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Bài 2:
a: =>x^2=60
=>\(x=\pm2\sqrt{15}\)
b: =>2^2x+3=2^3x
=>3x=2x+3
=>x=3
c: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}\cdot\dfrac{1}{2}=1\)
\(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}=2\)
=>1/2x-2=4
=>1/2x=6
=>x=12
\(=\dfrac{2^{19}\cdot3^9+3\cdot5\cdot2^{18}\cdot3^8}{2^9\cdot2^{10}\cdot3^9+2^{20}\cdot3^{10}}\)
\(=\dfrac{2^{19}\cdot3^9+3^9\cdot2^{18}\cdot5}{2^{19}\cdot3^9+2^{20}\cdot3^{10}}\)
\(=\dfrac{2^{18}\cdot3^9\left(2+5\right)}{2^{19}\cdot3^9\cdot\left(1+2\cdot3\right)}=\dfrac{1}{2}\)
\(A=\dfrac{2^{19}.27^3-15.\left(-4\right)^9.9^4}{6^9.2^{10}+\left(-12\right)^{10}}\)
\(A=\dfrac{2^{19}.3^9+3.5.2^{18}.3^{12}}{2^9.3^9.2^{10}+3^{10}.2^{20}}\)
\(A=\dfrac{2^{18}.3^9\left(2+3.5.3^3\right)}{2^{19}.3^9\left(1+3.2\right)}=\dfrac{2+5.3^4}{2.7}=\dfrac{407}{14}\)
Chúc bạn học tốt!!!
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) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
$B=\left(\dfrac1{2^2}-1\right)\left(\dfrac1{3^2}-1\right)\cdots\left(\dfrac1{99^2}-1\right)$
$=\left(-\dfrac{2^2-1}{2^2}\right)\left(-\dfrac{3^2-1}{3^2}\right)\cdots\left(-\dfrac{99^2-1}{99^2}\right)$
$=(-1)^{98}\prod_{k=2}^{99}\dfrac{(k-1)(k+1)}{k^2}$
$=\left(\prod_{k=2}^{99}\dfrac{k-1}{k}\right)\left(\prod_{k=2}^{99}\dfrac{k+1}{k}\right)$
$=\left(\dfrac12\cdot\dfrac23\cdot\dfrac34\cdots\dfrac{98}{99}\right)\left(\dfrac32\cdot\dfrac43\cdot\dfrac54\cdots\dfrac{100}{99}\right)$
$=\dfrac1{99}\cdot\dfrac{100}{2}$
$=\dfrac{50}{99}.$
a: \(=\dfrac{3}{8}\left(27+\dfrac{1}{5}-51-\dfrac{1}{5}\right)+19\)
\(=-24\cdot\dfrac{3}{8}+19=-9+19=10\)
b: \(=\left(35+\dfrac{1}{6}-46-\dfrac{1}{6}\right):\left(\dfrac{-4}{5}\right)\)
\(=\dfrac{-11\cdot5}{-4}=\dfrac{55}{4}\)
c: \(=\left(\dfrac{-15+8}{20}\right):\left[\dfrac{3}{7}+\dfrac{7}{3}\cdot\dfrac{12-5}{20}\right]\)
\(=\dfrac{-7}{20}:\left(\dfrac{3}{7}+\dfrac{49}{60}\right)\)
\(=-\dfrac{147}{523}\)
a: \(=\dfrac{2^{19}\cdot3^9+3^9\cdot5\cdot2^{18}}{2^{19}\cdot3^9+2^{10}}\)
\(=\dfrac{3^9\cdot2^{18}\cdot\left(2+5\right)}{2^{10}\cdot\left(2^9\cdot3^9+1\right)}=\dfrac{3^9\cdot7\cdot2^8}{6^9+1}\)
b: \(=\dfrac{\dfrac{-1}{8}-\dfrac{27}{64}\cdot4}{-2+\dfrac{9}{16}-\dfrac{3}{8}}=\dfrac{-29}{16}:\dfrac{-29}{16}=1\)