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Giải:
a) \(\dfrac{5^3.90.4^3}{25^2.3^2.2^{13}}\)
\(=\dfrac{5^3.5.3^2.2.\left(2^2\right)^3}{\left(5^2\right)^2.3^2.2^{13}}\)
\(=\dfrac{5^4.3^2.2.2^6}{5^4.3^2.2^{13}}\)
\(=\dfrac{5^4.3^2.2^7}{5^4.3^2.2^{13}}\)
\(=\dfrac{1}{2^6}=\dfrac{1}{64}\)
b) \(\dfrac{15^2.16^4-15^3.16^3}{12^2.20^3-20^2.12^3}\)
\(=\dfrac{15^2.16^3.16-15^2.16^3.15}{12^2.20^2.20-20^2.12^2.12}\)
\(=\dfrac{15^2.16^3.\left(16-15\right)}{12^2.20^2.\left(20-12\right)}\)
\(=\dfrac{\left(3.5\right)^2.\left(2^4\right)^3.1}{\left(3.2^2\right)^2.\left(2^2.5\right)^2.8}\)
\(=\dfrac{3^2.5^2.2^{12}}{3^2.\left(2^2\right)^2.\left(2^2\right)^2.5^2.2^3}\)
\(=\dfrac{3^2.5^2.2^{12}}{3^2.5^2.2^4.2^4.2^3}\)
\(=\dfrac{3^2.5^2.2^{12}}{3^2.5^2.2^{11}}\)
\(=2\)
c) \(\dfrac{2.3+4.6+14.21}{3.5+6.10+21.35}\)
\(=\dfrac{2.3+2.3.2.2+2.3.7.7}{3.5+3.5.2.2+3.5.7.7}\)
\(=\dfrac{2.3+2.3.4+2.3.49}{3.5+3.5.4+3.5.49}\)
\(=\dfrac{2.3.\left(1+4+49\right)}{3.5.\left(1+4+49\right)}\)
\(=\dfrac{2.3}{3.5}\)
\(=\dfrac{2}{5}\)
Chúc bạn học tốt!
A = \(\dfrac{72^3.5^{42}}{108^4}\)
= \(\dfrac{\left(2^3.3^2\right)^3.5^{42}}{\left(2^2.3^3\right)^4}\)
= \(\dfrac{2^9.3^6.5^{42}}{2^8.3^{12}}\)
= \(\dfrac{2.5^{42}}{3^6}\)
@trần phúc nguyên
B = \(\dfrac{45^3.20^4.18^2}{180^5}\)
= \(\dfrac{\left(3^2.5\right)^3\left(2^2.5\right)^4.\left(2.3^2\right)^2}{\left(2^2.3^2.5\right)^5}\)
= \(\dfrac{3^6.5^3.2^8.5^4.2^2.3^4}{2^{10}.3^{10}.5^5}\)
= \(\dfrac{3^{10}.5^7.2^{10}}{2^{10}.3^{10}.5^5}\)
= 25
@trần phúc nguyên
`(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`
a) Ta có: \(\dfrac{25^{28}+25^{24}+25^{20}+...+25^4+1}{25^{30}+25^{28}+...+25^2+1}\)
\(=\dfrac{25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+...+\left(25^4+1\right)}{25^{28}\left(25^2+1\right)+25^{24}\left(25^2+1\right)+...+\left(25^2+1\right)}\)
\(=\dfrac{\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}{\left(25^2+1\right)\left(25^{28}+25^{24}+...+1\right)}\)
\(=\dfrac{\left(25^4+1\right)\cdot\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}{\left(25^2+1\right)\left[25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+25^8\left(25^4+1\right)+\left(25^4+1\right)\right]}\)
\(=\dfrac{\left(25^4+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}\)
\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}\)
\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}\)
\(=\dfrac{1}{25^2+1}=\dfrac{1}{626}\)
Ib vs môi. Mọi giải cho ok. Mik vừa hỏi dạng này xong. Ib mik chỉ cách giải
A= \(\dfrac{10.11.\left(1+5.5+7.7\right)}{11.12.\left(1+5.5+7.7\right)}=\dfrac{10}{12}=\dfrac{5}{6}\)
