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XO
12 tháng 8 2020
Áp dụng tính chất a2 - b2 = a2 - ab + ab - b2 = a(a - b) + b(a - b) = (a + b)(a - b)
B =\(\left(200^{-2}-1\right)\left(199^{-2}-1\right)...\left(101^{-2}-1\right)=\left(\frac{1}{200^2}-1\right)\left(\frac{1}{199^2}-1\right)...\left(\frac{1}{101^2}-1\right)\)
\(=\frac{1-200^2}{200^2}.\frac{1-199^2}{199^2}...\frac{1-101^2}{101^2}=\frac{1^2-200^2}{200^2}.\frac{1^2-199^2}{199^2}....\frac{1^2-101^2}{101^2}\)
\(=\frac{\left(1-200\right)\left(1+200\right)}{200^2}.\frac{\left(1-199\right)\left(1+199\right)}{199^2}...\frac{\left(1-101\right)\left(1+101\right)}{101^2}\)
\(=-\left(\frac{199.201}{200^2}.\frac{198.200}{199^2}...\frac{100.102}{101^2}\right)=-\frac{199.201.198.200..100.102}{200.200.199.199...101.101}\)
\(=-\frac{\left(199.198...100\right)\left(201.200...102\right)}{\left(200.199...101\right).\left(200.199...101\right)}=-\frac{100.201}{200.101}=-\frac{201}{202}\)
F
12 tháng 8 2020
Bài giải
\(B=\left(200^{-2}-1\right)\left(199^{-2}-1\right)\left(198^{-2}-1\right)...\left(101^{-2}-1\right)\)
\(B=\left(\frac{1}{200^2}-1\right)\left(\frac{1}{199^2}-1\right)\left(\frac{1}{198^2}-1\right)...\left(\frac{1}{101^2}-1\right)\)
\(B=\left[\left(\frac{1}{200}\right)^2-1^2\right]\left[\left(\frac{1}{199}\right)^2-1^2\right]\left[\left(\frac{1}{198}\right)^2-1^2\right]...\left[\left(\frac{1}{101}\right)^2-1^2\right]\)
\(B=\left(\frac{1}{200}+1\right)\left(\frac{1}{200}-1\right)\left(\frac{1}{199}+1\right) \left(\frac{1}{199}-1\right)..\left(\frac{1}{101}-1\right)\left(\frac{1}{101}+1\right)\)
\(B=\frac{201}{200}\cdot\frac{-199}{200}\cdot\frac{200}{199}\cdot\frac{-198}{199}\cdot...\cdot\frac{-100}{101}\cdot\frac{102}{101}\)
\(B=\frac{201\cdot\left(-199\right)\cdot200\cdot\left(-198\right)\cdot...\cdot\left(-100\right)\cdot102}{200\cdot200\cdot199\cdot199\cdot...\cdot101\cdot101}=\frac{100\cdot201}{200\cdot101}=\frac{201}{202}\)
XO
1 tháng 9 2020
\(A=202\left(200^{-2}-1\right)\left(199^{-2}-1\right)\left(198^{-2}-1\right)...\left(101^{-2}-1\right)\)
\(=202\left(\frac{1}{200^2}-1\right)\left(\frac{1}{199^2}-1\right)\left(\frac{1}{198^2}-1\right)...\left(\frac{1}{101^2}-1\right)\)
\(=-202\left(1-\frac{1}{200^2}\right)\left(1-\frac{1}{199^2}\right)\left(1-\frac{1}{198^2}\right)...\left(1-\frac{1}{101^2}\right)\)
\(=-202\left(\frac{199.201}{200^2}\right).\left(\frac{198.200}{199^2}\right).\left(\frac{197.199}{198^2}\right)...\left(\frac{102.100}{101^2}\right)\)
\(=-202.\frac{199.201.198.200.197.199...100.102}{200^2.199^2.198^2...101^2}\)
\(=-202.\frac{\left(199.198.197...100\right)\left(201.200.199...102\right)}{\left(200.199.198...101\right)\left(200.199.198...101\right)}\)
\(=-202.\frac{1.201}{2.101}=-202.\frac{201}{202}=-201\)
5 tháng 10 2015
a/P=1-1/2+1/3-1/4+1/5-1/6+...+1/199-1/200
=(1+1/3+1/5+1/7+...+1/199)-(1/2+1/4+1/6+...+1/200)
=(1+1/2+1/3+1/4+1/5+1/6+...+1/99+1/200)-2(1/2+1/4+1/6+...+1/200)
=(1+1/2+1/3+1/4+1/5+1/6+...+1/99+1/200)-(1+1/2+1/3+...+1/100)
=1/101+1/102+1/103+...+1/200
DQ
0
JA
5 tháng 10 2015
a/ P=1-1/2+1/3-1/4+....+1/199-1/200
= 1+1/2+1/3+1/4+1/5+...+1/200 - 2.(1/2+1/4+...+1/200)
= 1+1/2+1/3+1/4+1/5+...+1/200 - 1-1/2-1/3-...-1/100
=1/101+1/102+...+1/200
b/ k-k/2+ k/3- k/4+...+k/199-k/200
=k+k/2+k/2+...+k/199+k/200 -2(k/2+k/4+k/6+...+k/200)
=k+k/2+k/2+...+k/199+k/200-k-k/2-k/3-...-k/100
=k/101+k/102+...+k.200
TT
0
TL
0
Q
18 tháng 7 2017
A=(200-2-1)(199-2-1)....(101-2-1)
\(A=\left(\frac{1}{200^2}-1\right)\left(\frac{1}{199^2}-1\right)\left(\frac{1}{198^2}-1\right).....\left(\frac{1}{101^2}-1\right)\)
\(A=\frac{1-200^2}{200^2}.\frac{1-199^2}{199^2}.\frac{1-198^2}{198^2}.....\frac{1-101^2}{101^2}\)
\(A=\frac{\left(1-200\right)\left(1+200\right)}{200^2}.\frac{\left(1-199\right)\left(1+199\right)}{199^2}.....\frac{\left(1-100\right)\left(1+100\right)}{100^2}.\frac{\left(1-101\right)\left(1+101\right)}{101^2}\)
\(A=\frac{-199.201}{200.200}.\frac{-198.200}{199.199}.\frac{-197.199}{198.198}.....\frac{-99.101}{100.100}.\frac{-100.102}{101.101}\)
\(A=\frac{199.201}{200.200}.\frac{198.200}{199.199}.\frac{197.199}{198.198}.....\frac{99.101}{100.100}.\frac{100.102}{101.101}\)
\(\Rightarrow A=\frac{200}{2.101}=\frac{201}{202}\)
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