\(\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+....\frac{1}{41\cdot45}\)
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\(b)\) Ta có: \(x-\frac{37}{45}=\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45\text{ }}\)
\(\Leftrightarrow x-\frac{37}{45}=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\)
\(\Leftrightarrow x-\frac{37}{45}=\frac{1}{5}-\frac{1}{45}\)
\(\Leftrightarrow x-\frac{37}{45}=1\)
\(\Leftrightarrow x=1+\frac{37}{45}\)
\(\Leftrightarrow x=\frac{82}{45}\)
Vậy \(x=\frac{82}{45}\)
A=\(\frac{2^{35}.9^{25}.5^{25}.13^{22}.7^{16}.5^{16}}{4^{17}.7^{17}.9^{26}.13^{22}.5^{22}.5^9.5^9}=\frac{2^{35}.5^1}{4^{17}.7^1.9}=\frac{2^{35}.5}{2^{34}.7^1.9}\)= \(\frac{2.5}{7.9}=\frac{10}{63}\)
\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(=\frac{1}{1.3}-\frac{1}{11.13}\)
\(=\frac{1}{3}-\frac{1}{143}\)
\(=\frac{140}{429}\)
\(\frac{1}{1.3.7}+\frac{1}{3.7.9}+\frac{1}{7.9.13}+\frac{1}{9.13.15}+\frac{1}{13.15.19}\)
\(=\frac{1}{2}\left(\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+...+\frac{1}{13.15}-\frac{1}{15.19}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.3}-\frac{1}{15.19}\right)=\frac{47}{285}\)
\(C=\frac{2}{1\cdot5}+\frac{2}{5\cdot9}+\frac{2}{9\cdot13}+\frac{2}{13\cdot17}+\frac{2}{17\cdot21}\)
\(C=\frac{2}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+....+\frac{1}{17}-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\frac{20}{21}\)
\(=\frac{10}{21}\)
= 7/4.(4/1.5 + 4/5.9 + 4/9.13 + 4/13.17 + 4/17.21)
= 7/4.(1-1/5+1/5-1/9+1/9-1/13+1/13-1/17+1/17-1/21)
= 7/4.(1-1/21)
= 7/4.20/21 = 5/3
Tk mk nha
Đặt biểu thức bằng A
4/7A=1-1/5+1/5-1/9+...+1/17_1/21
4/7A=1-1/21
4/7A=20/21
A=35/21=5/3
\(\frac{1}{1.3.7}=\frac{1}{6}\left(\frac{1}{1.3}-\frac{1}{3.7}\right)\)
\(\frac{1}{3.7.9}=\frac{1}{6}\left(\frac{1}{3.7}-\frac{1}{7.9}\right)\)
....
\(\frac{1}{13.15.19}=\frac{1}{6}\left(\frac{1}{13.15}-\frac{1}{15.19}\right)\)
Cộng các vế với nhau ta được
\(\frac{1}{1.3.7}+\frac{1}{3.7.9}+...+\frac{1}{13.15.19}=\frac{1}{6}\left(\frac{1}{1.3}-\frac{1}{15.19}\right)=\frac{37}{3.15.19}\)
lớp 6 thì có
Gọi A=\(\frac{1}{5.9}+\frac{1}{9.13}+....+\frac{1}{41.45}\)
=>4A=\(\frac{4}{5.9}+\frac{4}{9.13}+....+\frac{4}{41.45}\)
4A=\(\frac{9-5}{5.9}+\frac{13-9}{9.13}+....+\frac{45-41}{41.45}\)
4A\(=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+....+\frac{1}{41}-\frac{1}{45}\)
\(4A=\frac{1}{5}-\frac{1}{45}\)
\(4A=\frac{8}{45}\)
\(=>A=\frac{2}{45}\)