\(\frac{72^2}{24^2}=\frac{\left(2^3\cdot3^2\right)^2}{\left(2^3\cdot3\right)^2}=\frac{2^6\cdot3^4}{2^6\cdot3^2}=3^2=9\)

\(\frac{15^3}{27}=\frac{3^3\cdot5^3}{3^3}=5^3=125\)

72^2/24^2 = (72/24)^2 = 3^2 = 9

15^3/27 = 15^3/3^3 = (15/3)^3 = 5^3 = 125

\(\frac{72^2}{24^2}=\left(\frac{72}{24}\right)^2=3^2=9\)

\(\frac{15^3}{27^3}=\frac{15^3}{3^3}=\left(\frac{15}{3}\right)^3=5^3=125\)

\(\frac{72^2}{24^2}=\sqrt{\frac{72^2}{24^2}}=\frac{72}{24}=3\)

\(\frac{72^2}{24^2}\)=\(\frac{72^2}{24}\)=\(3^2\)=3

mình nghĩ là đề như vậy:

\(\frac{24}{8.16}-\frac{40}{16.24}+\frac{56}{24.32}-\frac{72}{32.40}=\frac{8+16}{8.16}-\frac{16+24}{16.24}+\frac{24+32}{24.32}-\frac{32+40}{32.40}\)

\(=\frac{8}{8.16}+\frac{16}{8.16}-\frac{16}{16.24}-\frac{24}{16.24}+\frac{24}{24.32}+\frac{32}{24.32}-\frac{32}{32.40}-\frac{40}{32.40}\)

\(=\frac{1}{16}+\frac{1}{8}-\frac{1}{24}-\frac{1}{16}+\frac{1}{32}+\frac{1}{24}-\frac{1}{40}-\frac{1}{32}\)

\(=\frac{1}{8}-\frac{1}{40}=\frac{1}{10}\)

72²/24 = 216

(-7,5)³/(2,5)³ = 27

15³/27 = 125

63 : 7 + 24 x 2 – ( 81 – 72)

= 63 : 7 + 24 x 2 – 9

= 9 + 24 x 2 – 9

= 9 + 48 – 9

= 48

\(\frac{72^2}{24^2}=\frac{72}{24}=3\)

\(\frac{\left(-7,5\right)^3}{\left(2,5\right)^3}=\frac{-7,5}{2,5}=-3\)

\(\frac{15^3}{27}=\frac{\left(3.5\right)^3}{\left(3.9\right)}=\frac{3^3.5^3}{3.9}=\frac{3^2.5^3}{9}=\frac{9.5^3}{9}=5^3=125\)

a) \(\frac{x}{6}^2=\frac{24}{25}\)

\(\Rightarrow x^2.25=6.24\)

\(\Rightarrow x^2.25=144\)

\(\Rightarrow x^2=144\div25\)

\(\Rightarrow x^2=5,76=2,4^2=\left(-2,4^2\right)\)

\(\Rightarrow x\in\left\{2,4;-2,4\right\}\)

Vậy \(x\in\left\{2,4;-2,4\right\}\)

b) \(\frac{72-x}{7}=\frac{x-40}{9}\)

\(\Rightarrow\left(72-x\right).9=\left(x-40\right).7\)

\(\Rightarrow648-9x=7x-280\)

\(\Rightarrow\left(-280\right)-648\) \(=-9x-7x\)

\(\Rightarrow-928=-16x\)

\(\Rightarrow x=58\)

Vậy \(x=58\)