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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
\(A=202\left(200^{-2}-1\right)\left(199^{-2}-1\right)\left(198^{-2}-1\right)...\left(101^{-2}-1\right)\)
\(=202\left(\frac{1}{200^2}-1\right)\left(\frac{1}{199^2}-1\right)\left(\frac{1}{198^2}-1\right)...\left(\frac{1}{101^2}-1\right)\)
\(=-202\left(1-\frac{1}{200^2}\right)\left(1-\frac{1}{199^2}\right)\left(1-\frac{1}{198^2}\right)...\left(1-\frac{1}{101^2}\right)\)
\(=-202\left(\frac{199.201}{200^2}\right).\left(\frac{198.200}{199^2}\right).\left(\frac{197.199}{198^2}\right)...\left(\frac{102.100}{101^2}\right)\)
\(=-202.\frac{199.201.198.200.197.199...100.102}{200^2.199^2.198^2...101^2}\)
\(=-202.\frac{\left(199.198.197...100\right)\left(201.200.199...102\right)}{\left(200.199.198...101\right)\left(200.199.198...101\right)}\)
\(=-202.\frac{1.201}{2.101}=-202.\frac{201}{202}=-201\)
NT
1
TT
0
DQ
0
TN
0
MG
14 tháng 11 2017
Ta có:\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}=\dfrac{100}{200}=\dfrac{1}{2}\)
Lại có:
\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}< \dfrac{1}{101}+\dfrac{1}{101}+...+\dfrac{1}{101}=\dfrac{100}{101}\)
Vậy ...
Những dãy trên đều có 100 số hạng.
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