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a) \(56^2+44^2+2.56.44=56^2+2.56.44+44^2=\left(56+44\right)^2=100^2=10000\)
b) \(36^2+64^2+72.64=36^2+2.36.64+64^2=\left(36+64\right)^2=100^2=10000\)
c) \(136^2+36^2-72.136=136^2-2.36.136+36^2=\left(136-36\right)^2=100^2=10000\)
a) $56^2+44^2+2.56.44=56^2+2.56.44+44^2=\left(56+44\right)^2=100^2=10000$562+442+2.56.44=562+2.56.44+442=(56+44)2=1002=10000
b) $36^2+64^2+72.64=36^2+2.36.64+64^2=\left(36+64\right)^2=100^2=10000$362+642+72.64=362+2.36.64+642=(36+64)2=1002=10000
c) $136^2+36^2-72.136=136^2-2.36.136+36^2=\left(136-36\right)^2=100^2=10000$
\(56^2+44^2+2.44.56\)
\(=\left(56+44\right)^2\)
\(=100^2=10000\)
\(2018^2-2017.2019\)
\(=2018^2-\left(2018-1\right)\left(2018+1\right)\)
\(=2018^2-\left(2018^2-1\right)=1\)
\(56^2+56.88+44^2\)
\(=56^2+2.56.44+44^2\)
\(=\left(56+44\right)^2\)
\(=100^2=10000\)
\(\frac{2018^3+1}{2018^2-2017}\)
\(=\frac{\left(2018+1\right)\left(2018^2-2018+1\right)}{2018^2-2017}\)
\(=\frac{2019\left(2018^2-2017\right)}{2018^2-2017}=2019\)
Chúc bạn học tốt.
Ta có:
\(1-\dfrac{1}{1+2+...+n}=1-\dfrac{1}{\dfrac{n\left(n+1\right)}{2}}=\dfrac{n\left(n+1\right)-2}{n\left(n+1\right)}=\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
\(\Rightarrow S=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\dfrac{99.102}{100.101}\)
\(=\dfrac{1.2.3...99}{2.3.4...100}.\dfrac{4.5.6...102}{3.4.5...101}=\dfrac{1}{100}.\dfrac{102}{3}=\dfrac{17}{50}\)
Ta thấy \(1-\dfrac{1}{n^2}=\dfrac{\left(n-1\right)\left(n+1\right)}{n^2}\) với mọi \(n>0\).
Từ đó \(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{100^2}\right)=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}...\dfrac{99.101}{100}=\left(\dfrac{1}{2}.\dfrac{2}{3}...\dfrac{99}{100}\right).\left(\dfrac{3}{2}.\dfrac{4}{3}...\dfrac{101}{100}\right)=\dfrac{1}{100}.\dfrac{101}{2}=\dfrac{101}{200}\).
Sửa đề: \(\dfrac{100+\dfrac{99}{2}+\dfrac{98}{3}+...+\dfrac{1}{100}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{101}}-2\)
\(=\dfrac{\left(\dfrac{99}{2}+1\right)+\left(\dfrac{98}{3}+1\right)+...+\left(\dfrac{1}{100}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{101}}-2\)
\(=\dfrac{\dfrac{101}{2}+\dfrac{101}{3}+...+\dfrac{101}{100}+\dfrac{101}{101}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}+\dfrac{1}{101}}-2\)
\(=\dfrac{101\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}+\dfrac{1}{101}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}+\dfrac{1}{101}}-2\)
\(=101-2=99\)
Vậy...
\(\dfrac{56^3+44^3}{100}-14.56\)
\(=\dfrac{\left(56+44\right)\left(56^2-56.44+44^2\right)}{100}-14.56\)
\(=\dfrac{100.\left(56^2-56.44+44^2\right)}{100}-14.56\)
\(=\)\(56^2-44.56+44^2-14.56\)
\(=\left(56^2-14.56\right)-\left(44.56-44^2\right)\)
\(=56\left(56-14\right)-44\left(56-44\right)\)