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Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)
\(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)
\(=100.\frac{2}{101}=\frac{200}{101}\)
$\textbf{a)}$
$\left(-\dfrac34+\dfrac27\right):\dfrac27+\left(-\dfrac14+\dfrac57\right):\dfrac23$
$=\left(-\dfrac{13}{28}\right)\cdot\dfrac72+\dfrac{13}{28}\cdot\dfrac32$
$=-\dfrac{13}{8}+\dfrac{39}{56}$
$=-\dfrac{13}{14}.$
$\textbf{b)}$
$\left(-\dfrac13\right)^2\cdot\dfrac4{11}+\dfrac7{11}\cdot\left(-\dfrac13\right)^2$
$=\dfrac19\left(\dfrac4{11}+\dfrac7{11}\right)$
$=\dfrac19.$
\(A = {1\over2}-{3\over4}+{5\over6}-{7\over12}={6\over12}-{9\over12}+{10\over12}-{7\over12}\)\(={0\over12}=0\)
$\textbf{a)}$
$A=\dfrac{\left(\dfrac23\right)^3\cdot\left(-\dfrac34\right)^2\cdot(-1)^{2019}}{36\cdot\dfrac15\cdot\left(\dfrac25\right)^2\cdot\left(-\dfrac5{12}\right)^3}$
$=\dfrac{\dfrac8{27}\cdot\dfrac9{16}\cdot(-1)}{36\cdot\dfrac15\cdot\dfrac4{25}\cdot\left(-\dfrac{125}{1728}\right)}$
$=\dfrac{-\dfrac16}{-\dfrac5{12}}$
$=\dfrac16\cdot\dfrac{12}5$
$=\dfrac25.$
$\textbf{b)}$
$B=\dfrac1{19}+\dfrac9{19\cdot29}+\dfrac9{29\cdot39}+\cdots+\dfrac9{2009\cdot2019}$
$=\dfrac1{19}+\left(\dfrac1{19}-\dfrac1{29}\right)+\left(\dfrac1{29}-\dfrac1{39}\right)+\cdots+\left(\dfrac1{2009}-\dfrac1{2019}\right)$
$=\dfrac1{19}+\dfrac1{19}-\dfrac1{2019}$
$=\dfrac2{19}-\dfrac1{2019}$
$=\dfrac{2\cdot2019-19}{19\cdot2019}$
$=\dfrac{4019}{38361}.$
$B=\left(\dfrac1{2^2}-1\right)\left(\dfrac1{3^2}-1\right)\cdots\left(\dfrac1{99^2}-1\right)$
$=\left(-\dfrac{2^2-1}{2^2}\right)\left(-\dfrac{3^2-1}{3^2}\right)\cdots\left(-\dfrac{99^2-1}{99^2}\right)$
$=(-1)^{98}\prod_{k=2}^{99}\dfrac{(k-1)(k+1)}{k^2}$
$=\left(\prod_{k=2}^{99}\dfrac{k-1}{k}\right)\left(\prod_{k=2}^{99}\dfrac{k+1}{k}\right)$
$=\left(\dfrac12\cdot\dfrac23\cdot\dfrac34\cdots\dfrac{98}{99}\right)\left(\dfrac32\cdot\dfrac43\cdot\dfrac54\cdots\dfrac{100}{99}\right)$
$=\dfrac1{99}\cdot\dfrac{100}{2}$
$=\dfrac{50}{99}.$
A) \(=\frac{\left(-1\right).2^{17}.5^6.3^{12}}{2^{16}.5^53^{13}}=\frac{10}{3}\)
B) Tương tự câu A bạn tự làm nha
Sửa câu A thành \(\frac{-10}{3}\)nha nãy viết vội quá
A) = \(\frac{\left(-1\right)\cdot2^{17}\cdot\left(5^2\right)^3\cdot\left(3^3\right)^4}{\left(2\cdot3\right)^{11}\cdot\left(2\cdot5\right)^5\cdot3^2}\)= \(\frac{\left(-1\right)\cdot2^{17}\cdot5^6\cdot3^{12}}{\left(2^{11}\cdot2^5\right)\cdot\left(3^{11}\cdot3^2\right)\cdot5^5}\)= \(\frac{\left(-1\right)\cdot2^{16}\cdot2\cdot5^5\cdot5\cdot3^{12}}{2^{16}\cdot3^{12}\cdot3\cdot5^5}\)
= \(\frac{\left(-1\right)\cdot2\cdot5}{3}=\frac{-10}{3}\)
B) = \(\frac{\left(3^2\right)^4\cdot\left(7^2\right)^3\cdot2^5}{\left(7\cdot3\right)^5\cdot3^4\cdot\left(2\cdot3\right)^3}\)= \(\frac{3^8\cdot7^6\cdot2^5}{7^5\cdot3^5\cdot3^4\cdot2^3\cdot3^3}\)= \(\frac{3^8\cdot7^5\cdot7\cdot2^3\cdot2^2}{7^5\cdot3^8\cdot3^4\cdot2^3}\)= \(\frac{7\cdot4}{3^4}\)= \(\frac{28}{81}\)