\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}\)

Ta lấy vễ trên chia vế dưới

\(=3.2=6\)

\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}\)

Ta lấy vế trên chia vế dưới

\(=2^3.3=24\)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.3^{32}}=3.2=6\)
\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}=3.2=6\)

\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)

Em đăng mỗi lần 1 câu thôi thì mn sẽ giúp em nhanh hơn nhé!

Bài 16: 

a: Xét ΔOEH và ΔOFH có 

OE=OF

\(\widehat{EOH}=\widehat{FOH}\)

OH chung

Do đó: ΔOEH=ΔOFH

Bài 15:

a: Xét ΔABM và ΔACN có 

AB=AC

\(\widehat{NAC}\) chung

AM=AN

Do đó: ΔABM=ΔACN

b: Xét ΔNBC và ΔMCB có

NB=MC

\(\widehat{NBC}=\widehat{MCB}\)

BC chung

Do đó: ΔNBC=ΔMCB

Suy ra: \(\widehat{NCB}=\widehat{MBC}\)

hay \(\widehat{OBC}=\widehat{OCB}\)

Bài 16: 

1) \(x^2+4x+4=\left(x+2\right)^2\)

2) \(x^2+6x+9=\left(x+3\right)^2\)

3) \(4x^2+4x+1=\left(2x+1\right)^2\)

4) \(9+12x+4x^2=\left(2x+3\right)^2\)

5) \(x^2-2x+1=\left(x-1\right)^2\)

6) \(x^2-8x+16=\left(x-4\right)^2\)

7) \(36-12x+x^2=\left(x-6\right)^2\)

8) \(4x^2-12xy+9y^2=\left(2x+3y\right)^2\)

9) \(9x^2-6x+1=\left(3x-1\right)^2\)

10) \(4x^2+12x+9=\left(2x+3\right)^2\)

11) \(x^2+3x+\dfrac{9}{4}=\left(x+\dfrac{3}{2}\right)^2\)

12) \(4x^2-6x+\dfrac{9}{4}=\left(2x-\dfrac{3}{2}\right)^2\)

Bài 16: 

1: \(x^2+4x+4=\left(x+2\right)^2\)

2: \(x^2+6x+9=\left(x+3\right)^2\)

3: \(4x^2+4x+1=\left(2x+1\right)^2\)

4: \(4x^2+12x+9=\left(2x+3\right)^2\)

5: \(x^2-2x+1=\left(x-1\right)^2\)

6: \(x^2-8x+16=\left(x-4\right)^2\)

7: \(36-12x+x^2=\left(6-x\right)^2\)

8: \(4x^2-12xy+9y^2=\left(2x-3y\right)^2\)

a) Vì \(-45< -16\) nên \(\left(-\dfrac{45}{17}\right)^{15}< \left(\dfrac{-16}{17}\right)^{15}\)

b) Vì \(21< 23\) nên \(\left(-\dfrac{8}{9}\right)^{21}< \left(-\dfrac{8}{9}\right)^{23}\)

c) \(27^{40}=3^{3^{40}}=3^{120}\)

\(64^{60}=8^{2^{60}}=8^{120}\)

Vì \(3< 8\) nên \(3^{120}< 8^{120}\) hay \(27^{40}< 64^{60}\)

Thời gian anh Kiên đi là

120 : 50 = 2,4 giờ = 2 giờ 24 phút

Anh Kiên khởi hành lúc

10 giờ 15 phút - 2 giờ 24 phút - 20 phút = 7 giờ 31 phút