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Ta có
\(\frac{7}{12}=\frac{3}{12}+\frac{4}{12}=\frac{1}{4}+\frac{1}{3}=\frac{20}{80}+\frac{20}{60}=\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}\)
\(>\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}\)
hay B>A
nhớ tick mình nha
Ta chia A làm hai phần mỗi phần 20 số hạng
\(C=\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\)với \(D=\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\)
Xét \(C=\frac{1}{41}+\frac{1}{42}+...+\frac{1}{61}\)\(>\frac{1}{61}+\frac{1}{61}+\frac{1}{61}+...+\frac{1}{61}\)\(=\frac{1}{61}.20=\frac{1}{3}\)
Xét \(D=\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}>\frac{1}{80}+\frac{1}{80}+...+\frac{1}{80}=\frac{1}{80}.20=\frac{1}{4}\)
Mà A = C + D > \(\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
=> A > B
Ta có:
7/12 = 4/12 + 3/12 = 1/3 + 1/4 = 20/60 + 20/80
1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 = (1/41 + 1/42 + 1/43 + ...+ 1/60) + (1/61 + 1/62 +...+ 1/79 + 1/80)
Do 1/41> 1/42 > 1/43 > ...>1/59 > 1/60
=> (1/41 + 1/42 + 1/43 + ...+ 1/60) > 1/60 + ...+ 1/60 = 20/60
và 1/61> 1/62> ... >1/79> 1/80
=> (1/61 + 1/62 +...+ 1/79 + 1/80) > 1/80 + ...+ 1/80 = 20/80
Vậy: 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 20/60 + 20/80 = 7/12
=> 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 7/12
=>A>B
A=\(\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)\) +\(\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\right)\)
Ta có : \(\frac{1}{41}>\frac{1}{60};\frac{1}{42}>\frac{1}{60};...;\frac{1}{60}=\frac{1}{60}\) => \(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}>\frac{20}{60}=\frac{1}{3}\)
\(\frac{1}{61}>\frac{1}{80};\frac{1}{62}>\frac{1}{80};...;\frac{1}{80}=\frac{1}{80}\) => \(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}>\frac{20}{80}=\frac{1}{4}\)
=> A > \(\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
Vậy a >\(\frac{7}{12}\)
\(\frac{7}{12}=\frac{3}{12}+\frac{4}{12}=\frac{1}{4}+\frac{1}{3}\)
ta có:\(A=\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}=\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\right)\)
ta có:\(\frac{1}{41}>\frac{1}{42}>\frac{1}{43}>...>\frac{1}{60}\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{59}+\frac{1}{60}>\frac{1}{60}+...+\frac{1}{60}=\frac{20}{60}=\frac{1}{3}\left(1\right)\)
\(\frac{1}{61}>\frac{1}{62}>\frac{1}{63}>...>\frac{1}{80}\Rightarrow\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}>\frac{1}{80}+\frac{1}{80}+...+\frac{1}{80}=\frac{20}{80}=\frac{1}{4}\left(2\right)\)
từ (1) (2) suy ra \(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
\(\Rightarrow A=\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}>\frac{7}{12}\left(đfcm\right)\)
Ta có:
7/12 = 4/12 + 3/12 = 1/3 + 1/4 = 20/60 + 20/80
1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 = (1/41 + 1/42 + 1/43 + ...+ 1/60) + (1/61 + 1/62 +...+ 1/79 + 1/80)
Do 1/41> 1/42 > 1/43 > ...>1/59 > 1/60
=> (1/41 + 1/42 + 1/43 + ...+ 1/60) > 1/60 + ...+ 1/60 = 20/60
và 1/61> 1/62> ... >1/79> 1/80
=> (1/61 + 1/62 +...+ 1/79 + 1/80) > 1/80 + ...+ 1/80 = 20/80
Vậy: 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 20/60 + 20/80 = 7/12
=> 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 7/12
\(A>B\)
2)lx^2+lx+1ll=x^2
=>x^2+lx+1l=x^2=>lx+1l=0=>x=-1
3)\(\frac{\left(-\frac{1}{2}\right)^n}{\left(-\frac{1}{2}\right)^{n-2}}=\left(-\frac{1}{2}\right)^{n-n-2}=\left(-\frac{1}{2}\right)^{-2}=4\)
1)\(A=\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}\)
\(\Rightarrow A=\left(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}+...+\frac{1}{80}\right)\)
\(\Rightarrow A=C+D\)
Ta có:\(\frac{1}{41}>\frac{1}{60};>\frac{1}{60}:\frac{1}{43}>\frac{1}{60};...;\frac{1}{59}>\frac{1}{60};\frac{1}{60}=\frac{1}{60}\)
\(\Rightarrow C=\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)
Ta thấy C có 20 số hạng
\(\Rightarrow C>\frac{1}{60}.20=\frac{1}{3}\)
Ta có:\(\frac{1}{61}>\frac{1}{80};\frac{1}{62}>\frac{1}{80};\frac{1}{63}>\frac{1}{80};...;\frac{1}{79}>\frac{1}{80};\frac{1}{80}=\frac{1}{80}\)
\(\Rightarrow D=\frac{1}{61}+\frac{1}{62}+\frac{1}{63}+...+\frac{1}{80}>\frac{1}{80}+\frac{1}{80}+\frac{1}{80}+...+\frac{1}{80}\)
Ta thấy D có 20 số hạng.
\(\Rightarrow D>\frac{1}{80}.20=\frac{1}{4}\)
\(\Rightarrow A=C+D>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
\(\Rightarrow A>B\)
