrac{1}{n}+\sqrt{\dfrac{1}{n^3}+\dfrac{1}{n^4}}}}{1-\dfrac{1}{\sqrt{n}}}=\dfrac{0}{1}=0\)

\(c=\lim\dfrac{\sqrt{2n^2-1+\dfrac{7}{n^2}}}{3+\dfrac{5}{n}}=\dfrac{+\infty}{3}=+\infty\)

\(d=\lim\dfrac{\sqrt{3+\dfrac{2}{n}}-1}{3-\dfrac{2}{n}}=\dfrac{\sqrt{3}-1}{3}\)

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

\(=lim\dfrac{\dfrac{2}{n}+\dfrac{1}{n^3}}{3-\dfrac{3}{n^2}+\dfrac{3}{n^3}}=0\)

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

\(\lim\dfrac{n^3-2n+1}{-3n-4}=\lim\dfrac{n^2-2+\dfrac{1}{n}}{-3-\dfrac{4}{n}}=\dfrac{+\infty}{-3}=-\infty\)

`@` `\text {Ans}`

`\downarrow`

`a)`

\(2^{n+3}\cdot5^{n+3}=20^9\div2^9\)

`=>`\(\left(2\cdot5\right)^{n+3}=\left(20\div2\right)^9\)

`=>`\(10^{n+3}=10^9\)

`=>`\(n+3=9\)

`=> n = 9 - 3`

`=> n= 6`

Vậy, `n=6`

`b)`

\(3^{n+5}-3^{n+4}=1458\)

`=> 3^n*3^5 - 3^n*3^4 = 1458`

`=> 3^n*(3^5 - 3^4) = 1458`

`=> 3^n*162 = 1458`

`=> 3^n = 1458 \div 162`

`=> 3^n = 9`

`=> 3^n = 3^2`

`=> n=2`

Vậy, `n=2.`

`c)`

\(5^{n+3}+5^{n+2}=3750\)

`=> 5^n*5^3 + 5^n*5^2 = 3750`

`=> 5^n*(5^3+5^2) = 3750`

`=> 5^n*150 = 3750`

`=> 5^n = 3750 \div 150`

`=> 5^n =25`

`=> 5^n = 5^2`

`=> n=2`

Vậy, `n=2.`

`d)`

\(\dfrac{2}{7}x+\dfrac{3}{14}x=\dfrac{1}{2}\)

`=> 1/2x = 1/2`

`=> x = 1/2 \div 1/2`

`=> x=1`

Vậy, `x=1`

`e)`

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

`=> (x+2)(x+3) = -3*(-2)`

`=> (x+2)(x+3) = -6`

`=> x(x+3) + 2(x+3) = -6`

`=> x^2 + 3x + 2x + 6 = -6`

`=> x^2 + 5x + 6 - 6 = 0`

`=> x^2 + 5x = 0`

`=> x(x+5) = 0`

`=>`\(\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy, `x \in {0; -5}`

`@` `\text {Kaizuu lv u}`

1:

\(\lim\limits_{n\rightarrow\infty}\dfrac{3n^5+3n^3-1}{n^3-2n}=\lim\limits_{n\rightarrow\infty}\dfrac{n^5\left(3+\dfrac{3}{n^2}-\dfrac{1}{n^5}\right)}{n^3\left(1-\dfrac{2}{n^2}\right)}\)

\(=\lim\limits_{n\rightarrow\infty}n^2\cdot3=+\infty\)

2: \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^7+3n^5-n}{3n^2-2n}=\lim\limits_{n\rightarrow\infty}\dfrac{3n^6+3n^4-1}{3n-2}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^6\left(3+\dfrac{3}{n^2}-\dfrac{1}{n^6}\right)}{n\left(3-\dfrac{2}{n}\right)}=\lim\limits_{n\rightarrow\infty}n^5=+\infty\)

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

\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)

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

\(c=\lim\dfrac{\sqrt{1+\dfrac{3}{n^3}-\dfrac{2}{n^4}}}{2-\dfrac{2}{n}+\dfrac{3}{n^2}}=\dfrac{1}{2}\)

\(d=\lim\dfrac{\sqrt{1-\dfrac{4}{n}}-\sqrt{4+\dfrac{1}{n^2}}}{\sqrt{3+\dfrac{1}{n^2}}-1}=\dfrac{1-2}{\sqrt{3}-1}=-\dfrac{1+\sqrt{3}}{2}\)

Lim 3.4n-2.13n/5n+6.13n