\(\frac{x-3}{-\left(x-3\right)}=-1\) rút gọn tử mẫu cái ra luôn mà

Nguyễn Bích Hà

Ko dịch được đề, đoán đại là \(\lim\limits\left(\sqrt[3]{n+1}-\sqrt[3]{n}\right)\) (hay là \(3\sqrt{n+1}-3\sqrt{n}\) ?)

\(\lim\limits\left(\sqrt[3]{n+1}-\sqrt[3]{n}\right)=lim\frac{\left(\sqrt[3]{n+1}-\sqrt[3]{n}\right)\left(\sqrt[3]{\left(n+1\right)^2}+\sqrt[3]{n\left(n+1\right)}+\sqrt[3]{n^2}\right)}{\sqrt[3]{\left(n+1\right)^2}+\sqrt[3]{n\left(n+1\right)}+\sqrt[3]{n^2}}\)

\(=lim\frac{1}{\sqrt[3]{\left(n+1\right)^2}+\sqrt[3]{n\left(n+1\right)}+\sqrt[3]{n^2}}=0\)

Giải giúp mk với

Giúp mk giải những câu này với khó quá trời

Câu 16:

\(\lim\limits_{x\rightarrow-\infty}\frac{\sqrt{x^4+2x^2+1}}{\left(2x-1\right)\left(x-3\right)}=\lim\limits_{x\rightarrow-\infty}\frac{\sqrt{\left(x^2+1\right)^2}}{2x^2-7x+3}=\lim\limits_{x\rightarrow-\infty}\frac{x^2+1}{2x^2-7x+3}=\lim\limits_{x\rightarrow-\infty}\frac{1+\frac{1}{x^2}}{2-\frac{7}{x}+\frac{3}{x^2}}=\frac{1}{2}\)

Câu 18:

\(\lim\limits_{x\rightarrow+\infty}\frac{x^2+3x}{x^2-5x}=\lim\limits_{x\rightarrow+\infty}\frac{1+\frac{3}{x}}{1-\frac{5}{x}}=1\)

\(\lim\limits_{x\rightarrow0^-}\frac{1}{\left|x\right|}=+\infty\)

\(\lim\limits_{x\rightarrow-\infty}\frac{\left(2x-1\right)\left(x+3\right)}{3x^3-2}=\lim\limits_{x\rightarrow-\infty}\frac{2x^2+5x-3}{3x^3-2}=\lim\limits_{x\rightarrow-\infty}\frac{\frac{2}{x}+\frac{5}{x}-\frac{3}{x^3}}{3-\frac{2}{x^3}}=\frac{0}{3}=0\)

\(\lim\limits_{x\rightarrow2^+}\frac{2x+5}{2-x}=\lim\limits_{x\rightarrow2^+}\frac{-2x-5}{x-2}=\frac{-9}{0}=-\infty\)

Bạn tự đánh giá đáp án nào đúng nhé

Câu 19:

\(\lim\limits_{x\rightarrow+\infty}\frac{3x+1}{\sqrt{x^2+3}+2x}=\lim\limits_{x\rightarrow+\infty}\frac{3+\frac{1}{x}}{\sqrt{1+\frac{3}{x^2}}+2}=\frac{3}{1+2}=1\)

Câu 20:

\(\lim\limits_{x\rightarrow2^+}f\left(x\right)=\lim\limits_{x\rightarrow2^+}\left(2x^2-3mx-1\right)=7-6m\)

\(f\left(2\right)=7-6m\)

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

Để hàm số có giới han khi \(x\rightarrow2\)

\(\Leftrightarrow f\left(2\right)=\lim\limits_{x\rightarrow2^+}f\left(x\right)=\lim\limits_{x\rightarrow2^-}f\left(x\right)\)

\(\Leftrightarrow7-6m=1\Rightarrow m=1\)

21.

