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Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)
\(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)
\(=100.\frac{2}{101}=\frac{200}{101}\)
a. 2\(\left(2^{-1}+3^{-1}\right):\left(2^{-1}-3^{-1}\right)+\left(2^{-1}:2^0\right):2^3\)
= \(\left(\frac12+\frac13\right):\left(\frac12-\frac13\right)+\left(\frac12:1\right):8\)
=\(\left(\frac36+\frac26\right):\left(\frac36-\frac26\right)+\left(\frac12\right):8\)
= \(\left(\frac56:\frac16\right)+\left(\frac{1}{2_{}}\right):8\)
= \(5+\frac12:8\)
= \(5+\frac{1}{16}\)
= \(\frac51+\frac{1}{16}=\frac{80}{16}+\frac{1}{16}=\frac{81}{16}\)
\(B=\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{99.100}\right)\)
\(B=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}...\frac{9898}{99.100}\)
\(B=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{98.101}{99.100}\)
\(B=\frac{1.2.3...98}{2.3.4...99}.\frac{4.5.6...101}{3.4.5...100}\)
\(B=\frac{1}{99}.\frac{101}{3}=\frac{101}{297}\)
\(a=\frac{6.2.3.4+6.3.4.5+6.4.5.6+...+6.98.99.100}{2.3.4+3.4.5+4.5.6+...+98.99.100}=6\)
thay vào p(x) suy ra a không là nghiệm của đa thức nhé bạn
$\textbf{a)}$
$\left(-\dfrac34+\dfrac27\right):\dfrac27+\left(-\dfrac14+\dfrac57\right):\dfrac23$
$=\left(-\dfrac{13}{28}\right)\cdot\dfrac72+\dfrac{13}{28}\cdot\dfrac32$
$=-\dfrac{13}{8}+\dfrac{39}{56}$
$=-\dfrac{13}{14}.$
$\textbf{b)}$
$\left(-\dfrac13\right)^2\cdot\dfrac4{11}+\dfrac7{11}\cdot\left(-\dfrac13\right)^2$
$=\dfrac19\left(\dfrac4{11}+\dfrac7{11}\right)$
$=\dfrac19.$
9/14