Giúp em vơi ạ
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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.
Bài 3:
a: \(15x^2y-10xy^2=5xy\left(3x-2y\right)\)
b: \(x^2+2xy+y^2-9=\left(x+y-3\right)\left(x+y+3\right)\)
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
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
\(\overrightarrow{AB}=\left(1;1\right)\Rightarrow AB=\sqrt{2}\)
Từ C hạ CH vuông góc AB \(\Rightarrow S_{ABC}=\dfrac{1}{2}CH.AB\Rightarrow CH=\dfrac{2S_{ABC}}{AB}=\dfrac{3}{\sqrt{2}}\)
Từ G hạ GK vuông góc AB, gọi M là trung điểm AB
Theo định lý Talet: \(\dfrac{GK}{CH}=\dfrac{GM}{CM}=\dfrac{1}{3}\Rightarrow d\left(G;AB\right)=GK=\dfrac{CH}{3}=\dfrac{\sqrt{2}}{2}\)
Phương trình AB có dạng:
\(1\left(x-2\right)-1\left(y+3\right)=0\Leftrightarrow x-y-5=0\)
G thuộc d nên tọa độ có dạng: \(G\left(a;3a-8\right)\)
\(d\left(G;AB\right)=\dfrac{\sqrt{2}}{2}=\dfrac{\left|a-\left(3a-8\right)-5\right|}{\sqrt{1^2+\left(-1\right)^2}}=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left|-2a+3\right|=1\Rightarrow\left[{}\begin{matrix}a=1\\a=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}G\left(1;-5\right)\\G\left(2;-2\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x_C=3x_G-\left(x_A+x_B\right)=...\\y_C=3y_G-\left(y_A+y_B\right)=...\end{matrix}\right.\)