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\(a,9.3^3.\frac{1}{81}.3^2=3^2.3^3.3^{\left(-4\right)}.3^2=3^{2+3-4+2}=3^3.\)
\(b,4.2^5:\left(2^3.\frac{1}{16}\right)=2^2.2^5:\left(2^3.2^{-4}\right)=2^{2+5}:2^{3-4}=2^7:2^{-1}=2^{7-\left(-1\right)}=2^8.\)
\(c,3^2.2^5.\left(\frac{2}{3}\right)^2=3^2.2^5.\frac{2^2}{3^2}=\left(\frac{3^2}{3^2}\right).\left(2^5.2^2\right)=1.2^{5+2}=2^7\)
\(d,\left(\frac{1}{3}\right)^2.\frac{1}{3}.9^2=\left(\frac{1}{3}\right)^2.\frac{1}{3}.\left(3^2\right)^2=\left(\frac{1}{3}\right)^{2+1}.3^4=\left(\frac{1}{3}\right)^3.\left(\frac{1}{3}\right)^{-4}=\left(\frac{1}{3}\right)^{3-4}=\left(\frac{1}{3}\right)^{-1}=3^1\)
a: \(5^3\cdot25^n=5^{3n}\)
\(\Leftrightarrow5^{3n}=5^3\cdot5^{2n}\)
=>3n=2n+3
hay n=3
b: \(a^{\left(2n+6\right)\left(3n-9\right)}=1\)
=>(2n+6)(3n-9)=0
=>n=-3 hoặc n=3
c: \(\dfrac{1}{3}\cdot3^n=7\cdot3^2\cdot3^4-2\cdot3^n\)
\(\Leftrightarrow3^n\cdot\dfrac{1}{3}+3^n\cdot2=7\cdot3^6\)
\(\Leftrightarrow3^n=3^7\)
hay n=7
c)
Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)
d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
\(a)\dfrac{1}{4}-\dfrac{3}{4}:\left(\dfrac{-5}{8}\right)\)
\(=\dfrac{1}{4}-\dfrac{3}{4}.\dfrac{-8}{5}\)
\(=\dfrac{1}{4}-\dfrac{-6}{5}\)
\(=\dfrac{5}{20}+\dfrac{24}{20}\)
\(=\dfrac{29}{20}\)
\(b)3-\left(\dfrac{-6}{7}\right)^0+\sqrt{\dfrac{1}{16}}:2\)
\(=3-1+\sqrt{\left(\dfrac{1}{4}\right)^2}:2\)
\(=2+\dfrac{1}{4}.\dfrac{1}{2}\)
\(=\dfrac{16}{8}+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
\(c)\dfrac{9^5.2^6}{4^3.3^8}=\dfrac{\left(3^2\right)^5.2^6}{\left(2^2\right)^3.3^8}=\dfrac{3^{10}.2^6}{2^6.3^8}=3^2=9\)
Theo đề bài ta có:
\(B=\dfrac{-1^2.-2^2.....-100^2}{1.2.2.3.....99.100}\)
\(B=\dfrac{1^2.2^2.....100^2}{1.2.2.3.....99.100}\)
\(B=\dfrac{1.1.2.2......100.100}{1.2.2.3.....99.100}\)
\(B=\dfrac{1.2.3......100}{1.2.3.......99}.\dfrac{1.2.3......100}{2.3.4......100}\)
\(B=100\)
`9.3^3 .1/81 .3^2 `
`= 9. 3^3/3^4 .3^2`
`= 9.1/3 .9 = 9/3 .3 = 3.3 =9 =3^2`
`4.2^5 :(2^3 .1/16)`
`= 4. 32 : (8/16)`
`= 4.32:1/2 = 4.32.2 = 256 = 2^8`
1/
\(=3^2.3^3.\dfrac{1}{3^4}.3^2=3^3\)
2/
\(=2^2.2^5:\left(2^3.\dfrac{1}{2^4}\right)=2^7.2=2^8\)