K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

D
datcoder
CTVVIP
14 tháng 8 2023

\(\log_{\dfrac{1}{4}}x>-2\\ \Rightarrow\left\{{}\begin{matrix}x>0\\\log_{\dfrac{1}{4}}x>\log_{\dfrac{1}{4}}16\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x>0\\x< 16\end{matrix}\right.\\ \Leftrightarrow0< x< 16\)

Chọn C.

26 tháng 8 2023

\(0,5^{3x-1}>0,25\)

\(\Leftrightarrow0,5^{3x-1}>0,5^2\)

\(\Leftrightarrow3x-1< 2\)

\(\Leftrightarrow3x< 3\)

\(\Leftrightarrow x< \dfrac{3}{3}\)

\(\Leftrightarrow x< 1\)

Vậy: \(\left(-\infty;1\right)\)

Chọn A

HQ
Hà Quang Minh
Giáo viên
26 tháng 8 2023

Ta có: 

\(f'\left(x\right)=6x^2-2x\\ g'\left(x\right)=3x^2+x\)

Theo đề bài, ta có: 

\(f'\left(x\right)>g'\left(x\right)\\ \Leftrightarrow6x^2-2x>3x^2+x\\ \Leftrightarrow3x^2-3x>0\\ \Leftrightarrow3x\left(x-1\right)>0\\ \Leftrightarrow\left[{}\begin{matrix}x>1\\x< 0\end{matrix}\right.\)

Vậy tập nghiệm của bất phương trình là \(\left(-\infty;0\right)\cup\left(1;+\infty\right)\)

Chọn D.

NV
13 tháng 5 2020

\(\Delta y=4\sqrt{2\left(x+\Delta x\right)-6}-4\sqrt{2x-6}=\frac{8\Delta x}{\sqrt{2x+2\Delta x-6}+\sqrt{2x-6}}\)

\(f'\left(x\right)=\lim\limits_{\Delta\rightarrow0}\frac{\Delta y}{\Delta x}=\lim\limits_{\Delta x\rightarrow0}\frac{8\Delta x}{\Delta x\left(\sqrt{2x+2\Delta x-6}+\sqrt{2x-6}\right)}\)

\(=\lim\limits_{\Delta x\rightarrow0}\frac{8}{\sqrt{2x+2\Delta x-6}+\sqrt{2x-6}}=\frac{8}{2\sqrt{2x-6}}=\frac{4}{\sqrt{2x-6}}\)

b/ \(f'\left(5\right)=\frac{4}{\sqrt{2.5-6}}=2\) ; \(f\left(5\right)=4\sqrt{2.5-6}=8\)

Pt tiếp tuyến: \(y=2\left(x-5\right)+8=2x-2\)

c/ \(f'\left(x\right)>4\Leftrightarrow\frac{4}{\sqrt{2x-6}}>4\Leftrightarrow\frac{1}{\sqrt{2x-6}}>1\)

\(\Leftrightarrow\sqrt{2x-6}< 1\Leftrightarrow2x-6< 1\Rightarrow x< \frac{7}{2}\)

\(\Rightarrow3< x< \frac{7}{2}\)

18 tháng 2 2021

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

2 tháng 3 2021

cảm ơn bạn nhé , giờ mới trả lời được bucminh

 

9 tháng 2 2021

Hic nan qua :( Lam vay

P/s: Anh Lam check all ho em nhung bai em lam nhe :( Em cam on

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

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

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

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

 

Chọn B

18 tháng 5 2022

Tại sao z ạ ? 

24 tháng 11 2023

a: \(\lim\limits_{n\rightarrow+\infty}\dfrac{n^5+n^2-n+2}{\left(2n^3-1\right)\left(n^2+n+1\right)}\)

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

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

\(=\dfrac{1}{2\cdot1}=\dfrac{1}{2}\)

b: \(\lim\limits_{n\rightarrow+\infty}\dfrac{\sqrt{n^2-n+2}}{n+2}\)

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

\(=\lim\limits_{n\rightarrow+\infty}\dfrac{\sqrt{1-\dfrac{1}{n}+\dfrac{2}{n^2}}}{1+\dfrac{2}{n}}=\dfrac{\sqrt{1-0+0}}{1+0}=\dfrac{1}{1}=1\)

c: \(\lim\limits_{n\rightarrow+\infty}\dfrac{n-\sqrt[3]{n^2-n^3}}{n^2+n+1}\)

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

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

d: \(\lim\limits_{n\rightarrow+\infty}\left(n-\sqrt{n^2+n+1}\right)\)

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

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

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

Tham khảo:

undefined