\(\left(a\cdot b\right)^{2n}=...\)Tại sao?
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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.
PH
1
28 tháng 10 2018
\(\left(x-1\right)\left(x+1\right)=x^2+x-x-1=x^2-1\)
\(\Rightarrow x\left(x+1\right)\left(x-1\right)=x\left(x^2-1\right)\)(ĐPCM)
SORRY VỲ GIẢI BẠN CHƯA HIỂU BÀI TRƯỚC
AH
Akai Haruma
Giáo viên
25 tháng 6 2019
Lời giải:
\(A=\frac{1}{1(2n-1)}+\frac{1}{3(2n-3)}+...+\frac{1}{(2n-3).3}+\frac{1}{(2n-1).1}\)
\(2nA=\frac{1+(2n-1)}{1(2n-1)}+\frac{3+(2n-3)}{3(2n-3)}+....+\frac{(2n-3)+3}{(2n-3).3}+\frac{(2n-1)+1}{(2n-1).1}\)
\(2nA=\frac{1}{2n-1}+1+\frac{1}{2n-3}+\frac{1}{3}+...+\frac{1}{3}+\frac{1}{2n-3}+1+\frac{1}{2n-1}\)
\(=\left(\frac{1}{2n-1}+\frac{1}{2n-3}+...+\frac{1}{3}+1\right)+\left(1+\frac{1}{3}+...+\frac{1}{2n-3}+\frac{1}{2n-1}\right)\)
\(=2\left(1+\frac{1}{3}+...+\frac{1}{2n-1}\right)\)
\(\Rightarrow A=\frac{1}{n}\left(1+\frac{1}{3}+...+\frac{1}{2n-1}\right)\)
\(=a^{2n}\cdot b^{2n}\)