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

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

a.

\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2-ax+2021}-x+1\right)\)

\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{\left(\sqrt{x^2-ax+2021}-x\right)\left(\sqrt{x^2-ax+2021}+x\right)}{\sqrt{x^2-ax+2021}+x}+1\right)\)

\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{-ax+2021}{\sqrt{x^2-ax+2021}+x}+1\right)\)

\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{x\left(-a+\dfrac{2021}{x}\right)}{x\left(\sqrt{1-\dfrac{a}{x}+\dfrac{2021}{x^2}}+1\right)}+1\right)\)

\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{-a+\dfrac{2021}{x}}{\sqrt{1-\dfrac{a}{x}+\dfrac{2021}{x^2}}+1}+1\right)\)

\(=\dfrac{-a+0}{\sqrt{1+0+0}+1}+1=-\dfrac{a}{2}+1\)

\(\Rightarrow a^2=-\dfrac{a}{2}+1\Rightarrow2a^2+a-2=0\)

Pt trên có 2 nghiệm pb nên có 2 giá trị a thỏa mãn

b.

\(\lim\limits_{x\rightarrow-1}f\left(x\right)=\lim\limits_{x\rightarrow-1}\dfrac{x^3+1}{x+1}\)

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

\(=1+1+1=3\)

\(f\left(-1\right)=3a\)

Hàm gián đoạn tại điểm \(x_0=-1\) khi:

\(\lim\limits_{x\rightarrow-1}f\left(x\right)\ne f\left(-1\right)\Rightarrow3\ne3a\)

\(\Rightarrow a\ne1\)

Trình bày công thức các thứ khá dài nên tôi thử nói hướng, nếu bạn hiểu đc và làm đc thì ok còn nếu k hiểu thì bảo mình, mình làm full cho

Bây giờ phân tích mẫu trước, ra (x-1)²(x+2)

Để cái lim này nó ra đc 1 số thực thì tử và mẫu cùng phải triệt tiêu (x-1)² đi, tức là tử phải chia hết (x-1)², tức là tử cũng phải có nghiệm kép x=1

Do đó \(\left\{{}\begin{matrix}f\left(1\right)=0\\f'\left(1\right)=0\end{matrix}\right.\)

Mình cảm ơn bạn ạ.

Tại vì thật ra mình cũng biết là cái tử nó phải bằng 0 rồi, nhưng cho bằng 0 xong mình không biết tính \(a^2+b^2\) thế nào.

Mong bạn giúp đỡ ạ !

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

\(\Rightarrow a=-3\)

Đặt \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{n\left(n+1\right)}=A\)

\(\Leftrightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{n}-\frac{1}{n+1}\)

\(\Leftrightarrow A=\frac{n+1}{n+1}-\frac{1}{n+1}=\frac{n}{n+1}\)

a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2+1-x^2}{\sqrt{x^2+1}-x}+\lim\limits_{x\rightarrow-\infty}\dfrac{3x^3-1-x^3}{\sqrt[3]{\left(3x^3-1\right)^2}+x\sqrt[3]{3x^3-1}+x^2}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{1}{x}}{-\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}-\dfrac{x}{x}}+\lim\limits_{x\rightarrow-\infty}\dfrac{-\dfrac{1}{x^2}}{\dfrac{\sqrt[3]{\left(3x^3-1\right)^2}}{x^2}+\dfrac{x\sqrt[3]{3x^3-1}}{x^2}+\dfrac{x^2}{x^2}}=0\)

b/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+x-x^2}{\sqrt{x^2+x}+x}+\lim\limits_{x\rightarrow+\infty}\dfrac{x^3-x^3+x^2}{x^2+x\sqrt[3]{x^3-x^2}+\sqrt[3]{\left(x^3-x^2\right)^2}}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x}{x}}{\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}}+\dfrac{x}{x}}+\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x^2}{x^2}}{\dfrac{x^2}{x^2}+\dfrac{x\sqrt[3]{x^3-x^2}}{x^2}+\dfrac{\sqrt[3]{\left(x^3-x^2\right)^2}}{x^2}}\)

\(=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)

c/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{2x-1-2x-1}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{4x^2-1}+\sqrt[3]{\left(2x+1\right)^2}}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{-\dfrac{2}{x^{\dfrac{2}{3}}}}{\dfrac{\sqrt[3]{\left(2x-1\right)^2}}{x^{\dfrac{2}{3}}}+\dfrac{\sqrt[3]{4x^2-1}}{x^{\dfrac{2}{3}}}+\dfrac{\sqrt[3]{\left(2x+1\right)^2}}{x^{\dfrac{2}{3}}}}=0\)

