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7 tháng 1 2023

30 collection

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31 Do you go on social media every day?

32 I don't watch much TV, exept a bit on Youtube

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33 .........forget to give me those stamps

34 ........... fluent E speaker

35 ........ singing an E song?

36 ........ Lan is a poor student, she studies very well

NV
4 tháng 3 2022

4.

\(\lim\limits_{x\rightarrow8}f\left(x\right)=\lim\limits_{x\rightarrow8}\dfrac{\sqrt[3]{x}-2}{x-8}=\lim\limits_{x\rightarrow8}\dfrac{x-8}{\left(x-8\right)\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}=\lim\limits_{x\rightarrow8}\dfrac{1}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}\)

\(=\dfrac{1}{4+4+4}=\dfrac{1}{12}\)

\(f\left(8\right)=3.8-20=4\)

\(\Rightarrow\lim\limits_{x\rightarrow8}f\left(x\right)\ne f\left(8\right)\)

\(\Rightarrow\) Hàm gián đoạn tại \(x=8\)

5.

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

\(=\lim\limits_{x\rightarrow0^+}\left(\dfrac{2}{\sqrt[]{1+2x}+1}-\dfrac{3}{1+\sqrt[3]{1+3x}+\sqrt[3]{\left(1+3x\right)^2}}\right)=\dfrac{2}{1+1}-\dfrac{3}{1+1+1}=0\)

\(f\left(0\right)=\lim\limits_{x\rightarrow0^-}f\left(x\right)=\lim\limits_{x\rightarrow0^-}\left(3x^2-2x\right)=0\)

\(\Rightarrow\lim\limits_{x\rightarrow0^+}f\left(x\right)=\lim\limits_{x\rightarrow0^-}f\left(x\right)=f\left(0\right)\)

\(\Rightarrow\) Hàm liên tục tại \(x=0\)

NV
4 tháng 3 2022

6.

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

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

\(=\lim\limits_{x\rightarrow0^+}\dfrac{\dfrac{-x^2}{\sqrt[]{4x+1}+2x+1}+\dfrac{x^2\left(8x+12\right)}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{6x+1}+\sqrt[3]{\left(6x+1\right)^2}}}{x^2}\)

\(=\lim\limits_{x\rightarrow0^+}\left(\dfrac{-1}{\sqrt[]{4x+1}+2x+1}+\dfrac{8x+12}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{6x+1}+\sqrt[3]{\left(6x+1\right)^2}}\right)\)

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

\(f\left(0\right)=\lim\limits_{x\rightarrow0^-}f\left(x\right)=\lim\limits_{x\rightarrow0^-}\left(2-3x\right)=2\)

\(\Rightarrow\lim\limits_{x\rightarrow0^+}f\left(x\right)\ne\lim\limits_{x\rightarrow0^-}f\left(x\right)\)

\(\Rightarrow\) Hàm gián đoạn tại \(x=0\)

6 tháng 7 2021

mình làm những bài bn chưa lm nhé

9B

10A

bài 2

have repainted

bàii 3

ride - walikking

swimming

watch

6 tháng 7 2021

Dù sao cũng cảm ơn bạn 🥰

c: \(\sqrt{3+\sqrt{8}}=\sqrt{2}+1\)

d: \(\sqrt{11+4\sqrt{6}}=2\sqrt{2}+3\)

e: \(\sqrt{14-6\sqrt{5}}=3-\sqrt{5}\)

4 tháng 9 2021

chị làm chi tiết cho em được ko ạ 

 

25 tháng 10 2021

12 A

13 C

14 C

15 C

16 B

14 A

18 C

19 A

20 D

a: \(\sqrt{2x+3}=5\)

\(\Leftrightarrow2x+3=25\)

hay x=11

b: \(\sqrt{\left(x-2\right)^2}=8\)

\(\Leftrightarrow\left|x-2\right|=8\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=8\\x-2=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-6\end{matrix}\right.\)

2 tháng 10 2021

a) \(\sqrt{3+2x}=5\left(đk:x\ge-\dfrac{3}{2}\right)\)

\(\Leftrightarrow3+2x=25\Leftrightarrow x=11\left(tm\right)\)

b) \(\sqrt{\left(x-2\right)^2}=8\)

\(\Leftrightarrow\left|x-2\right|=8\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=8\\x-2=-8\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-6\end{matrix}\right.\)

c) \(đk:x\le3\)

\(\Leftrightarrow\sqrt{3-x}-3\sqrt{3-x}+5\sqrt{3-x}=6\)

\(\Leftrightarrow3\sqrt{3-x}=6\)

\(\Leftrightarrow\sqrt{3-x}=2\Leftrightarrow3-x=4\Leftrightarrow x=-1\left(tm\right)\)

d) \(đk:x\ge0\)

\(\Leftrightarrow4\sqrt{x}-6\sqrt{x}+4\sqrt{x}=5\)

\(\Leftrightarrow2\sqrt{x}=5\Leftrightarrow\sqrt{x}=\dfrac{5}{2}\Leftrightarrow x=\dfrac{25}{4}\left(tm\right)\)

e) \(đk:x\ge-5\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\Leftrightarrow\sqrt{x+5}=2\Leftrightarrow x+5=4\Leftrightarrow x=-1\left(tm\right)\)

f) \(đk:x\ge-2\)

\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)

\(\Leftrightarrow2\sqrt{x+2}=6\Leftrightarrow\sqrt{x+2}=3\Leftrightarrow x+2=9\Leftrightarrow x=7\left(tm\right)\)

7 tháng 9 2023

Máy mình lỗi hay đúng là không có câu hỏi nào vậy nhỉ?

7D

8C

9D

10B

11C