\(a)4^7:2^5\\ =\left(2^2\right)^7:2^5\\ =2^{14}:2^5\\ =2^9\\ b)3^{10}:9^3\\ =3^{10}:\left(3^2\right)^3\\ =3^{10}:3^6\\ =3^4\\ c)27^9:3^{10}\\ \left(3^3\right)^9:3^{10}\\ =3^{27}:3^{10}\\ =3^{17}\\ d)25^5:5^3\\ =\left(5^2\right)^5:5^3\\ =5^{10}:5^3\\ =5^7\\ e)36^7:6^4\\ =\left(6^2\right)^7:6^4\\ =6^{14}:6^4\\ =6^{10}\\ g)4^3\cdot8^4\\ =\left(2^2\right)^3\cdot\left(2^3\right)^4\\ =2^6\cdot3^{12}\\ =2^{18}\)

đâu

Olm chào em,hiện tại yêu cầu của em chưa cụ thể, vậy em vui lòng đăng lại câu hỏi để được sự trợ giúp tốt nhất từ olm cho tài khoản vip em nhé. 

M là trung điểm của đoạn nào vậy bạn?

\(\left(\dfrac{2}{3}\right)^8:\left(\dfrac{4}{9}\right)^3\\ =\left[\left(\dfrac{2}{3}\right)^2\right]^4:\left(\dfrac{4}{9}\right)^3\\ =\left(\dfrac{4}{9}\right)^4:\left(\dfrac{4}{9}\right)^3\\ =\dfrac{4}{9}\)

\(27^3:3^2\\ =\left(3^3\right)^3:3^2\\ =3^9:3^2\\ =3^7\\ =2187\)

\(\left(-\dfrac{3}{5}\right)^4:\left(\dfrac{5}{2}\right)^4\\ =\left(-\dfrac{3}{5}:\dfrac{5}{2}\right)^4\\ =\left(-\dfrac{6}{25}\right)^4\)

\(\left(\dfrac{3}{5}\right)^{12}:\left(\dfrac{9}{25}\right)^5\\ =\left(\dfrac{3}{5}\right)^{12}:\left[\left(\dfrac{3}{5}\right)^2\right]^5\\ =\left(\dfrac{3}{5}\right)^{12}:\left(\dfrac{3}{5}\right)^{10}\\ =\left(\dfrac{3}{5}\right)^2\\ =\dfrac{9}{25}\)

\(\left(\dfrac{2}{3}\right)^8:\left(\dfrac{4}{9}\right)^3=\left(\dfrac{4}{9}\right)^4:\left(\dfrac{4}{9}\right)^3=\dfrac{4}{9}\)

\(27^3:3^2=3^9:3^2=3^7=2187\)

a) Ta có: 

\(64^8=\left(2^6\right)^8=2^{6\cdot8}=2^{48}\)

\(16^{12}=\left(2^4\right)^{12}=2^{4\cdot12}=2^{48}\)

\(\Rightarrow64^8=16^{12}\)

b) Ta có:

\(\left(\dfrac{1}{16}\right)^{10}=\left[\left(\dfrac{1}{2}\right)^4\right]^{10}=\left(\dfrac{1}{2}\right)^{4\cdot10}=\left(\dfrac{1}{2}\right)^{40}\)

Mà: 50 > 40 => `(1/2)^50<(1/2)^40` 

c) Ta có: 

\(\left(\dfrac{9}{16}\right)^{100}=\left[\left(\dfrac{3}{4}\right)^2\right]^{100}=\left(\dfrac{3}{4}\right)^{200}\)

Mà: `3/4>2/3=>(3/4)^200>(2/3)^200`

\(^{^{ }}\)a,64^8=16^12

b,(1/16)^10<(1/2)^50

c,(2/3)^200>(9/16)^100

CỦA BẠN ĐÂY NẾU SAI THÌ CHO MÌNH XIN LỖI NHÉ

a; 25 x 5³ x \(\dfrac{1}{625}\) x 5²

   = 5² x 5³ x \(\dfrac{1}{5^4}\) x 5²

= 5⁵ x \(\dfrac{1}{5^4}\) x 5²

= 5 x 5²

= 5³

a) 

\(25\cdot5^3\cdot\dfrac{1}{625}\cdot5^2\\ =\left(5^2\cdot5^3\cdot5^2\right)\cdot\dfrac{1}{625}\\ =5^7\cdot\dfrac{1}{5^4}\\ =5^3\)

b) 

