Bài 2: 

a: Xét ΔBHA vuông tại H và ΔBHD vuông tại H có 

BH chung

HA=HD

Do đó: ΔBHA=ΔBHD

b: Ta có: ΔBHA=ΔBHD

nên \(\widehat{ABH}=\widehat{DBH}\)

hay BH là tia phân giác của góc ABD

Bài 3:

Kẻ Oz//Ax thì Oz//By

Do đó \(\widehat{zOA}=\widehat{OAx}=32^0;\widehat{zOB}=180^0-\widehat{OBy}=58^0\)

\(\Rightarrow\widehat{AOB}=\widehat{zOA}+\widehat{zOB}=32^0+58^0=90^0\)

Vậy OA⊥OB

4/18 và 3/18

28/24 và 9/24

3/36 và 4/36

\(\frac{2}{9}\)Và \(\frac{1}{6}\)

\(\frac{2}{9}=\frac{2\times2}{9\times2}=\frac{4}{18}\)

\(\frac{1}{6}=\frac{1\times3}{6\times3}=\frac{3}{18}\)

\(\frac{7}{6}\)Và \(\frac{3}{8}\)

\(\frac{7}{6}=\frac{7\times4}{6\times4}=\frac{28}{24}\)

\(\frac{3}{8}=\frac{3\times3}{8\times3}=\frac{9}{24}\)

\(\frac{1}{12}\)Và \(\frac{1}{9}\)

\(\frac{1}{12}=\frac{1\times3}{12\times3}=\frac{3}{36}\)

\(\frac{1}{9}=\frac{1\times4}{9\times4}=\frac{4}{36}\)

de mak bn

đề bài là j bạn?? 

so sánh à bạn 

là tính tổng á

Bài 2: 

Ta có: \(\widehat{A}>\widehat{B}>\widehat{C}\)

nên BC>AC>AB

hay OH<OI<OK

6)  \(\dfrac{8^6}{256}=\dfrac{\left(2^3\right)^6}{2^8}=\dfrac{2^{18}}{2^8}=2^{10}=1024\)

7) \(\left(\dfrac{1}{2}\right)^{15}.\left(\dfrac{1}{4}\right)^{20}=\left(\dfrac{1}{2}\right)^{15}.\left[\left(\dfrac{1}{2}\right)^2\right]^{20}=\left(\dfrac{1}{2}\right)^{15}.\left(\dfrac{1}{2}\right)^{40}=\left(\dfrac{1}{2}\right)^{55}=\dfrac{1}{2^{55}}\)

8)  \(\left(\dfrac{1}{9}\right)^{25}\div\left(\dfrac{1}{3}\right)^{30}=\left(\dfrac{1}{3}\right)^{50}\div\left(\dfrac{1}{3}\right)^{30}=\left(\dfrac{1}{3}\right)^{20}=\dfrac{1}{3^{20}}\)

9)\(\left(\dfrac{1}{16}\right)^3\div\left(\dfrac{1}{8}\right)^2=\left(\dfrac{1}{2}\right)^{12}\div\left(\dfrac{1}{2}\right)^6=\left(\dfrac{1}{2}\right)^6=\dfrac{1}{64}\)

10)  \(\dfrac{27^2.8^5}{6^2.32^3}=\dfrac{3^6.2^{15}}{3^2.2^2.2^{15}}=\dfrac{3^4}{2^2}=\dfrac{81}{4}\)