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A = \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{125}\)
\(\frac{7}{4}-y\)x \(\frac{5}{6}=\frac{1}{2}+\frac{1}{3}\)
\(\frac{7}{4}-y\)x \(\frac{5}{6}=\frac{5}{6}\)
\(y\) x \(\frac{5}{6}=\frac{7}{4}-\frac{5}{6}\)
\(y\) x \(\frac{5}{6}=\frac{11}{12}\)
y == \(\frac{11}{12}:\frac{5}{6}\)
y == \(\frac{11}{10}\)
Ta có \(\frac{7}{4}-\frac{5}{6}y=\frac{5}{6}\)
\(\frac{7}{4}=\frac{5}{6}y+\frac{5}{6}\)
\(\frac{7}{4}=\frac{5}{6}\left(y+1\right)\)
\(\frac{7}{4}:\frac{5}{6}=y+1\)
\(\frac{7}{4}\cdot\frac{6}{5}=\frac{21}{10}=y+1\)
\(y=\frac{21}{10}-1=\frac{11}{10}\)
\(A=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{25.125}\)
\(A=\frac{1}{100}.\left(1-\frac{1}{101}\right)+\frac{1}{100}.\left(\frac{1}{2}-\frac{1}{102}\right)+\frac{1}{100}.\left(\frac{1}{3}-\frac{1}{103}\right)+...+\frac{1}{100}.\left(\frac{1}{25}-\frac{1}{125}\right)\)
\(A=\frac{1}{100}.\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+...+\frac{1}{25}-\frac{1}{125}\right)\)
\(A=\frac{1}{100}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)\)
\(B=\frac{1}{1.26}+\frac{1}{2.27}+\frac{1}{3.28}+...+\frac{1}{100.125}\)
\(B=\frac{1}{25}.\left(1-\frac{1}{26}\right)+\frac{1}{25}.\left(\frac{1}{2}-\frac{1}{27}\right)+\frac{1}{25}.\left(\frac{1}{3}-\frac{1}{28}\right)+...+\frac{1}{25}.\left(\frac{1}{100}-\frac{1}{125}\right)\)
\(B=\frac{1}{25}.\left(1-\frac{1}{26}+\frac{1}{2}-\frac{1}{27}+\frac{1}{3}-\frac{1}{28}+...+\frac{1}{100}-\frac{1}{125}\right)\)
\(B=\frac{1}{25}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\frac{1}{26}-\frac{1}{27}-\frac{1}{28}-...-\frac{1}{125}\right)\)
\(B=\frac{1}{25}.\left(1+\frac{1}{2}+...+\frac{1}{25}+\frac{1}{26}+\frac{1}{27}+...+\frac{1}{100}-\frac{1}{26}-\frac{1}{27}-...-\frac{1}{100}-\frac{1}{101}-...-\frac{1}{125}\right)\)\(B=\frac{1}{25}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)\)
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 )
\(3\frac{4}{7}4\frac{2}{5}2\frac{1}{2}\)
\(=\frac{275}{7}\)
\(3\frac{1}{5}+\frac{3}{4}-\frac{1}{2}:\frac{2}{7}=\frac{16}{5}+\frac{3}{4}-\frac{7}{4}=\frac{11}{5}\)
\(3\frac{4}{7}4\frac{2}{5}2\frac{1}{2}=\frac{275}{7}\)
\(3\frac{1}{5}+\frac{3}{4}+\frac{1}{2}:\frac{2}{7}\)
\(=3\frac{1}{5}+\frac{3}{4}+\frac{7}{4}\)
\(=\frac{11}{5}\)
đặt A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{999.1000}+1\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}+1\)
\(=1-\frac{1}{1000}+1\)
\(=\frac{1999}{1000}\)
\(\frac{13}{25}< \frac{133}{153}\)
\(\frac{13}{15}>\frac{1333}{1555}\)
mk nha nhất
\(\frac{13}{25}< \frac{133}{153}\)
\(\frac{13}{15}>\frac{1333}{1555}\)