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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 => ĐPCM
Ta có : 1/41 + 1/42 + ... + 1/60 > 1/60 * 20 = 1/3 .
1/61 + 1/62 + ... + 1/80 > 1/80 * 20 = 1/4 .
⇒ 1/41 + 1/42 + ... + 1/80 > 1/3 + 1/4 = 4/12 + 3/12 .
= 7/12 .
Do đó : A > 7/12 .
Vậy bài toán được chứng minh .
bạn ơi cái câu <1 số hạng cuối cùng là j thế?
Ta có:
A=9999931999−5555571997
A=9999931998.999993−5555571996.555557
A=(9999932)999.999993 − (5555572)998.555557
A=\(\overline{\left(....9\right)}^{999}\) . 999993 - \(\overline{\left(...1\right)}.\text{555557}\)
A=\(\overline{\left(...7\right)}-\overline{\left(...7\right)}\)
A= \(\overline{\left(...0\right)}\)
Vì A có tận cùng là 0 nên \(A⋮5\)
Sửa đề là chứng minh nha bạn.
Ta có: \(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{80}>\dfrac{1}{41}+\dfrac{1}{41}+\dfrac{1}{41}+...+\dfrac{1}{41}\)(40 phân số \(\dfrac{1}{41}\))
\(=\dfrac{1.40}{41}=\dfrac{40}{41}>\dfrac{7}{12}\) (*)
Từ (*) suy ra: \(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{80}>\dfrac{7}{12}^{\left(đpcm\right)}\)
Đặt \(A=\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+\dfrac{1}{44}+...+\dfrac{1}{80}\)
\(=\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{60}\right)+\) \(\left(\dfrac{1}{61}+\dfrac{1}{62}+...+\dfrac{1}{80}\right)\)
Nhận xét:
\(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}\) \(=\dfrac{1}{3}\)
\(\dfrac{1}{61}+\dfrac{1}{62}+...+\dfrac{1}{80}>\dfrac{1}{80}+\dfrac{1}{80}+...+\dfrac{1}{80}\) \(=\dfrac{1}{4}\)
\(\Rightarrow A>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}>\dfrac{1}{12}\)
Vậy \(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{80}>\dfrac{1}{12}\) (Đpcm)
a, Ta có :
\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{1\cdot2\cdot3\cdot4}+...+\dfrac{1}{1\cdot2\cdot3\cdot...\cdot100}\\ < \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}< 1\\ \Rightarrow M< 1\\ \RightarrowĐpcm\)
Đặt \(A=\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{80}\)
\(=\left(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{60}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}+...+\dfrac{1}{80}\right)\)
Mặt khác:
\(\dfrac{7}{12}=\dfrac{20}{60}+\dfrac{20}{80}\)
mà \(\left\{{}\begin{matrix}\dfrac{20}{60}< \left(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{60}\right)\\\dfrac{20}{80}< \left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}+...+\dfrac{1}{80}\right)\end{matrix}\right.\)
⇒ \(\dfrac{7}{12}< A\) (1)
Ta có:
\(\dfrac{5}{6}=\dfrac{20}{40}+\dfrac{20}{60}\)
mà \(\left\{{}\begin{matrix}\dfrac{20}{40}>\left(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{60}\right)\\\dfrac{20}{60}>\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}+...+\dfrac{1}{80}\right)\end{matrix}\right.\)
⇒ \(A< \dfrac{5}{6}< 1\)(2)
Từ (1) và (2)
⇒ \(\dfrac{7}{12}< A< 1\) (đpcm)