\(\lim\limits_{x\rightarrow4}\frac{\sqrt{2x+1}-3}{x^2-5x+4}=\lim\limits_{x\rightarrow4}\frac{\left(\sqrt{2x+1}-3\right)\left(\sqrt{2x+1}+3\right)}{\left(x-4\right)\left(x-1\right)\left(\sqrt{2x+1}+3\right)}\)

\(=\lim\limits_{x\rightarrow4}\frac{2x-8}{\left(x-4\right)\left(x-1\right)\left(\sqrt{2x+1}+3\right)}=\lim\limits_{x\rightarrow4}\frac{2}{\left(x-1\right)\left(\sqrt{2x+1}+3\right)}=\frac{2}{3.\left(3+3\right)}=\frac{1}{9}\)

22.

Khi \(x\rightarrow3\) \(\Rightarrow\left\{{}\begin{matrix}x-1>0\\x+2>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left|x-1\right|=x-1\\\left|x+2\right|=x+2\end{matrix}\right.\)

\(\lim\limits_{x\rightarrow3}\frac{\left|x-1\right|-2}{5-\left|x+2\right|}=\lim\limits_{x\rightarrow3}\frac{x-1-2}{5-\left(x+2\right)}=\lim\limits_{x\rightarrow3}\frac{x-3}{-\left(x-3\right)}=-1\)

23.

\(\lim\limits_{x\rightarrow2^+}\left[\frac{-40}{x^2+4x-12}+\frac{5}{x-2}\right]=\lim\limits_{x\rightarrow2^+}\left[\frac{-40}{\left(x+6\right)\left(x-2\right)}+\frac{5}{x-2}\right]\)

\(=\lim\limits_{x\rightarrow2^+}\left[\frac{-40+5\left(x+6\right)}{\left(x+6\right)\left(x-2\right)}\right]=\lim\limits_{x\rightarrow2^+}\frac{5x-10}{\left(x+6\right)\left(x-2\right)}=\lim\limits_{x\rightarrow2^+}\frac{5}{x+6}=\frac{5}{2+6}=\frac{5}{8}\)

24.

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

\(=\lim\limits_{x\rightarrow2^-}\frac{4-x}{\left(2-x\right)\left(3-x\right)}=\frac{2}{0}=+\infty\)

25.

\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{x^2+5x}+\sqrt{x^2+8}\right)=\lim\limits_{x\rightarrow-\infty}x\left[-\sqrt{1+\frac{5}{x}}-\sqrt{1+\frac{8}{x^2}}\right]\)

\(=-\infty.\left(-1-1\right)=+\infty\)

26.

\(\lim\limits_{x\rightarrow+\infty}\frac{\left(2x+5\right)^3\left(2-x\right)^4}{x^7+1}=\lim\limits_{x\rightarrow+\infty}\frac{x^3\left(2+\frac{5}{x}\right)^3.x^4.\left(\frac{2}{x}-1\right)^4}{x^7\left(1+\frac{1}{x^7}\right)}=\lim\limits_{x\rightarrow+\infty}\frac{\left(2+\frac{5}{x}\right)^3\left(\frac{2}{x}-1\right)^4}{1+\frac{1}{x^7}}\)

\(=\frac{2^3.\left(-1\right)^4}{1}=8\)

27.

\(\lim\limits_{x\rightarrow3}\frac{\sqrt{2x+3}-\sqrt[3]{7x+6}}{x-3}=\lim\limits_{x\rightarrow3}\frac{\sqrt{2x+3}-3+3-\sqrt[3]{7x+6}}{x-3}\)

\(=\lim\limits_{x\rightarrow3}\frac{\frac{2\left(x-3\right)}{\sqrt{2x+3}+3}-\frac{7\left(x-3\right)}{9+3\sqrt[3]{7x+6}+\sqrt[3]{\left(7x+6\right)^2}}}{x-3}\)

\(=\lim\limits_{x\rightarrow3}\left(\frac{2}{\sqrt{2x+3}+3}-\frac{7}{9+3\sqrt[3]{7x+6}+\sqrt[3]{\left(7x+6\right)^2}}\right)\)

\(=\frac{2}{\sqrt{9}+3}-\frac{7}{9+3\sqrt[3]{27}+\sqrt[3]{27^2}}=\frac{2}{27}\)

Ok bạn nhé

Cảm ơn bạn nhiều!

Bạn ơi tại sao từ đầu = thứ hai mà lại ra kết quả bằng -1 được

Mk hỏi ở cau22 đó