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\(\dfrac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)
\(=\dfrac{1.2+2.4+3.6+4.8+5.10}{3.4+3.4.2.2+3.4.3.3+3.4.2.8+3.4.5.5}\)
\(=\dfrac{1.2.\left(4+3^2+2.8+5^2\right)}{3.4.\left(4+3^2+2.8+5^2\right)}\)
\(=\dfrac{1}{6}\)
\(\dfrac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)
\(=\dfrac{1.2+1.2.2^2+1.2.3^2+1.2.4^2+1.2.5^2}{3.4+3.4.2^2+3.4.3^2+3.4.4^2+3.4.5^2}\)
\(=\dfrac{1.2\left(1+2^2+3^2+4^2+5^2\right)}{3.4\left(1+2^2+3^2+4^2+5^2\right)}\\ =\dfrac{2}{12}=\dfrac{1}{6}\)
\(A=2017:\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2017.2018}\right)\)
\(=2017:\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\right)\)
\(=2017:\left(1-\dfrac{1}{2018}\right)\)
\(=2017:\dfrac{2017}{2018}\)
\(=2017\cdot\dfrac{2018}{2017}\)
\(=2018\)
#NgDat
\(A=2017:\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2017\cdot2018}\right)\)
\(A=2017:\left(\dfrac{1}{1}\cdot\dfrac{1}{2}+\dfrac{1}{2}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{1}{4}+...+\dfrac{1}{2017}\cdot\dfrac{1}{2018}\right)\)
\(A=2017:\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\right)\)
\(A=2017:\left(\dfrac{1}{1}-\dfrac{1}{2018}\right)\)
\(A=2017:\left(\dfrac{2018}{2018}-\dfrac{1}{2018}\right)\)
\(A=2017:\dfrac{2017}{2018}\)
\(A=2018.\)
\(P=\dfrac{\left(1+2+3+...+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63\cdot1,2-21\cdot3,6\right)}{1-2+3-4+5-6+...+99-100}\)
đề là vậy nhé mn
để ý chút thấy liền ah : 63.1,2-21.3,6=63.1,2-21.3.1,2= 63.1,2- 63.1,2=0
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Ta có P = \(\dfrac{\left(1+2+3+...+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+5-...+99-100}\)= \(\dfrac{\left(1+2+3+...+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)0}{1-2+3-4+5-...+99-100}\)= \(\dfrac{0}{1-2+3-4+5-6+...+99-100}=0\)