\(5^2\cdot3^5\cdot\left(\dfrac{3}{5}\right)^2\\ =5^2\cdot3^5\cdot\dfrac{3^2}{5^2}\\ =3^5\cdot3^2\\ =3^7\)

c) 

\(\left(-\dfrac{1}{7}\right)^4\cdot49^2\\ =\dfrac{\left(-1\right)^4}{7^4}\cdot\left(7^2\right)^2\\ =\dfrac{1}{7^4}\cdot7^4\\ =1\)

d) 

\(\left(\dfrac{1}{16}\right)^2:\left(\dfrac{1}{2}\right)^4\cdot\left(-\dfrac{1}{8}\right)^3\\ =\left[\left(\dfrac{1}{2}\right)^4\right]^2:\left(\dfrac{1}{2}\right)^4\cdot\left[\left(-\dfrac{1}{2}\right)^3\right]^3\\ =\left(\dfrac{1}{2}\right)^8:\left(\dfrac{1}{2}\right)^4\cdot\left(-\dfrac{1}{2}\right)^9\\ =\left(\dfrac{1}{2}\right)^4\cdot\left(-\dfrac{1}{2}\right)^9\\ =\left(\dfrac{1}{2}\right)^4\cdot-\left(\dfrac{1}{2}\right)^9\\ =-\left(\dfrac{1}{2}\right)^{13}\)

a; \(x\) : (- \(\dfrac{1}{3}\))³ = \(\dfrac{1}{9}\)

    \(x\) : (\(-\dfrac{1}{27}\)) = \(\dfrac{1}{9}\)

    \(x\)               = \(\dfrac{1}{9}\) x (- \(\dfrac{1}{27}\))

    \(x\)               = - \(\dfrac{1}{243}\)

    Vậy  \(x\)    =  - \(\dfrac{1}{243}\)

b; (\(\dfrac{4}{5}\))⁵ x \(x\) = (\(\dfrac{4}{5}\))⁷

               \(x\) = (\(\dfrac{4}{5}\))⁷ : (\(\dfrac{4}{5}\))⁵

               \(x\) = \(\dfrac{4^7}{5^7}\) : \(\dfrac{4^5}{5^5}\)

               \(x\) = \(\dfrac{4^7}{5^7}\) x \(\dfrac{5^5}{4^5}\)

               \(x\) = \(\dfrac{4^2}{5^2}\)

               \(x\) = \(\dfrac{16}{25}\)

Vậy  \(x\) = \(\dfrac{16}{25}\)

a; 5 - (- \(\dfrac{5}{11}\) ) + (\(\dfrac{1}{3}\))² : 3

  =   5 + \(\dfrac{5}{11}\) + \(\dfrac{1}{9}\) : 3

  = \(\dfrac{55}{11}\) + \(\dfrac{5}{11}\) + \(\dfrac{1}{9}\) x \(\dfrac{1}{3}\)

  =   \(\dfrac{55}{11}\) + \(\dfrac{5}{11}\) + \(\dfrac{1}{27}\)

  =  \(\dfrac{60}{11}\) + \(\dfrac{1}{27}\)

  =   \(\dfrac{1620}{297}\) + \(\dfrac{11}{297}\)

 = \(\dfrac{1631}{297}\)

b; 2³ + 3 x (\(\dfrac{1}{2}\))⁰ + (- 2)² : \(\dfrac{1}{2}\)

=    8 + 3 x 1 + 4 : \(\dfrac{1}{2}\)

=    8 + 3 + 4 x \(\dfrac{2}{1}\)

= 8 + 3 + 8

= 11 + 8

=  19

a: ΔABC cân tại A

mà AD là đường cao

nên D là trung điểm của BC

ΔADB vuông tại D

=>\(DA^2+DB^2=AB^2\)

=>\(DB=\sqrt{5^2-4^2}=3\left(cm\right)\)

b: Xét ΔHDB vuông tại D và ΔHEA vuông tại E có

\(\widehat{DHB}=\widehat{EHA}\)(hai góc đối đỉnh)

Do đó: ΔHDB~ΔHEA

=>\(\dfrac{HD}{HE}=\dfrac{HB}{HA}\)

=>\(HD\cdot HA=HB\cdot HE\)