thiếu by her nhx câu đó vẫn đúng mà

Đổi:

cos10^o = sin(90^o - 10^o) = sin80^o

cos40^o = sin(90^o - 40^o) = sin50^o

Vì 30 < 50 < 70 < 80

=> sin30^o < sin50^o < sin70^o < sin80^o

=> sin30^o < cos40^o < sin70^o < cos10^o

1. A lot of cards have been sold during the season.

2. This switch mustn't be touched.

3. Exercises won't be corrected by the teacher tomorrow.

4. Cloth puppets used to be made.

5. How many ways can this problem be solved?

6. This dress is used only on occasions.

7. Refreshment will be served.

8. A lot of the work is being gone.

9. My car can't be repaired.

10. A great deal of coffee is drank in China.

1. was hardly locking

2. have sat

3. was sitting

4. would forgive

5. was hung

6. am attending

7. D. for ten years

8. discovered

9. was running

10. still lived

11. will be playing

12. shall have been working

13. am

14. have been

15. leave

Điều kiện: \(\left\{{}\begin{matrix}x^2-2x+1\ne0\\x^2-1\ne0\\x^2+2x+1\ne0\\5x^2+5\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2\ne0\\\left(x-1\right)\left(x+1\right)\ne0\\\left(x+1\right)^2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)

\(E=\left(\dfrac{3x+2}{x^2-2x+1}-\dfrac{6}{x^2-1}-\dfrac{3x-2}{x^2+2x+1}\right).\dfrac{x^2+2x+1}{5x^2+5}\)

\(=\left[\dfrac{3x+2}{\left(x-1\right)^2}-\dfrac{6}{\left(x-1\right)\left(x+1\right)}-\dfrac{3x-2}{\left(x+1\right)^2}\right].\dfrac{\left(x+1\right)^2}{5x^2+5}\)

\(=\dfrac{\left(3x+2\right)\left(x+1\right)^2-6\left(x-1\right)\left(x+1\right)-\left(3x-2\right)\left(x-1\right)^2}{\left(x-1\right)^2.\left(x+1\right)^2}.\dfrac{\left(x+1\right)^2}{5x^2+5}\)

\(=\dfrac{\left(3x+2\right)\left(x^2+2x+1\right)-6\left(x^2-1\right)-\left(3x-2\right)\left(x^2-2x+1\right)}{\left(x-1\right)^2.\left(x+1\right)^2}.\dfrac{\left(x+1\right)^2}{5x^2+5}\)

\(=\dfrac{\left(3x^3+6x^2+3x+2x^2+4x+2\right)-\left(6x^2-6\right)-\left(3x^3-6x^2+3x-2x^2+4x-2\right)}{\left(x-1\right)^2.\left(x+1\right)^2}.\dfrac{\left(x+1\right)^2}{5x^2+5}\)

\(=\dfrac{3x^3+8x^2+7x+2-6x^2+6-3x^3+8x^2-7x+2}{\left(x-1\right)^2.\left(x+1\right)^2}.\dfrac{\left(x+1\right)^2}{5x^2+5}\)

\(=\dfrac{10x^2+10}{\left(x-1\right)^2.\left(x+1\right)^2}.\dfrac{\left(x+1\right)^2}{5x^2+5}\)

\(=\dfrac{10\left(x^2+1\right)}{\left(x-1\right)^2.\left(x+1\right)^2}.\dfrac{\left(x+1\right)^2}{5\left(x^2+1\right)}\)

\(=\dfrac{2}{\left(x-1\right)^2}\)

Since the start of rain the protesters have melted away.

a, 24+25+26+27+28+29+30+31 (có (31-24)+1=8 (phần tử))

= (24+31)+(25+30)+(26+29)+(27+28) (có 8:2=4 (cặp))

= 55+55+55+55+55

= 5.55

= 275

b. 2+4+6+8+10+...+100 (có (100-2):2+1=50 (phần tử))

= (2+100)+(4+98)+(6+96)+...+(50+52) (có 50:2=25 (cặp))

= 102+102+102+...+102

= 102.25

= 2550

Đề phải là: \(A=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{199}{200}\)

\(A=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{199}{200}\)

\(=\dfrac{1}{200}\)

Question 1: A (won't)

Question 2: C (since)

Question 3: D (weren't)

Question 4: B (know)

Question 5: B (happy)

Đề 1:

1. Tìm x:

a. \(\sqrt{x}=3\Leftrightarrow x=3^2\Leftrightarrow x=9\)

Vậy \(x=9.\)

b. \(\sqrt{x^2}+4=x+2\)

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

Điều kiện: \(x-2\ge0\Leftrightarrow x\ge2,\) khi đó phương trình tương đương:

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

\(\Leftrightarrow\left[{}\begin{matrix}0=-2\\2x=2\end{matrix}\right.\)

\(\Leftrightarrow x=1\) (loại)

Vậy phương trình vô nghiệm.

c. Điều kiện: \(x\ge0\)

\(\sqrt{x}=x\Leftrightarrow x=x^2\Leftrightarrow x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

So với điều kiện thấy \(x=0\) hoặc \(x=1\) thỏa mãn.

Vậy \(x\in\left\{0;1\right\}.\)

2. Tìm x:

a. \(\sqrt{x}>1\Leftrightarrow x>1^2\Leftrightarrow x>1\)

Vậy \(x>1.\)

b. \(\sqrt{x}< 3\Leftrightarrow0\le x< 3^2\Leftrightarrow0\le x< 9\)

Vậy \(0\le x< 9.\)

Đề 2:

So sánh

a. Ta có: \(4< 5\Rightarrow\sqrt{4}< \sqrt{5}.\) Do đó: \(2< \sqrt{5}\)

b. Ta có: \(25>5\Rightarrow\sqrt{25}>\sqrt{5}\Leftrightarrow5>\sqrt{5}\Leftrightarrow5-3>\sqrt{5}-3\Leftrightarrow2>\sqrt{5}-3\)

2. Tìm x:

a. \(\sqrt{x}< \sqrt{2}\Leftrightarrow0\le x< 2\)

Vậy \(0\le x< 2.\)

b. Điều kiện: \(\left\{{}\begin{matrix}x\ge0\\2-x\ge0\end{matrix}\right.\Leftrightarrow0\le x\le2\)

\(\sqrt{x}>\sqrt{2-x}\Leftrightarrow x>2-x\Leftrightarrow2x>2\Leftrightarrow x>1\)

So với điều kiện, ta được: \(1< x\le2.\)

Vậy \(1< x\le2.\)

c. \(x^2=2\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)

Vậy \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}.\)

d. \(x^2=5\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)

Vậy \(x\in\left\{\sqrt{5};-\sqrt{5}\right\}.\)