\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{4x+5}-2x-3}{\left(x+1\right)^2}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{4x+5-\left(2x+3\right)^2}{\sqrt{4x+5}+2x+3}\cdot\dfrac{1}{\left(x+1\right)^2}\)

\(=\lim\limits_{x\rightarrow-1}\left(\dfrac{4x+5-4x^2-12x-9}{\left(\sqrt{4x+5}+2x+3\right)\cdot\left(x+1\right)^2}\right)\)

\(=\lim\limits_{x\rightarrow-1}\left(\dfrac{-4x^2-8x-4}{\left(\sqrt{4x+5}+2x+3\right)\cdot\left(x+1\right)^2}\right)\)

\(=\lim\limits_{x\rightarrow-1}\left(\dfrac{-4\left(x^2+2x+1\right)}{\left(x+1\right)^2\cdot\left(\sqrt{4x+5}+2x+3\right)}\right)\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{-4}{\sqrt{4x+5}+2x+3}\)

\(=\dfrac{-4}{\sqrt{-4+5}-2+3}=\dfrac{-4}{1+1}=-\dfrac{4}{2}=-2\)

Sau mỗi phút, số lượng virus tăng lên gấp 3 lần trước đó

=>Số lượng con vius có sau 11 phút sẽ tăng thêm \(3^{11}\)(lần)

=>Sau 11 phút, số lượng con virus là:

\(5\cdot3^{11}=885735\left(con\right)\)

Lời giải:

\(I=\lim\limits_{x\to 1}\frac{\sqrt{3x+1}(\sqrt[3]{2-x}-1)+(\sqrt{3x+1}-2)}{x-1}=\lim\limits_{x\to 1}\frac{\sqrt{3x+1}.\frac{1-x}{\sqrt[3]{(2-x)^2}+\sqrt[3]{2-x}+1}+\frac{3(x-1)}{\sqrt{3x+1}+2}}{x-1}\)

\(=\lim\limits_{x\to 1}\left[ \frac{-\sqrt{3x+1}}{\sqrt[3]{(2-x)^2}+\sqrt[3]{2-x}+1}+\frac{3}{\sqrt{3x+1}+2} \right]=\) $\frac{1}{12}$

\(\text{Ta có:} \ 3n \ \vdots \ 3 \Rightarrow 3n+2 \ \text{chia 3 dư 2} \\ \text{Mà một số chính phương khi chia 3 chỉ dư 0 hoặc 1} \\ \Rightarrow \text{Không tồn tại số tự nhiên} \ n \ \text{thỏa mãn}\)

\(\text{Ta có:} \ u_n=\dfrac{\sqrt{n}}{n+9} \Rightarrow \dfrac{1}{u_n}=\dfrac{n+9}{\sqrt n}=\sqrt n+\dfrac{9}{\sqrt n} \\ \text{Áp dụng BĐT Cauchy, ta có:} \\ \sqrt n + \dfrac{9}{\sqrt n} \geq 6 \ \text{hay} \ \dfrac{1}{u_n} \geq 6 \\ \Rightarrow u_n \leq \dfrac{1}{6} \\ \text{Vậy dãy} \ (u_n) \ \text{bị chặn trên bởi} \ \dfrac{1}{6}\)

\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt[3]{x+6}-\sqrt{x+2}}{x^2-4}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\sqrt[3]{x+6}-2+2-\sqrt{x+2}}{\left(x-2\right)\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\dfrac{x+6-8}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+\dfrac{4-x-2}{2+\sqrt{x+2}}}{\left(x-2\right)\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(\dfrac{1}{\sqrt[3]{\left(x+6\right)^2}+2\sqrt[3]{x+6}+4}-\dfrac{1}{2+\sqrt{x+2}}\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\dfrac{1}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-\dfrac{1}{2+\sqrt{x+2}}}{x+2}\)

\(=\dfrac{\dfrac{1}{\sqrt[3]{\left(2+6\right)^2}+2\cdot\sqrt[3]{2+6}+4}-\dfrac{1}{2+\sqrt{2+2}}}{2+2}\)

\(=\dfrac{\dfrac{1}{\sqrt[3]{64}+2\cdot\sqrt[3]{8}+4}-\dfrac{1}{2+2}}{4}\)

\(=\dfrac{\dfrac{1}{4+2\cdot2+4}-\dfrac{1}{4}}{4}=\left(\dfrac{1}{16}-\dfrac{1}{4}\right):4=\left(\dfrac{1}{16}-\dfrac{4}{16}\right)\cdot\dfrac{1}{4}=\dfrac{-3}{64}\)

a: \(\lim\limits_{x\rightarrow-3}\dfrac{\sqrt{x+5}-\sqrt{x^2+x-4}}{-x^2+x+12}\)

