Tính :
\(\frac{72^2}{24^2};\frac{\left(-7,5\right)^3}{\left(2,5\right)^3};\frac{15^3}{27}\)
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\(\frac{72^2}{24^2}=\frac{\left(2^3\cdot3^2\right)^2}{\left(2^3\cdot3\right)^2}=\frac{2^6\cdot3^4}{2^6\cdot3^2}=3^2=9\)
\(\frac{15^3}{27}=\frac{3^3\cdot5^3}{3^3}=5^3=125\)
\(\frac{72^2}{24^2}=\sqrt{\frac{72^2}{24^2}}=\frac{72}{24}=3\)
mình nghĩ là đề như vậy:
\(\frac{24}{8.16}-\frac{40}{16.24}+\frac{56}{24.32}-\frac{72}{32.40}=\frac{8+16}{8.16}-\frac{16+24}{16.24}+\frac{24+32}{24.32}-\frac{32+40}{32.40}\)
\(=\frac{8}{8.16}+\frac{16}{8.16}-\frac{16}{16.24}-\frac{24}{16.24}+\frac{24}{24.32}+\frac{32}{24.32}-\frac{32}{32.40}-\frac{40}{32.40}\)
\(=\frac{1}{16}+\frac{1}{8}-\frac{1}{24}-\frac{1}{16}+\frac{1}{32}+\frac{1}{24}-\frac{1}{40}-\frac{1}{32}\)
\(=\frac{1}{8}-\frac{1}{40}=\frac{1}{10}\)
63 : 7 + 24 x 2 – ( 81 – 72)
= 63 : 7 + 24 x 2 – 9
= 9 + 24 x 2 – 9
= 9 + 48 – 9
= 48
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 \(x=58\)
\(\frac{72^2}{24^2}=\frac{3^2.24^2}{24^2}=3^2=9\)
\(\frac{\left(-7,5\right)^2}{2,5^2}=\frac{\left(7,5\right)^2}{\left(2,5\right)^2}=\frac{\left(2,5\right)^2.3^2}{\left(2,5\right)^2}=9\)
\(\frac{15^2}{27}=\frac{3^2.5^2}{27}=\frac{9.25}{27}=\frac{25}{3}\)
\(\frac{72^2}{24^2}=\left(\frac{72}{24}\right)^2=3^2=9\)
\(\frac{\left(-7,5\right)^3}{\left(2,5\right)^3}=\left(\frac{-7,5}{2,5}\right)^3=\left(-3\right)^3=-27\)
\(\frac{15^3}{27}=\frac{15^3}{3^3}=\left(\frac{15}{3}\right)^3=5^3=125\)
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