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A=\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.2^9.3^9}{\left(2^2\right)^6.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)
\(=\frac{2.6}{3.7}\)\(=\frac{4}{7}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right).\left(2n+1\right)}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\)
\(2A=1-\frac{1}{2n+1}\)
\(A=\frac{1}{2}-\frac{1}{\left(2n+1\right).2}< \frac{1}{2}\)
Vậy:...
- Hok tốt ~
\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\)
=>\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\)
=>\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}+\frac{1}{2n+1}\)
=>\(2A=1-\frac{1}{2n-1}\)
=>\(2A=\frac{2n}{2n+1}\)
=>\(A=\frac{2n}{4n+2}=\frac{2n}{2\left(n+1\right)}=\frac{n}{n+1}< \frac{1}{2}\)
zậy A<1/2
Ta có: \(1-\frac{4}{1}=-3=-\frac{2.1+1}{2.1-1}\)
\(-3.\left(1-\frac{4}{9}\right)=-3.\frac{5}{9}=-\frac{5}{3}=-\frac{2.2+1}{2.2-1}\)
\(-\frac{5}{3}.\left(1-\frac{1}{25}\right)=-\frac{5}{3}.\frac{21}{25}=-\frac{7}{5}=-\frac{2.3+1}{2.3-1}\)
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Vậy kết quả cuối cùng của biểu thức là: \(-\frac{2n+1}{2n-1}\)
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{\left(2x-1\right)\left(2x+1\right)}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{\left(2x-1\right)\left(2x+1\right)}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{\left(2x-1\right)}-\frac{1}{\left(2x+1\right)}\)
\(2A=1-\frac{1}{2x+1}=\frac{2x}{2x+1}\)
\(A=\frac{x}{2x+1}\)
Mà \(A=\frac{49}{99}\) \(\Leftrightarrow\frac{x}{2x+1}=\frac{49}{99}\Leftrightarrow x=49\)
\(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{\left(2n-1\right)\left(2n+1\right)}\)
\(=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\right)\)
\(=2.\left(1-\frac{1}{2n+1}\right)\)
\(=2.\left(\frac{2n}{2n+1}\right)\)
\(=\frac{4n}{2n+1}\)
Tham khảo nhé~