\(\frac{1^2}{2^2-1}\cdot\frac{3^2}{4^2-1}\cdot\frac{5^2}{6^2-1}....\frac{n^2}{\left(n+1\right)^2-1}\)
Rút gọn biểu thức sau:
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21 tháng 11 2016
\(C=\frac{1^2}{2^2-1}.\frac{3^2}{4^2-1}.\frac{5^2}{6^2-1}....\frac{n^2}{\left(n+1\right)^2-1}\)
\(=\frac{1^2}{1.3}.\frac{3^2}{3.5}.\frac{5^2}{5.7}.....\frac{n^2}{n.\left(n+2\right)}\)
\(=\frac{1}{n+2}\)
19 tháng 6 2019
\(b,\)\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=2^{64}-1-2^{64}=-1\)
DL
19 tháng 6 2019
a) Đặt \(A=\left(\frac{1}{2}+1\right).\left(\frac{1}{4}+1\right).\left(\frac{1}{16}+1\right)...\left(1+\frac{1}{2^{2n}}\right)\)
Rút gọn: \(A=\frac{2+1}{2}.\frac{4+1}{4}.\frac{16+1}{16}...\frac{2^{2.n}+1}{2^{2.n}}=\frac{2^{2.0}+1}{2^{2.0}}.\frac{2^{2.1}+1}{2^{2.1}}.\frac{2^{2.2}+1}{2^{2.2}}...\frac{2^{2.n}+1}{2^{2.n}}\)
\(\Rightarrow A=\frac{\left(2^{2.0}+1\right).\left(2^{2.1}+1\right).\left(2^{2.2}+1\right)...\left(2^{2.n}+1\right)}{2^{2.0}.2^{2.1}.2^{2.2}...2^{2.n}}.\)
b) Đặt \(B=\left(2+1\right).\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2-1\right).\left(2+1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^2-1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2^4-1\right).\left(2^4+1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^8-1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2^{16}-1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}=\left(2^{32}-1\right).\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=2^{64}-1-2^{64}=-1\)Vậy B =-1.
L
29 tháng 5 2019
\(\left(\frac{1}{2+2.\sqrt{a}}+\frac{1}{2-2.\sqrt{a}}-\frac{a^2+1}{1-a^2}\right).\left(1+\frac{1}{a}\right)\)
\(=\left(\frac{2-2.\sqrt{a}+2+2.\sqrt{a}}{\left(2+2.\sqrt{a}\right)\left(2-2.\sqrt{a}\right)}-\frac{a^2+1}{\left(1-a\right).\left(1+a\right)}\right).\left(\frac{a+1}{a}\right)\)
\(=\left(\frac{4}{4-4a}-\frac{a^2+1}{\left(1-a\right).\left(1+a\right)}\right).\left(\frac{a+1}{a}\right)=\frac{\left(1+a\right)}{\left(1-a\right).\left(1+a\right)}\cdot\frac{a+1}{a}=\frac{1+a}{\left(1-a\right).a}=\frac{a+1}{a-a^2}\)
Xét dạng tổng quát :
\(\frac{n^2}{\left(n+1\right)^2-1}=\frac{n^2}{n^2+2n+1-1}=\frac{n^2}{n\left(n+2\right)}=\frac{n}{n+2}\)
Khi đó ta có biến đổi của biểu thức đã cho :
\(\frac{1}{3}\cdot\frac{3}{5}\cdot\frac{5}{7}\cdot...\cdot\frac{n}{n+2}=\frac{1}{n+2}\)