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45-x/3=56/7
x/3= 45-56/7
x/3 = 315-56/ 7
x/3 = 37
x= 37.3
x= 111
a) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{x\times\left(x+3\right)}=\dfrac{99}{200}\)
Ta có: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=\dfrac{1}{100}\)
\(\Rightarrow x+1=100\)
\(x=100-1\)
\(x=99\)
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}\)
\(\Leftrightarrow A=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(\Leftrightarrow A=\frac{3}{2}\left(1-\frac{1}{101}\right)\)
\(\Leftrightarrow A=\frac{3}{2}.\frac{100}{101}\)
\(\Leftrightarrow A=\frac{150}{101}\)
Ko còn bài nào giải làm tạm bài này chứ mk cũng ko muốn làm n bài dễ
a)\(45-\frac{x}{3}=\frac{56}{7}\)
\(\Leftrightarrow\frac{x}{3}=45-\frac{56}{7}\)
\(\Leftrightarrow\frac{x}{3}=37\)
\(\Rightarrow x=171\)
b)\(x-\frac{7}{85}=\frac{4}{17}\)
\(\Leftrightarrow x=\frac{4}{17}+\frac{7}{85}\)
\(\Rightarrow x=\frac{27}{85}\)
a)45—x/3=56/7
45—x/3=8
x/3=45—8
x/3=37
x=37x3
x=111
\(\frac{3}{1x3}+\frac{3}{3x5}+...+\frac{3}{49x51}=\frac{3}{2}\left(\frac{2}{1x3}+\frac{2}{3x5}+...+\frac{2}{49x51}\right)=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\frac{50}{51}=\frac{25}{17}\)
\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{2017.2019}\)
\(=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{3}{2}.\left(1-\frac{1}{2019}\right)\)
\(=\frac{3}{2}.\frac{2018}{2019}\)
\(=\frac{1009}{673}\)
\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}.....+\frac{3}{2017.2019}\)
\(=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{2017.2019}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{3}+....+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{2019}\right)\)
\(=\frac{3}{2}.\frac{2018}{2019}=\frac{1009}{673}\)
\(A=\frac{3}{1.3}+\frac{3}{3.5}+.....+\frac{3}{19.21}\)
\(A=\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+......+\frac{2}{19.21}\right)\)
\(A=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{19}-\frac{1}{21}\right)\)
\(A=\frac{3}{2}.\left(1-\frac{1}{21}\right)\)
\(A=\frac{3}{2}.\frac{20}{21}\)
\(A=\frac{10}{7}\)
Ta có:
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{19.21}\)
\(\Rightarrow A=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{19}-\frac{1}{21}\right)\)
\(\Rightarrow A=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{21}\right)=\frac{2}{3}.\frac{20}{21}=\frac{40}{63}\)