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\(B=\frac{1^2}{2^2-1}.\frac{3^2}{4^2-1}.\frac{5^2}{6^2-1}...\frac{\left(2n+1\right)^2}{\left(2n+2\right)^2-1}\)
\(=\frac{1^2}{\left(2-1\right)\left(2+1\right)}.\frac{3^2}{\left(4-1\right)\left(4+1\right)}...\frac{\left(2n+1\right)^2}{\left(2n+2-1\right)\left(2n+2+1\right)}\)
\(=\frac{1}{1.3}.\frac{3^2}{3.5}...\frac{\left(2n+1\right)^2}{\left(2n+1\right)\left(2n+3\right)}\)
\(=\frac{1}{2n+3}\)
a/ \(\left(2n^3-5n^2+1\right):\left(2n-1\right)=n^2-2n-1\)
b/ \(x\ne0;\pm2\)
\(\left(\frac{x^2}{x\left(x^2-4\right)}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\right):\left(\frac{x^2-4+10-x^2}{x+2}\right)\)
\(=\left(\frac{x}{x^2-4}-\frac{2\left(x+2\right)}{x^2-4}+\frac{x-2}{x^2-4}\right):\left(\frac{6}{x+2}\right)\)
\(=\left(\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\right).\left(\frac{x+2}{6}\right)\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{\left(x+2\right)}{6}=-\frac{1}{x-2}=\frac{1}{2-x}\)
c/
\(\left(3x-1\right)^2+2\left(3x-1\right)\left(3x+4\right)+\left(3x+4\right)^2\)
\(=\left(3x-1+3x+4\right)^2\)
\(=\left(6x+3\right)^2\)
a. Ta có: \(\frac{1}{2^2}\)< \(\frac{1}{1.3}\)
\(\frac{1}{4^2}\)< 1/(3.5)
1/(6^2) <1/(5.7)
...
1/(2n)^2 < 1/(2n-1)(2n+1)
=> 1/2^2 +1/4^2 + 1/6^2 +...+1/(2n)^2 < 1/(1.3) +...+1/(2n-1)(2n+1)
=> 2(1/2^2 +1/4^2 + 1/6^2 +...+1/(2n)^2) < (1/1 - 1/3 +1/3 - 1/5 + 1/5 - 1/7 +...+ 1/(2n-1) - 1/(2n+1)
=>2(1/2^2 +1/4^2 + 1/6^2 +...+1/(2n)^2) < 1 - 1/(2n+1) = 2n/(2n+1)
=> 1/2^2 +1/4^2 + 1/6^2 +...+1/(2n)^2 < 2n/(2n+1) . 1/2
Vì 2n/2n+1 < 1 => 2n/(2n+1) . 1/2 < 1/2
=> 1/2^2 +1/4^2 + 1/6^2 +...+1/(2n)^2 <1/2
Câu b tương tự
\(\frac{150}{5.8}+\frac{150}{8.11}+\frac{150}{11.14}+.....+\frac{150}{47.50}\)
\(=50.\left(\frac{3}{5.8}+\frac{5}{8.11}+.....+\frac{3}{47.50}\right)\)
\(=50.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{47}-\frac{1}{50}\right)\)
\(=50.\left(\frac{1}{5}-\frac{1}{50}\right)\)
\(=50.\frac{9}{50}=9\)
\(b,\)\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=2^{64}-1-2^{64}=-1\)
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.
\(B=\frac{1^2}{2^2-1}.\frac{3^2}{4^2-1}....\frac{\left(2n+1\right)^2}{\left(2n+2\right)^2-1}\)
\(=\frac{1^2}{\left(2-1\right)\left(2+1\right)}.\frac{3^2}{\left(4-1\right)\left(4+1\right)}...\frac{\left(2n+1\right)^2}{\left(2n+2-1\right)\left(2n+2+1\right)}\)
\(=\frac{1^2}{1.3}.\frac{3^2}{3.5}...\frac{\left(2n+1\right)^2}{\left(2n+1\right)\left(2n+3\right)}\)
\(=\frac{2n+1}{2n+3}\)
P/S: tham khảo nha, mk ko chắc là đúng