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Ta có:
\(A=\dfrac{2^3.3^3.\left(2+3\right)}{2^4.3^3.\left(2-1\right)}=\dfrac{5}{2.1}=\dfrac{5}{2}\)
Giải:
Ta có: \(A=\dfrac{2^4.3^3+2^3.3^4}{2^5.3^3-2^4.3^3}.\)
\(=\dfrac{2^4.3^3+2^3.3^3.3}{2^5.3^3-2^4.3^3}.\)
\(=\dfrac{3^3\left(2^4+2^3.3\right)}{3^3\left(2^5-2^4\right)}.\)
\(=\dfrac{16+24}{32-16}\).
\(=\dfrac{40}{16}=\dfrac{5}{2}.\)
Vậy \(A=\dfrac{5}{2}.\)
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\(A=\frac{2^4.3^3+2^3.3^4}{2^5.3^3-2^4.3^2}\)
\(A=\frac{2^3.3^3\left(2+3\right)}{2^4.3^2\left(2.3-1\right)}\)
\(A=\frac{2^3.3^3.5}{2^4.3^2.5}\)
\(A=\frac{3}{2}\)
\(A=\frac{2^4.3^3+2^3.3^4}{2^5.3^3-2^4.3^2}\)
\(A=\frac{2^3.3^3\left(2+3\right)}{2^4.3^2\left(2.3-1\right)}\)
\(A=\frac{2^3.3^3.5}{2^4.3^2.5}\)
\(A=\frac{3}{2}\)
`(4^2. 5.11)/(44.20)`
`=(4.11.4.5)/(4.11.4.5)`
`=1`
`(13.15.16)/(18.65.7)`
`=(13.15.16)/(2.3.3.13.5.7)`
`=8/21`
`(7.2.8.5^2)/(14.2.5)`
`=(14.2.4.5.5)/(14.2.5)`
`=4.5`
`=20`
`(2^3. 3^3. 5)/(3.2^3. 5^3)`
`=(2^3. 3.5.3^2)/(2^3. 3.5.5^2)`
`=(3^2)/(5^2)`
`=9/25`
**Quy đồng:
`(4^2. 5.11)/(44.20)=1=525/525`
`(13.15.16)/(18.65.7)=8/21=200/525`
`(7.2.8.5^2)/(14.2.5)=20=840/525`
`(2^3. 3^3. 5)/(3.2^3. 5^3)=9/25=189/525`
\(\dfrac{2^3.3^4}{2^3.3^2.5}=\dfrac{1.3^2}{1.1.5}=\dfrac{9}{5}\)
a: \(=\dfrac{-4\cdot13\cdot9\cdot5}{3\cdot4\cdot5\cdot2\cdot13}=\dfrac{3}{2}\)
b: \(=\dfrac{1}{2}\cdot\dfrac{1}{3}\cdot5=\dfrac{5}{6}\)
\(A=\dfrac{2^4.3^3+2^3.3^4}{2^5.3^4-2^6.3^3}=\dfrac{2^3.3^3.\left(2+3\right)}{2^5.3^3.\left(3-2\right)}=\dfrac{2^3.3^3.5}{2^5.3^3.1}\)
\(=\dfrac{5}{2^2}=\dfrac{5}{4}\)
\(=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{18}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot2^3\cdot7^3}\)
\(=\dfrac{2^{12}\cdot3^4\cdot2}{2^{12}\cdot3^5\left(2^6\cdot3+1\right)}-\dfrac{5^{10}\cdot7^3\cdot\left(-6\right)}{5^9\cdot7^3\cdot9}\)
\(=\dfrac{2}{193}+\dfrac{5\cdot2}{3}=\dfrac{1936}{579}\)
\(\dfrac{3}{2}\)
3/2