Giúp em với ạ.
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UH
0
giup em voi a em dang can gap a
GL
2 tháng 10 2023
1. Going in the rain is interesting.
2. Don't forget to learn by heart new words.
3. If he doesn't apologize me, I won't forgive him.
4. My grandmother is the most helpful (person) in my village.
5. What is the price of that hat?
6. Your book is different from my book.
6 tháng 9 2021
Bài 3:
a: \(15x^2y-10xy^2=5xy\left(3x-2y\right)\)
b: \(x^2+2xy+y^2-9=\left(x+y-3\right)\left(x+y+3\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 9 2021
ĐKXĐ cho căn thức: \(x\ge-\dfrac{1}{2}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{3x+1-\sqrt{2x+1}}{x^2-x}=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{3}{x}+\dfrac{1}{x^2}-\sqrt{\dfrac{2}{x^3}+\dfrac{1}{x^4}}}{1-\dfrac{1}{x}}=\dfrac{0}{1}=0\)
\(\Rightarrow y=0\) là TCN
\(\lim\limits_{x\rightarrow0}\dfrac{3x+1-\sqrt{2x+1}}{x^2-x}=\lim\limits_{x\rightarrow0}\dfrac{9x^2+4x}{x\left(x-1\right)\left(3x+1+\sqrt{2x+1}\right)}=\lim\limits_{x\rightarrow0}\dfrac{9x+4}{\left(x-1\right)\left(3x+1+\sqrt{2x+1}\right)}\)
\(=\dfrac{4}{-1\left(1+1\right)}\) hữu hạn
\(\Rightarrow x=0\) không phải tiệm cận
\(\lim\limits_{x\rightarrow1}\dfrac{3x+1-\sqrt{2x+1}}{x\left(x-1\right)}=\dfrac{4-\sqrt{3}}{0}=+\infty\Rightarrow x=1\) là TCĐ
Đồ thị hàm số có 2 tiệm cận
DT
Đỗ Thanh Hải
CTVVIP
6 tháng 12 2021
1 is washing
2 aren't watching
3 am having
4 is studying
5 are staying
6 are rising
7 are wautubg
8 are becoming
e; (1\(\dfrac{1}{3}\) - 25%\(x\) - \(\dfrac{5}{12}\)) - 2\(x\) = 1,6 : \(\dfrac{3}{5}\)
\(\dfrac{4}{3}\) - 0,25\(x\) - \(\dfrac{5}{12}\) - 2\(x\) = \(\dfrac{8}{3}\)
(\(\dfrac{4}{3}\) - \(\dfrac{5}{12}\)) - (0,25\(x\) + 2\(x\)) = \(\dfrac{8}{3}\)
\(\dfrac{11}{12}\) - 2,25\(x\) = \(\dfrac{8}{3}\)
2,25\(x\) = \(\dfrac{11}{12}\) - \(\dfrac{8}{3}\)
2,25\(x\) = - \(\dfrac{7}{4}\)
\(x\) = - \(\dfrac{7}{4}\) : 2,25
\(x\) = - \(\dfrac{7}{9}\)
Vậy \(x=-\dfrac{7}{9}\)
f; \(\dfrac{1}{2}\).(\(x-\dfrac{2}{3}\)) - \(\dfrac{1}{3}\) (2\(x\) - 3) = \(x\)
\(\dfrac{1}{2}\)\(x\) - \(\dfrac{1}{3}\) - \(\dfrac{2}{3}x\) + 1= \(x\)
(\(\dfrac{1}{2}\) - \(\dfrac{2}{3}\))\(x\) + (1 - \(\dfrac{1}{3}\)) = \(x\)
- \(\dfrac{1}{6}\)\(x\) + \(\dfrac{2}{3}\) = \(x\)
\(x\) + \(\dfrac{1}{6}\)\(x\) = \(\dfrac{2}{3}\)
\(\dfrac{7}{6}\)\(x\) = \(\dfrac{2}{3}\)
\(x\) = \(\dfrac{2}{3}\) : \(\dfrac{7}{6}\)
\(x\) = \(\dfrac{4}{7}\)
Vậy \(x\) = \(\dfrac{4}{7}\)