Check lai ho minh nhe :v

cảm ơn bạn nhé , giờ mới trả lời được 

a: \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x+7}-\sqrt{5-x^2}}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x+7}-2+2-\sqrt{5-x^2}}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x+7-8}{\sqrt[3]{\left(x+7\right)^2}+2\cdot\sqrt[3]{x+7}+4}+\dfrac{4-5+x^2}{2+\sqrt{5-x^2}}}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x-1}{\sqrt[3]{\left(x+7\right)^2}+2\cdot\sqrt[3]{x+7}+4}+\dfrac{x^2-1}{2+\sqrt{5-x^2}}}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(\dfrac{1}{\sqrt[3]{\left(x+7\right)^2}+2\cdot\sqrt[3]{\left(x+7\right)}+4}+\dfrac{x+1}{2+\sqrt{5-x^2}}\right)}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{1}{\sqrt[3]{\left(x+7\right)^2}+2\cdot\sqrt[3]{x+7}+4}+\dfrac{x+1}{2+\sqrt{5-x^2}}\)

\(=\dfrac{1}{\sqrt[3]{\left(1+7\right)^2}+2\cdot\sqrt[3]{1+7}+4}+\dfrac{1+1}{2+\sqrt{5-1^2}}\)

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

\(=\dfrac{1}{12}+\dfrac{1}{2}=\dfrac{7}{12}\)

b: \(\lim\limits_{x\rightarrow5}\dfrac{x-5}{\sqrt{x}-\sqrt{5}}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\left(\sqrt{x}-\sqrt{5}\right)\left(\sqrt{x}+\sqrt{5}\right)}{\sqrt{x}-\sqrt{5}}\)

\(=\lim\limits_{x\rightarrow5}\sqrt{x}+\sqrt{5}=\sqrt{5}+\sqrt{5}=2\sqrt{5}\)

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

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{2}{\sqrt{1+x}+1}+\dfrac{1}{4+2\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}\right)=\dfrac{13}{12}\)

2/ \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x+7}-\sqrt{x+3}}{x^2-3x+2}=\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x+7}-2-\left(\sqrt{x+3}-2\right)}{\left(x-1\right)\left(x-2\right)}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{x-1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}-\dfrac{x-1}{\sqrt{x+3}+2}}{\left(x-1\right)\left(x-2\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}-\dfrac{1}{\sqrt{x+3}+2}}{x-2}=\dfrac{1}{6}\)

3/ \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x^2+7}-\sqrt{5-x^2}}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x^2+7}-2+2-\sqrt{5-x^2}}{x^2-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{\left(x^2-1\right)}{\sqrt[3]{\left(x^2+7\right)^2}+2\sqrt[3]{x^2+7}+4}+\dfrac{x^2-1}{2+\sqrt{5-x^2}}}{x^2-1}\)

\(=\lim\limits_{x\rightarrow1}\left(\dfrac{1}{\sqrt[3]{\left(x^2+7\right)^2}+2\sqrt[3]{x^2+7}+4}+\dfrac{1}{2+\sqrt{5-x^2}}\right)=\dfrac{1}{3}\)

4/ \(\lim\limits_{x\rightarrow-2}\dfrac{\sqrt{x+11}-\sqrt[3]{8x+43}}{2x^2+3x-2}=\lim\limits_{x\rightarrow-2}\dfrac{\sqrt{x+11}-3-\left(\sqrt[3]{8x+43}-3\right)}{\left(2x-1\right)\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow-2}\dfrac{\dfrac{x+2}{\sqrt{x+11}+3}-\dfrac{8\left(x+2\right)}{\sqrt[3]{\left(8x+43\right)^2}+3\sqrt[3]{8x+43}+9}}{\left(2x-1\right)\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow-2}\dfrac{\dfrac{1}{\sqrt{x+11}+3}-\dfrac{8}{\sqrt[3]{\left(8x+43\right)^2}+3\sqrt[3]{8x+43}+9}}{2x-1}=\dfrac{7}{270}\)

5/ \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[n]{1+ax}-\sqrt[m]{1+bx}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\sqrt[n]{1+ax}-1-\left(\sqrt[m]{1+bx}-1\right)}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{ax}{\sqrt[n]{\left(1+ax\right)^{n-1}}+\sqrt[n]{\left(1+ax\right)^{n-2}}+...+1}-\dfrac{bx}{\sqrt[m]{\left(1+bx\right)^{m-1}}+\sqrt[m]{\left(1+ax\right)^{m-2}}+...+1}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{a}{\sqrt[n]{\left(1+ax\right)^{n-1}}+\sqrt[n]{\left(1+ax\right)^{n-2}}+...+1}-\dfrac{b}{\sqrt[m]{\left(1+bx\right)^{m-1}}+\sqrt[m]{\left(1+ax\right)^{m-2}}+...+1}\)

\(=\dfrac{a}{n}-\dfrac{b}{m}\)

6/ \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}-1}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}-\sqrt{1+4x}+\sqrt{1+4x}-1}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\left(\sqrt[3]{1+6x}-1\right)+\sqrt{1+4x}-1}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\dfrac{6x}{\sqrt[3]{\left(1+6x\right)^2}+\sqrt[3]{1+6x}+1}+\dfrac{4x}{\sqrt{1+4x}+1}}{x}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{6\sqrt{1+4x}}{\sqrt[3]{\left(1+6x\right)^2}+\sqrt[3]{1+6x}+1}+\dfrac{4}{\sqrt{1+4x}+1}\right)=4\)