\(3,\\ a,\dfrac{3x}{2\cdot5}+\dfrac{3x}{5\cdot8}+\dfrac{3x}{8\cdot11}+\dfrac{3x}{11\cdot14}=\dfrac{1}{21}\\ \Leftrightarrow x\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}\right)=\dfrac{1}{21}\\ \Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}\right)=\dfrac{1}{21}\\ \Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\\ \Leftrightarrow\dfrac{3}{7}x=\dfrac{1}{21}\\ \Leftrightarrow x=\dfrac{1}{21}\cdot\dfrac{7}{3}=\dfrac{1}{9}\)
\(b,\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\\ \Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)-\left(x+1\right)\left(\dfrac{1}{13}+\dfrac{1}{14}\right)=0\\ \Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\\ \Leftrightarrow x=-1\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\right)\)
\(4,\)
\(a,\) Gọi \(d=ƯCLN\left(n+1,2n+3\right) \)
\(\Rightarrow n+1⋮d;2n+3⋮d\\ \Rightarrow\left(2n+3\right)-2\left(n+1\right)⋮d\\ \Rightarrow1⋮d\Rightarrow d=1\\ \RightarrowƯCLN\left(n+1,2n+3\right)=1\)
\(\Rightarrow\dfrac{n+1}{2n+3}\) là phân số tối giản
\(b,\) Gọi \(d=ƯCLN\left(2n+3,4n+8\right)\)
\(\Rightarrow2n+3⋮d;4n+8⋮d\\ \Rightarrow4n+8-2\left(2n+3\right)⋮d\\ \Rightarrow2⋮d\Rightarrow d\inƯ\left(2\right)=\left\{1;2\right\}\\ d=2\Rightarrow2n+3⋮2\left(L\right)\\ \Rightarrow d=1\\ \RightarrowƯCLN\left(2n+3,4n+8\right)=1\)
\(\Rightarrow\dfrac{2n+3}{4n+8}\) là phân số tối giản
Bài 5:
a: \(\dfrac{5^2\cdot6^{11}\cdot16^2+6^2\cdot12^6\cdot15^2}{2\cdot6^{12}\cdot10^4-81^2\cdot960^3}\)
\(=\dfrac{5^2\cdot2^{11}\cdot3^{11}\cdot2^8+2^2\cdot3^2\cdot2^{12}\cdot3^6\cdot3^2\cdot5^2}{2\cdot2^{12}\cdot3^{12}\cdot2^4\cdot5^4-3^8\cdot2^{18}\cdot3^3\cdot5^3}\)
\(=\dfrac{5^2\cdot2^{19}\cdot3^{11}+2^{14}\cdot3^{10}\cdot5^2}{2^{17}\cdot3^{12}\cdot5^4-2^{18}\cdot3^{11}\cdot5^3}\)
\(=\dfrac{2^{14}\cdot3^{10}\cdot5^2\left(2^5\cdot3+1\right)}{2^{17}\cdot3^{11}\cdot5^3\cdot\left(3\cdot5-2\right)}\)
\(=\dfrac{97}{2^3\cdot3\cdot5\cdot13}=\dfrac{97}{1560}\)
\(1,\\ a,11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)=\dfrac{146}{13}-\dfrac{18}{7}-\dfrac{68}{13}=6-\dfrac{18}{7}=\dfrac{24}{7}\\ b,\left(6-2\dfrac{4}{5}\right)\cdot3\dfrac{1}{8}-1\dfrac{3}{5}:\dfrac{1}{4}\\ =\left(6-\dfrac{14}{5}\right)\cdot\dfrac{25}{8}-\dfrac{8}{5}\cdot4\\ =\dfrac{16}{5}\cdot\dfrac{25}{8}-\dfrac{32}{5}\\ =10-\dfrac{32}{5}=\dfrac{18}{5}\\ c,\dfrac{-7}{25}\cdot\dfrac{11}{13}+\dfrac{-7}{25}\cdot\dfrac{2}{13}-\dfrac{18}{25}=\dfrac{-7}{25}\left(\dfrac{11}{13}+\dfrac{2}{13}\right)-\dfrac{18}{25}=-\dfrac{7}{25}-\dfrac{18}{25}=-1\\ d,\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{20}\right)\\ =\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{19}{20}=\dfrac{1}{20}\)
