Lời giải:

Coi số cam là 1 phần thì số táo là 2 phần. Số đào là 3 phần.

Tổng số phần bằng nhau: $1+2+3=6$ (phần) 

Số cam là: $24:6\times 1=4$ (quả) 

Số táo là: $24:6\times 2=8$ (quả) 

Số đào là: $24-4-8=12$ (quả)

bài nào vậy bn:?

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

lỗi

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