\(\dfrac{8^2.5^4}{2^5.25}\)
HELP MEEEE
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a: \(=\dfrac{2^9\cdot5^9\cdot3^{40}}{2^{12}\cdot5^{10}\cdot3^{20}}=\dfrac{3^{20}}{5\cdot2^3}\)
b: \(=\dfrac{-3^8\cdot2^{10}\cdot5^6}{2^9\cdot\left(-1\right)\cdot3^6\cdot5^7}=\dfrac{-2}{5}\cdot3^2=-\dfrac{18}{5}\)
c: \(=\dfrac{3^{186}\cdot5^{100}}{5^{100}\cdot3^{187}}=\dfrac{1}{3}\)
\(\frac{2^{12}\cdot5^7+4^6\cdot25^3}{8^5\cdot25^3+\left(2^2\cdot5\right)^6}\)
\(=\frac{2^{12}.5^7+2^{12}\cdot5^6}{2^{15}\cdot5^6+2^{12}\cdot5^6}\)
\(=\frac{2^{12}\cdot5^6\cdot\left(1\cdot5+1\right)}{2^{12}\cdot5^6\cdot\left(2^3\cdot1+1\right)}\)
\(=\frac{6}{9}\)
\(=\frac{2}{3}\)
`a)x^3=343=7^3`
`=>x=7`
Vậy `x=7`
`b)(x-2,5)^4=(x-2,5)^2`
`=>(x-2,5)^2[(x-2,5)^2-1]=0`
`+)(x-2,5)^2=0<=>x=2,5`
`+)(x-2,5)^2=1`
`TH1:x-2,5=1<=>x=3,5`
`th2:x-2,5=-1<=>x=1,5`
Vậy `x=0` hoặc `x=1,5` hoặc `x=3,5
1/1x2 + 2/2x4 + 3/4x7 + 4/7x11 +...+ 8/29x37 + 9/37x46
=2-1/1x2+4-2/2x4+...+46-37/37x46
=1-1/2+1/2-1/4+...+1/37-1/46
=1-1/46
=45/46
a: \(=\dfrac{2^5\cdot3^5\cdot2^{12}\cdot2^{16}\cdot5^{16}}{2^{30}\cdot3^{10}\cdot5^{16}}=\dfrac{2^{33}\cdot3^5}{2^{30}\cdot3^{10}}=\dfrac{8}{243}\)
c: \(=\dfrac{4^7\cdot3^{12}\cdot5^4+3^{12}\cdot5^6\cdot4^7}{2^{14}\cdot3^{14}\cdot5^4+2^{14}\cdot3^{14}\cdot5^6}\)
\(=\dfrac{2^{14}\cdot3^{12}\cdot5^4\left(1+25\right)}{2^{14}\cdot3^{14}\cdot5^4\left(1+25\right)}=\dfrac{1}{9}\)
\(\dfrac{8^2\cdot5^4}{2^5\cdot25}=\dfrac{8\cdot8\cdot5^2\cdot5^2}{2^3\cdot2^2\cdot5^2}=\dfrac{2^3\cdot2^3\cdot5^2\cdot5^2}{2^3\cdot2^2\cdot5^2}=2\cdot5^2=2\cdot25=50\)
tks nhìu ạ :)))))))