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\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{79}{80}<\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}...\frac{78}{79}.\frac{79}{80}=\frac{1}{80}<\frac{1}{9}\)
\(\text{Vậy }A<\frac{1}{9}\)
c) C = ( 1 - 2 ) + ( 3 - 4 ) + ... + ( 79 - 80 )
C = ( -1 ) + ( -1 ) + ... + ( -1 )
C = ( -1 ) x ( 80 - 1 + 1 ) : 2
C = ( -1 ) x 80 : 2
C = ( -40 )
1/h=1/2(1/a+1/b)=1/2a+1/2b=(a+b)/2ab
=>(a+b/)2ab-1/h=0
quy dong len ta co
(a+b)h/2abh-2ab/2abh=0=> (ah+bh-2ab)/2abh=0 =>ah+bh-2ab=0
=>ah+bh-ab-ab=0
=>a(h-b)-b(a-h)=0
=>a(h-b)=b(a-h)
=>a/b=(a-h)(h-b)
ta co : A= ( 8^9+12/8^9+7) -1
= 5/8^9+7
B=(8^10+4/8^10-1)-1
=5/8^10-1
VI 8^9+7 < 8^10-1 NEN 5/8^9+7 > 5/8^10-1
VAY A > B
Ta có : A = ( 8^9+12/8^9+7) - 1
= 5/8^9 + 7
B = (8^10+4/8^10-1) - 1
= 5/8^10-1
VI 8^9 + 7 < 8^10 - 1 nên 5/8^9+7 > 5/8^10-1
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow A< 1\)
b) \(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow3B=1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{99}\)
\(\Rightarrow3B-B=1-\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow2B=1-\left(\frac{1}{3}\right)^{100}< 1\)
\(\Rightarrow2B< 1\)
\(\Rightarrow B< \frac{1}{2}\)
\(\frac{1.3.5...79}{2.4.6...80}\)= \(\frac{1.3.5...79}{\left(1.2\right).\left(2.2\right).\left(3.2\right)...\left(40.2\right)}\).\(\frac{1.3.5...79}{\left(1.2.3.4...40\right).\left(2.2.2.2...2.2\right)}\)=\(\frac{1.3.5...79}{\left(1.3.5...39\right).\left(2.4.6...40\right).2^{40}}\)<1/9