\(=\lim\limits_{x\rightarrow+\infty}x.\dfrac{x^2+5-x^2}{\sqrt{x^2+5}+x}=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{5x}{x}}{\sqrt{\dfrac{x^2}{x^2}+\dfrac{5}{x^2}}+\dfrac{x}{x}}=\dfrac{5}{2}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{9x^2+7x+1-9x^2}{\sqrt{9x^2+7x+1}+3x}=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{7x}{x}+\dfrac{1}{x}}{\sqrt{\dfrac{9x^2}{x^2}+\dfrac{7x}{x^2}+\dfrac{1}{x^2}}+\dfrac{3x}{x}}=\dfrac{7}{3+3}=\dfrac{7}{6}\)

\(\Rightarrow P=7-6=1\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{5+x^2-7-x^2}{\sqrt{5+x^2}+\sqrt{7+x^2}}=\lim\limits_{x\rightarrow-\infty}\dfrac{-\dfrac{2}{x}}{-\sqrt{\dfrac{5}{x^2}+\dfrac{x^2}{x^2}}-\sqrt{\dfrac{7}{x^2}+\dfrac{x^2}{x^2}}}=0\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{a^2x^2-a^2x^2+2x}{ax+\sqrt{a^2x^2-2x}}=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{2x}{x}}{\dfrac{ax}{x}+\sqrt{\dfrac{a^2x^2}{x^2}-\dfrac{2x}{x^2}}}=\dfrac{2}{a+a}=\dfrac{1}{a}\)

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

\(\lim\limits_{x\rightarrow-\infty}\dfrac{x^2+ax+5-x^2}{\sqrt{x^2+ax+5}-x}=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{ax}{x}+\dfrac{5}{x}}{-\sqrt{\dfrac{x^2}{x^2}+\dfrac{ax}{x^2}+\dfrac{5}{x^2}}-\dfrac{x}{x}}=\dfrac{-a}{2}\)

\(-\dfrac{a}{2}=5\Rightarrow a=-10\)

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

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{4x^4}{x^6}}{\dfrac{8x^6}{x^6}}=0\)

\(u_{n+1}-1=u_n\left(u_n-1\right)\Leftrightarrow\dfrac{1}{u_{n+1}-1}=\dfrac{1}{u_n-1}-\dfrac{1}{u_n}\Rightarrow\dfrac{1}{u_n}=\dfrac{1}{u_n-1}-\dfrac{1}{u_{n+1}-1}\)

Lan luot the i vo n:

\(\dfrac{1}{u_1}=\dfrac{1}{u_1-1}-\dfrac{1}{u_2-1}\)

\(\dfrac{1}{u_2}=\dfrac{1}{u_2-1}-\dfrac{1}{u_3-1}\)

...

\(\dfrac{1}{u_n}=\dfrac{1}{u_n-1}-\dfrac{1}{u_{n+1}-1}\)

Cong ve voi ve:

\(\dfrac{1}{u_1}+\dfrac{1}{u_2}+...+\dfrac{1}{u_n}=\dfrac{1}{u_1-1}-\dfrac{1}{u_{n+1}-1}\)

Do dãy tăng và ko bị chặn trên <bạn thay vô là biết>

\(\Rightarrow\lim\limits\left(u_{n+1}-1\right)=+\infty\Rightarrow\lim\limits\sum\limits^n_{i=1}\dfrac{1}{u_i}=\lim\limits\left(\dfrac{1}{u_1-1}-\dfrac{1}{u_{n+1}-1}\right)=1\)

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https://hoc24.vn/cau-hoi/.334447965337