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đề chính xác là
a/ \(13-2\times\left(36-5\times x\right)=1\)
b/ \(53-10\times\left(40-7\times x\right)=3\)
c/ \(\frac{3}{4}-\frac{1}{6}\div x=0,375\)
a,Đặt \(A=\frac{1}{1\times4}+\frac{1}{4\times7}+...+\frac{1}{97\times100}\)
\(\Rightarrow3A=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{97\times100}\)
\(\Rightarrow3A=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)
\(\Rightarrow3A=1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow A=\frac{99}{300}\)
b, \(\frac{1}{2}\times\frac{2}{3}\times...\times\frac{99}{100}=\frac{1\times2\times...\times99}{2\times3\times...\times1000}=\frac{1}{100}\)
c, \(\frac{3}{4}\times\frac{8}{9}\times...\times\frac{99}{100}=\frac{1.3}{2.2}\times\frac{2.4}{3.3}\times...\times\frac{9.11}{10.10}=\frac{1.2.....9}{2.3.....10}\times\frac{3.4.....11}{2.3.....10}=\frac{1}{10}\times\frac{11}{2}=\frac{11}{20}\) (dấu . là dấu nhân)
\(P=\frac{1}{5x8}+\frac{1}{8x11}+.....+\frac{1}{602x605}\)
\(\Rightarrow3P=\frac{3}{5x8}+\frac{3}{8x11}+......+\frac{3}{602x605}\)
\(\Rightarrow3P=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-.....+\frac{1}{602}-\frac{1}{605}\)
\(\Rightarrow3P=\frac{1}{5}-\frac{1}{605}\)
\(\Rightarrow3P=\frac{24}{121}\)
\(\Rightarrow P=\frac{24}{121}:3\)
\(\Rightarrow P=\frac{8}{121}\)
\(=\frac{1}{2}\times\frac{2}{3}\times....\times\frac{2003}{2004}\)
\(=\frac{1\times2\times3\times...\times2003}{2\times3\times4\times...\times2014}\)
\(=\frac{1}{2014}\)
1/2 . 1/3 . 1/4 . 1/5 . 1/6 . ( x - 1,010 ) = 1/360 - 1/720
1/2 . 1/3 . 1/4 . 1/5 . 1/6 . ( x - 1,010) = 1/720
( x - 1,010 ) . 1/2 . 1/3 . 1/4 . 1/5 . 1/6 = 1/720
( x - 1,010 ) . 1/720 = 1/720
x - 1,010 = 1/720 : 1/720
x - 1,010 = 1
x = 1 + 1,010
x = 2,01
2/
a) \(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\)
\(=\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+....+\frac{1}{17}-\frac{1}{21}\right)\)
\(=1-\frac{1}{21}=\frac{20}{21}\)
b) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot..\cdot\frac{2016}{2017}\)
\(=\frac{1}{2017}\)
c) \(A=2000-5-5-5-..-5\)(có 200 số 5)
\(A=2000-\left(5\cdot200\right)\)
\(A=2000-1000\)
\(A=1000\)