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.

Áp dụng tính chất a² - b² = a² - ab + ab - b² = a(a - b) + b(a - b) = (a + b)(a - b)

B =\(\left(200^{-2}-1\right)\left(199^{-2}-1\right)...\left(101^{-2}-1\right)=\left(\frac{1}{200^2}-1\right)\left(\frac{1}{199^2}-1\right)...\left(\frac{1}{101^2}-1\right)\)

\(=\frac{1-200^2}{200^2}.\frac{1-199^2}{199^2}...\frac{1-101^2}{101^2}=\frac{1^2-200^2}{200^2}.\frac{1^2-199^2}{199^2}....\frac{1^2-101^2}{101^2}\)

\(=\frac{\left(1-200\right)\left(1+200\right)}{200^2}.\frac{\left(1-199\right)\left(1+199\right)}{199^2}...\frac{\left(1-101\right)\left(1+101\right)}{101^2}\)

\(=-\left(\frac{199.201}{200^2}.\frac{198.200}{199^2}...\frac{100.102}{101^2}\right)=-\frac{199.201.198.200..100.102}{200.200.199.199...101.101}\)

\(=-\frac{\left(199.198...100\right)\left(201.200...102\right)}{\left(200.199...101\right).\left(200.199...101\right)}=-\frac{100.201}{200.101}=-\frac{201}{202}\)

                                          Bài giải

\(B=\left(200^{-2}-1\right)\left(199^{-2}-1\right)\left(198^{-2}-1\right)...\left(101^{-2}-1\right)\)

\(B=\left(\frac{1}{200^2}-1\right)\left(\frac{1}{199^2}-1\right)\left(\frac{1}{198^2}-1\right)...\left(\frac{1}{101^2}-1\right)\)

\(B=\left[\left(\frac{1}{200}\right)^2-1^2\right]\left[\left(\frac{1}{199}\right)^2-1^2\right]\left[\left(\frac{1}{198}\right)^2-1^2\right]...\left[\left(\frac{1}{101}\right)^2-1^2\right]\)

\(B=\left(\frac{1}{200}+1\right)\left(\frac{1}{200}-1\right)\left(\frac{1}{199}+1\right) \left(\frac{1}{199}-1\right)..\left(\frac{1}{101}-1\right)\left(\frac{1}{101}+1\right)\)

\(B=\frac{201}{200}\cdot\frac{-199}{200}\cdot\frac{200}{199}\cdot\frac{-198}{199}\cdot...\cdot\frac{-100}{101}\cdot\frac{102}{101}\)

\(B=\frac{201\cdot\left(-199\right)\cdot200\cdot\left(-198\right)\cdot...\cdot\left(-100\right)\cdot102}{200\cdot200\cdot199\cdot199\cdot...\cdot101\cdot101}=\frac{100\cdot201}{200\cdot101}=\frac{201}{202}\)

\(A=202\left(200^{-2}-1\right)\left(199^{-2}-1\right)\left(198^{-2}-1\right)...\left(101^{-2}-1\right)\)

\(=202\left(\frac{1}{200^2}-1\right)\left(\frac{1}{199^2}-1\right)\left(\frac{1}{198^2}-1\right)...\left(\frac{1}{101^2}-1\right)\)

\(=-202\left(1-\frac{1}{200^2}\right)\left(1-\frac{1}{199^2}\right)\left(1-\frac{1}{198^2}\right)...\left(1-\frac{1}{101^2}\right)\)

\(=-202\left(\frac{199.201}{200^2}\right).\left(\frac{198.200}{199^2}\right).\left(\frac{197.199}{198^2}\right)...\left(\frac{102.100}{101^2}\right)\)

\(=-202.\frac{199.201.198.200.197.199...100.102}{200^2.199^2.198^2...101^2}\)

\(=-202.\frac{\left(199.198.197...100\right)\left(201.200.199...102\right)}{\left(200.199.198...101\right)\left(200.199.198...101\right)}\)

\(=-202.\frac{1.201}{2.101}=-202.\frac{201}{202}=-201\)

Để Đó mình lo cho:

Ta có:

\(\frac{1}{101}+\frac{1}{102}+.......+\frac{1}{199}+\frac{1}{200}=A\)\(A\)

=>101A=\(101\times\left(\frac{1}{101}+\frac{1}{102}+......+\frac{1}{199}+\frac{1}{200}\right)\)

=>101A=\(\left(101\times\frac{1}{101}\right)+\left(101\times\frac{1}{102}\right)+........+\left(101\times\frac{1}{199}\right)+\left(101\times\frac{1}{200}\right)\)

=>101A=\(1+\frac{101}{102}+.....+\frac{101}{199}+\frac{101}{200}>1\)

=>101A>1

=>A>1

là A>1 nhé

Ai giải giúp mik câu này ik:

23.Chứng minh rằng:

\(\frac{a}{n\left(n+a\right)}\)\(=\)\(\frac{1}{n}-\frac{1}{n+a}\)

n, a\(\in\)N^*

A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)

Lại có B = \(\frac{1}{101.200}+\frac{1}{102.199}+...+\frac{1}{200.101}\)

=> 301B = \(\frac{301}{101.200}+\frac{301}{102.199}+...+\frac{301}{200.101}\) 

=> 301B = \(\frac{1}{101}+\frac{1}{200}+\frac{1}{102}+\frac{1}{199}+...+\frac{1}{200}+\frac{1}{101}=2\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)\)

=> B = \(\frac{2}{301}\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)\)

Khi đó \(\frac{A}{B}=\frac{\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)}{\frac{2}{301}\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)}=\frac{1}{\frac{2}{301}}=\frac{301}{2}=150,5\)

cái này dễ lắm chỉ là chưa để ý thôi:

a,1/101>1/102>...>1/199>1/200

=>1/101+1/102+...+1/199+1/200<100*1/101=100/101<1

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