tính nhanh
\(\frac{1}{1\times101}+\frac{2}{2x102}+..........................+\frac{25}{100x25}\)
ai đầu tiên mk k cho
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\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
\(=\frac{1\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}{4.\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
\(=\frac{1}{4}+\frac{3}{4}=1\)
\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
\(=\frac{1}{4}+\frac{3}{4}\)
=1
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+..+\frac{1}{1+2+3+...+50}\)
Ta có :
\(A=\frac{2}{2\left(1+2\right)}+\frac{2}{2\left(1+2+3\right)}+...+\frac{2}{2\left(1+2+..+50\right)}\)
\(A=\frac{2}{6}+\frac{2}{12}+...+\frac{2}{2550}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{50.51}\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{51}\right)\)
\(A=2\times\frac{49}{102}\)
\(A=\frac{49}{51}\)
đề bài mk chỉ cho 50 thôi ko có 51 đâu
nên mk cho bạn 1k thôi nhé
\(A=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{25.125}\)
\(A=\frac{1}{100}.\left(1-\frac{1}{101}\right)+\frac{1}{100}.\left(\frac{1}{2}-\frac{1}{102}\right)+\frac{1}{100}.\left(\frac{1}{3}-\frac{1}{103}\right)+...+\frac{1}{100}.\left(\frac{1}{25}-\frac{1}{125}\right)\)
\(A=\frac{1}{100}.\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+...+\frac{1}{25}-\frac{1}{125}\right)\)
\(A=\frac{1}{100}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)\)
\(B=\frac{1}{1.26}+\frac{1}{2.27}+\frac{1}{3.28}+...+\frac{1}{100.125}\)
\(B=\frac{1}{25}.\left(1-\frac{1}{26}\right)+\frac{1}{25}.\left(\frac{1}{2}-\frac{1}{27}\right)+\frac{1}{25}.\left(\frac{1}{3}-\frac{1}{28}\right)+...+\frac{1}{25}.\left(\frac{1}{100}-\frac{1}{125}\right)\)
\(B=\frac{1}{25}.\left(1-\frac{1}{26}+\frac{1}{2}-\frac{1}{27}+\frac{1}{3}-\frac{1}{28}+...+\frac{1}{100}-\frac{1}{125}\right)\)
\(B=\frac{1}{25}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\frac{1}{26}-\frac{1}{27}-\frac{1}{28}-...-\frac{1}{125}\right)\)
\(B=\frac{1}{25}.\left(1+\frac{1}{2}+...+\frac{1}{25}+\frac{1}{26}+\frac{1}{27}+...+\frac{1}{100}-\frac{1}{26}-\frac{1}{27}-...-\frac{1}{100}-\frac{1}{101}-...-\frac{1}{125}\right)\)\(B=\frac{1}{25}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)\)
Ta thấy biểu thức trong ngoặc của hai vế A và B giống nhau
Vậy A : B = \(\frac{1}{100}:\frac{1}{25}=\frac{1}{4}\)
\(A=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{25.125}\)
\(\Rightarrow A=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{24.25}\right)+\left(\frac{1}{101.102}+\frac{1}{102.103}+...+\frac{1}{124.125}\right)\)
\(A=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{24}-\frac{1}{25}\right)+\left(\frac{1}{101}-\frac{1}{102}+\frac{1}{102}-\frac{1}{103}+...+\frac{1}{124}-\frac{1}{125}\right)\)
\(A=\left(1-\frac{1}{25}\right)+\left(\frac{1}{101}-\frac{1}{125}\right)\)
\(A=\frac{24}{25}+\frac{24}{12625}\)
Bạn tự tính luôn nha trog máy tính của mình là : 0,961... ( k làm thành phân số được )
A = \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{125}\)
\(B=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\)
\(B=1+\frac{1+1+1+...+1}{2+2^2+2^3+...+2^{2016}}=1+\frac{1+1+1+...+1}{2^{1+2+3+...+2016}}\)
\(B=1+\frac{2016}{2^{2033136}}=1\frac{2016}{2^{2033136}}\)
đặt A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{999.1000}+1\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}+1\)
\(=1-\frac{1}{1000}+1\)
\(=\frac{1999}{1000}\)