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\(\left(x+\dfrac{1}{3}\right)\times\dfrac{9}{14}\times\dfrac{7}{3}-\dfrac{1}{3}=1:\dfrac{9}{5}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}-\dfrac{1}{3}=\dfrac{5}{9}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}=\dfrac{5}{9}+\dfrac{1}{3}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}=\dfrac{8}{9}\\ \Rightarrow x+\dfrac{1}{3}=\dfrac{8}{9}:\dfrac{3}{2}\\ \Rightarrow x+\dfrac{1}{3}=\dfrac{16}{27}\\ \Rightarrow x=\dfrac{16}{27}-\dfrac{1}{3}\\ \Rightarrow x=\dfrac{7}{27}\)
\(A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(\Leftrightarrow A=\frac{1}{1.3}-\frac{1}{11.13}=\frac{1}{3}-\frac{1}{143}=\frac{140}{429}\)
\(A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+\frac{4}{7.9.11}+\frac{4}{9.11.13}\)
\(\Rightarrow A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(\Rightarrow A=\frac{1}{1.3}-\frac{1}{11.13}=\frac{1}{3}-\frac{1}{143}=\frac{140}{429}\)
a) \(\left(\frac{1}{3}+\frac{1}{5}\right)+\left(\frac{1}{6}-\frac{1}{5}\right)=\left(\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)=\frac{1}{2}\)
b) \(\frac{3}{16}\times\frac{7}{5}+\frac{3}{5}\times\frac{9}{16}=\frac{21}{80}+\frac{27}{80}=\frac{48}{80}=\frac{3}{5}\)
c) \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2020\times2021}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2020}-\frac{1}{2021}\)
\(=1-\frac{1}{2021}=\frac{2020}{2021}\)
d) \(\frac{1}{1\times3}+\frac{1}{3\times5}+...+\frac{1}{2021\times2023}=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+...+\frac{2}{2021\times2023}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2021}-\frac{1}{2023}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{2023}\right)=\frac{1}{2}\times\frac{2022}{2023}=\frac{1011}{2023}\)
e) \(\frac{3}{2}\times\frac{1}{7}\times\frac{5}{4}+\frac{15}{2}\times\frac{6}{7}\times\frac{1}{4}==\frac{15}{56}+\frac{80}{56}=\frac{95}{56}\)
a) x= \(\frac{-5}{12}\)
b) x = \(\frac{2}{5}\)
c) x =\(\frac{-87}{140}\)
d) x = \(\frac{109}{140}\)
e) x = \(\frac{13}{63}\)
\(\frac{\left(72\times1995-1996\right)\times\left(18-36\div2\right)\times\left(148\div5-7\right)}{1+3+5+...+1993+1995}\)
= \(\frac{\left(72\times1995-1996\right)\times\left(18-18\right)\times\left(148\div5-7\right)}{1+3+5+...+1993+1995}\)
= \(\frac{\left(72\times1995-1996\right)\times0\times\left(148\div5-7\right)}{1+3+5+...+1993+1995}\)
= \(\frac{0}{1+3+5+...+1993+1995}\)
= 0 : (1 + 3 + 5 + ... + 1993 + 1995)
= 0
(72x1995-1996)x(18-36:2)x(148:5-7) / 1+3+5+....+1993+1995
=(72x1995-1996)x0x(148:5-7) / 1+3+5+...+1993+1995
=0 / 1+3+5+...+1993+1995
=0
Tích nha
(1-1/2) x (1-1/3) x (1-1/4) x (1-1/5)
=1/2*2/3*3/4*4/5
=1/5.
1-1/2 x 1-1/3 x 1-1/4 x 1-1/5
=> = \(\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{4}\cdot\frac{1}{5}=\frac{1}{120}\)
đúng rồi nha > - <
\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{\left(2x+1\right)\cdot\left(2x+3\right)}\)
\(=\frac{1}{3}\left(\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{3}{\left(2x+1\right)\left(2x+3\right)}\right)\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{\left(2x+1\right)}-\frac{1}{\left(2x+3\right)}\right)\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{\left(2x+3\right)}\right)\)
\(=\frac{1}{3}\left(\frac{2x+3}{3\left(2x+3\right)}-\frac{3}{3\left(2x+3\right)}\right)\)
\(=\frac{1}{3}\left(\frac{2x+3-3}{3\left(2x+3\right)}\right)\)
\(=\frac{1}{3}\left(\frac{2x}{6x+9}\right)\)
\(=\frac{2x}{3\left(6x+9\right)}=\frac{2x}{18x+27}\)