giúp m vs ak $\lim_{x\rightarrow+\infty}(\sqrt{x^2-x}-\sqrt[3]{x^3+1})=\lim_{x\rightarrow+\infty}\frac{-a.x}{b.x(\sqrt{1-\frac1x}+1)}$ $\frac ab$,51.Giới hạn với a;b là các số tự nhiên và là phân,số tối giản. Tính $P=a+b^2$

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Accepted Answer

Ta có:
$\displaystyle \begin{array}{{>{\displaystyle}l}}
\ \ \ \ \ \lim _{x ightarrow \infty }\left(\sqrt{x^{2} -x} -\sqrt[3]{x^{3} +1} ight)\
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=\lim _{x ightarrow \infty }\left(\sqrt{x^{2} -x} -\sqrt[3]{x^{3}\left( 1-\frac{1}{x^{3}} ight)} \ \ ight)\
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=\lim _{x ightarrow \infty }\left(\sqrt{x^{2} -x} -x.\sqrt[3]{1-\frac{1}{x^{3}}} \ \ ight)\
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=\lim _{x ightarrow \infty }\left(\sqrt{x^{2} -x} \ -\ x\ ight) \ \ \ \ \ \ \left( Vì\ \lim _{x ightarrow \infty }\frac{1}{x^{3}} =0\Longrightarrow \lim _{x ightarrow \infty }\left(\sqrt[3]{1-\frac{1}{x^{3}}} ight) =1 ight)\
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=\lim _{x ightarrow \infty }\frac{\left(\sqrt{x^{2} -x} \ -\ x\ ight)\left(\sqrt{x^{2} -x} \ +\ x\ ight)}{\left(\sqrt{x^{2} -x} \ +\ x\ ight)}\
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=\lim _{x ightarrow \infty }\frac{x^{2} -x\ -\ x^{2}}{\left(\sqrt{x^{2} -x} \ +\ x\ ight)}\
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=\lim _{x ightarrow \infty }\frac{-x\ }{\left(\sqrt{x^{2} -x} \ +\ x\ ight)}\
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=\lim _{x ightarrow \infty }\frac{-x\ }{\left(\sqrt{x^{2}\left( 1-\frac{1}{x} ight)} \ +\ x\ ight)}\
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=\lim _{x ightarrow \infty }\frac{-x\ }{\left( x.\sqrt{\left( 1-\frac{1}{x} ight)} \ +\ x\ ight)}\
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=\lim _{x ightarrow \infty }\frac{-x\ }{x.\left(\sqrt{\left( 1-\frac{1}{x} ight)} \ +\ 1 ight) \ }\
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=\lim _{x ightarrow \infty }\frac{-x\ }{x.2\ } \ \ \ \ \ \ \ \ \ \ \ \ \ \left( Vì\ \lim _{x ightarrow \infty }\left(\sqrt{\left( 1-\frac{1}{x} ight)} \ +\ 1 ight) \ =1+1=2 ight)\
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=\lim _{x ightarrow \infty }\frac{-1\ }{2\ }\
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=-\frac{1}{2}
\end{array}$

Ta có:
$\displaystyle \begin{array}{{>{\displaystyle}l}}
\ \ \ \ \ \ \lim _{x ightarrow +\infty }\frac{-a.x}{b.x\left(\sqrt{1-\frac{1}{x}} +1 ight)}\
\
=\lim _{x ightarrow +\infty }\frac{-a.x}{b.x.2}\
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=\lim _{x ightarrow +\infty }\frac{-a}{2b}
\end{array}$

Ta có:
$\displaystyle \begin{array}{{>{\displaystyle}l}}
\lim _{x ightarrow \infty }\left(\sqrt{x^{2} -x} -\sqrt[3]{x^{3} +1} ight) =\lim _{x ightarrow +\infty }\frac{-a.x}{b.x\left(\sqrt{1-\frac{1}{x}} +1 ight)}\
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\Leftrightarrow -\frac{1}{2} =-\frac{a}{2b}\
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Mà\ \frac{a}{b} \ tối\ giản\ \Longrightarrow \begin{cases}
a=1 & \
b=2 &
\end{cases}\
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\Longrightarrow P=a+b^{2} =1+2^{2} =5
\end{array}$