Question

[Hà Trí Kiên]

Tính  

Answer 1

[『Kuroba ム Tsuki Ryoo ︵²ᵏ⁷』]

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\(\dfrac{8^{14}}{4^4\cdot64^5}\)

`=`\(\dfrac{2^{42}}{2^8\cdot2^{30}}\)

`=`\(\dfrac{2^{42}}{2^{38}}=2^4=16\)

\(\dfrac{9^{10}\cdot27^7}{81^7\cdot3^{15}}\)

`=`\(\dfrac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}\)

`=`\(\dfrac{3^{41}}{3^{43}}\)

`=`\(\dfrac{1}{3^2}=\dfrac{1}{9}\)

\(\left(\dfrac{3}{10}\right)^4\cdot\left(0,3\right)^5\cdot\left(\dfrac{10}{3}\right)^{10}\)

`=`\(\left(\dfrac{3}{10}\right)^4\cdot\left(\dfrac{3}{10}\right)^5\cdot\left(\dfrac{10}{3}\right)^{10}\)

`=`\(\left(\dfrac{3}{10}\right)^9\cdot\left(\dfrac{10}{3}\right)^{10}=\dfrac{10}{3}\)

\(\dfrac{\left(4^3\right)^2\cdot9^4}{6^7\cdot8^2}\)

`=`\(\dfrac{\left(2^6\right)^2\cdot3^8}{2^7\cdot3^7\cdot2^9}\)

`=`\(\dfrac{2^{12}\cdot3^8}{2^{16}\cdot3^7}\)

`=`\(\dfrac{3}{2^4}=\dfrac{3}{16}\)

\(\dfrac{4^8\cdot9^4}{6^6\cdot8^3}\)

`=`\(\dfrac{2^{16}\cdot3^8}{2^6\cdot3^6\cdot2^9}\)

`=`\(\dfrac{2^{16}\cdot3^8}{2^{15}\cdot3^6}=2\cdot3^2=18\)

\(3^6\cdot\left(\dfrac{1}{3}\right)^6\cdot81^2\cdot\dfrac{1}{27^2}\)

`=`\(\left(3\cdot\dfrac{1}{3}\right)^6\cdot\dfrac{81^2}{27^2}\)

`=`\(1^6\cdot\dfrac{3^8}{3^6}=1\cdot3^2=9\)

`\text {#KaizuulvG}`