Tính giá trị biểu thức :
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{99}\right)\cdot\left(1-\frac{1}{100}\right)\)
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DD
6 tháng 8 2015
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)...\left(1+\frac{1}{99}\right)=\frac{3}{2}.\frac{4}{3}...\frac{100}{99}=\frac{100}{2}=50\)
6 tháng 8 2015
= \(\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\cdot\cdot\cdot\frac{99}{98}\cdot\frac{100}{99}=\frac{3.4.5....99.100}{2.3.4....98.99}=\frac{100}{2}=50\)
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.
17 tháng 8 2018
\(A=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)
\(A=\frac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot...\cdot99}\)
\(A=\frac{100}{2}\)
\(A=50\)
Ta có:
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{99}\right).\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{98}{99}.\frac{99}{100}\) \(=\frac{1.2.3...98.99}{2.3.4...99.100}=\frac{1}{100}\)
nha