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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
2T
1
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
Bài 1 :
a) \(-\frac{3}{4}.31\frac{11}{23}-0,75.8\frac{12}{23}\)
\(=\left(-1\right).\frac{3}{4}.31\frac{11}{23}-\frac{3}{4}.8\frac{12}{23}\)
\(=\frac{3}{4}.\left[\left(-1\right).31\frac{11}{23}-8\frac{12}{23}\right]\)
\(=\frac{3}{4}.\left(-31\frac{11}{23}-8\frac{12}{23}\right)\)
\(=\frac{3}{4}.\left(-40\right)=-30\)
b) \(4\frac{5}{9}:\left(-\frac{5}{7}\right)+5\frac{4}{9}:\left(-\frac{5}{7}\right)\)
\(=4\frac{5}{9}.\frac{7}{5}+5\frac{4}{9}.\frac{7}{5}\)
\(=\left(4\frac{5}{9}+5\frac{4}{9}\right).\frac{7}{5}\)
\(=10.\frac{7}{5}=14\)
c) \(\left(\frac{12}{35}-\frac{6}{7}+\frac{18}{14}\right):\frac{6}{-7}-\frac{-2}{5}-\left(-2021\right)^0\)
\(=\left(\frac{12}{35}-\frac{30}{35}+\frac{45}{35}\right):\frac{6}{-7}+\frac{2}{5}-1\)
\(=27.\frac{-7}{6}-\frac{3}{5}\)
\(=\frac{-321}{10}\)
Bài 2 :
a) \(\left(2x-3\right)\left(\frac{3}{4}x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\\frac{3}{4}x+1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=3\\\frac{3}{4}x=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{-4}{3}\end{cases}}\)
b) \(\frac{2}{3}x+\left(\frac{5}{7}\right)^0=\frac{3}{10}\)
\(\Rightarrow\frac{2}{3}x+1=\frac{3}{10}\)
\(\Rightarrow\frac{2}{3}x=\frac{-7}{10}\)
\(\Rightarrow x=\frac{-7}{10}:\frac{2}{3}=\frac{-21}{20}\)
c) \(\frac{3}{7}+\frac{1}{7}:x=\frac{3}{14}\)
\(\Rightarrow\frac{1}{7}:x=\frac{3}{14}-\frac{3}{7}=\frac{-3}{14}\)
\(\Rightarrow x=\frac{1}{7}:\frac{-3}{14}=\frac{-2}{3}\)
d) \(\frac{x+5}{2005}+\frac{x+6}{2004}+\frac{x+7}{2003}=-3\)
\(\Rightarrow\left(\frac{x+5}{2005}+1\right)+\left(\frac{x+6}{2004}+1\right)+\left(\frac{x+7}{2003}+1\right)=-3+1+1+1\)
\(\Rightarrow\frac{x+2010}{2005}+\frac{x+2010}{2004}+\frac{x+2010}{2003}=0\)
\(\Rightarrow\left(x+2010\right).\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)=0\)
\(\Rightarrow x+2010=0\left(\text{ do }\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\ne0\right)\)
=> x = -2010