\(=\lim\limits_{x\rightarrow-3}\dfrac{\sqrt{x^2+x-4}-\sqrt{x+5}}{x^2-x-12}\)

\(=\lim\limits_{x\rightarrow-3}\dfrac{\dfrac{x^2+x-4-x-5}{\sqrt{x^2+x-4}+\sqrt{x+5}}}{\left(x-4\right)\left(x+3\right)}\)

\(=\lim\limits_{x\rightarrow-3}\dfrac{\dfrac{\left(x^2-9\right)}{\sqrt{x^2+x-4}+\sqrt{x+5}}}{\left(x-4\right)\left(x+3\right)}\)

\(=\lim\limits_{x\rightarrow-3}\left(\dfrac{\left(x-3\right)\left(x+3\right)}{\sqrt{x^2+x-4}+\sqrt{x+5}}\cdot\dfrac{1}{\left(x-4\right)\left(x+3\right)}\right)\)

\(=\lim\limits_{x\rightarrow-3}\left(\dfrac{x-3}{\left(x-4\right)\left(\sqrt{x^2+x-4}+\sqrt{x+5}\right)}\right)\)

\(=\dfrac{-3-3}{\left(-3-4\right)\left(\sqrt{9-3-4}+\sqrt{-3+5}\right)}\)

\(=\dfrac{-6}{\left(-7\right)\left(\sqrt{2}+\sqrt{2}\right)}=\dfrac{6}{7\cdot2\sqrt{2}}=\dfrac{3}{7\sqrt{2}}=\dfrac{3\sqrt{2}}{14}\)

b: \(\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-2+2-\sqrt[3]{8-x}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{2\cdot\left(\sqrt{x+1}-1\right)+\dfrac{8-8+x}{4+2\cdot\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{2\left(\dfrac{x}{\sqrt{x+1}+1}\right)+\dfrac{x}{4+2\cdot\sqrt[3]{8-x}+\left(\sqrt[3]{8-x}\right)^2}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{2}{\sqrt{x+1}+1}+\dfrac{1}{4+2\cdot\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}\)

\(=\dfrac{2}{\sqrt{0+1}+1}+\dfrac{1}{4+2\cdot\sqrt[3]{8-0}+\sqrt[3]{\left(8-0\right)^2}}\)

\(=\dfrac{2}{1+1}+\dfrac{1}{4+2\cdot2+4}\)

\(=1+\dfrac{1}{12}=\dfrac{13}{12}\)

c: \(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}\right)\)

\(=\lim\limits_{x\rightarrow2}\left(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}\right)\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x-3+x-1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{2x-4}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{2}{\left(x-1\right)\left(x-3\right)}=\dfrac{2}{\left(2-1\right)\left(2-3\right)}=-2\)

\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{x^2+8}-\sqrt{x+4}}{2x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{x^2+8}-2+2-\sqrt{x+4}}{2x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\left(\dfrac{x^2+8-8}{\sqrt[3]{\left(x^2+8\right)^2}+2\cdot\sqrt[3]{x^2+8}+4}\right)+\dfrac{4-x-4}{2+\sqrt{x+4}}}{2x}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{\dfrac{x^2}{\sqrt[3]{\left(x^2+8\right)^2}+2\cdot\sqrt[3]{x^2+8}+4}-\dfrac{x}{2+\sqrt{x+4}}}{2x}\right)\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{\dfrac{x}{\sqrt[3]{\left(x^2+8\right)^2}+2\cdot\sqrt[3]{x^2+8}+4}-\dfrac{1}{2+\sqrt{x+4}}}{2}\right)\)

\(=\left(\dfrac{\dfrac{0}{\sqrt[3]{\left(0+8\right)^2}+2\cdot\sqrt[3]{0+8}+4}-\dfrac{1}{2+\sqrt{0+4}}}{2}\right)\)

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

\(\lim\limits_{x\rightarrow6}f\left(x\right)=\lim\limits_{x\rightarrow6}\dfrac{3x^2-23x+30}{x-6}\)

\(=\lim\limits_{x\rightarrow6}\dfrac{3x^2-18x-5x+30}{x-6}\)

\(=\lim\limits_{x\rightarrow6}\dfrac{\left(x-6\right)\left(3x-5\right)}{x-6}=\lim\limits_{x\rightarrow6}3x-5=3\cdot6-5=13\)

\(f\left(6\right)=a\)

Để hàm số liên tục  tại x=6 thì \(f\left(6\right)=\lim\limits_{x\rightarrow6}f\left(x\right)\)

=>a=13