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\(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{\left(2n-1\right)\left(2n+1\right)}\)
\(=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\right)\)
\(=2.\left(1-\frac{1}{2n+1}\right)\)
\(=2.\left(\frac{2n}{2n+1}\right)\)
\(=\frac{4n}{2n+1}\)
Tham khảo nhé~
A=\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.2^9.3^9}{\left(2^2\right)^6.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)
\(=\frac{2.6}{3.7}\)\(=\frac{4}{7}\)
\(a,\frac{x-1}{21}=\frac{3}{x+1}\)
\(\Leftrightarrow\left[x-1\right]\left[x+1\right]=63\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
\(\Leftrightarrow x^2=8^2\)
\(\Leftrightarrow x=\pm8\)
\(b,\frac{7}{x}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{21}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{7}{15}\)
\(\Leftrightarrow x=15\)
Vậy x = 15
Bài cuối tương tự
a) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.........+\frac{2}{x\left(x+1\right)}=\frac{1998}{2000}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.......+\frac{2}{x\left(x+1\right)}=\frac{1998}{2000}\)
\(\Leftrightarrow2.\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+......+\frac{1}{x\left(x+1\right)}\right]=\frac{1998}{2000}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{x\left(x+1\right)}=\frac{999}{2000}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+......+\frac{1}{x}-\frac{1}{x+1}=\frac{999}{2000}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{999}{2000}\)\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2000}\)
\(\Leftrightarrow x+1=2000\)\(\Leftrightarrow x=1999\)
Vậy \(x=1999\)
b) \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+......+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
\(\Leftrightarrow\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{15.2}{93}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+......+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Leftrightarrow\frac{1}{2x+3}=\frac{1}{93}\)\(\Leftrightarrow2x+3=93\)
\(\Leftrightarrow2x=90\)\(\Leftrightarrow x=45\)
Vậy \(x=45\)
muốn tìm n thì phải có 2 về chứ bạn