4.36.25 = (4.25).36 = 100.36 = 3600

I.

4. B

5. C

6. B

7. A

8. B

9. A

10. D

II.

1. C

2. A

3. A

4. A

5. C

6. C

7. A

8. C

9. C

32

112 16

112 \(⋮̸\)16

3. C

4. B

a. Ta có: \(AD\perp BC\) \(\Rightarrow\) \(\widehat{ADB}=\widehat{ADC}=90^o\)

Ta có: AD là tia phân giác góc A

\(\Rightarrow\widehat{BAD}=\widehat{CAD}\)

Xét \(\Delta\)ADB và \(\Delta\)ADC có:

      \(\widehat{ADB}=\widehat{ADC}=90^o\)

cgv AD: Cạnh chung

gnk \(\widehat{BAD}=\widehat{CAD}\)

\(\Rightarrow\)\(\Delta ADB=\Delta ADC\) (cgv - gnk)

b. Ta có: \(\Delta\)ADB = \(\Delta\)ADC 

\(\Rightarrow\) \(\left\{{}\begin{matrix}AB=AC\\DB=DC\end{matrix}\right.\)

DB = DC \(\Rightarrow\) D là trung điểm BC

disagree