a: Xét ΔCDA vuông tại A và ΔCBA vuông tại A có

CA chug

DA=BA

Do đó:ΔCDA=ΔCBA

b: Ta có: ΔCDB cân tại C

mà CA là đường cao

nên CA là đường phân giác

c: Xét ΔCEI vuông tại E và ΔCFI vuôg tại F có

CI chung

\(\widehat{ECI}=\widehat{FCI}\)

Do đó:ΔCEI=ΔCFI

Suy ra: CE=CF

Xét ΔCDB có CE/CD=CF/CB

nên EF//DB

1. What time does Nga get up in the morning?

2. You can play games in the afternoon but you must do your homework in the evening.

3. Does Lan walk or ride a bike to school?

1, What time does Nga get up in the morning?

2, You can play games in the afternoon but you must do your homework in the evening.

3, Does Lan walk or ride her bike to school?

1. I am not tall enough to reach the ceiling

2. She is sociable enough to make friends easily

3. He is looking at himself in the mirror at the moment 

4. I have to stop smoking

5. Mr. Kerry used to run 5 miles every morning when he was a teenager

6. We really like her sense of humor

7. He ought not to play computer games

8. Hilary doesn't hace to wear uniform at school

1. I am not tall enough to reach the ceiling

2. She is sociable enough to make friends easily

3. He is looking at himself in the mirror at the moment 

4. I have to stop smoking

5. Mr. Kerry used to run 5 miles every morning when he was a teenager

6. We really like her sense of humor

7. He ought not to play computer games

8. Hilary doesn't hace to wear uniform at school

\(1,=\dfrac{x+3-x}{x\left(x+3\right)}=\dfrac{3}{x\left(x+3\right)}\\ 2,=\dfrac{20x+2x-10}{2x-5}=\dfrac{22x-10}{2x-5}\\ 3,=\dfrac{x+5+x-7+x-4}{x-2}=\dfrac{3\left(x-2\right)}{x-2}=3\\ 4,=\dfrac{x^2+3x-x^2-2x-1}{x\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x\left(x+1\right)}\\ 5,=\dfrac{3xy+y^3-3xy-x^3}{x^2y^2}=\dfrac{y^3-x^3}{x^2y^2}\\ 6,=\dfrac{x^2-xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{x}{x+y}\)

\(7,=\dfrac{x^2+6x-2x+4}{x\left(x+2\right)\left(x-2\right)}=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)\left(x-2\right)}=\dfrac{x+2}{x\left(x-2\right)}\\ 8,=\dfrac{4x^2+8x+3x-6-32}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x^2+11x-38}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-2\right)\left(4x+19\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x+19}{x+2}\\ 9,=\dfrac{-5x-1-25x^2+15x}{x\left(5x-1\right)\left(5x+1\right)}=\dfrac{-\left(5x-1\right)^2}{x\left(5x-1\right)\left(5x+1\right)}=\dfrac{1-5x}{x\left(5x+1\right)}\\ 10,=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x-4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}=\dfrac{1}{x}\)