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A = 1/(5.6) + 1/(6.7) + ... + 1/(24.25)
= 1/5 - 1/6 + 1/6 - 1/7 + ... + 1/24 - 1/25
= 1/5 - 1/25
= 4/25
B = 2/(1.3) + 2/(3.5) + 2/(5.7) + ... + 2/(99.101)
= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/99 - 1/101
= 1 - 1/101
= 100/101
`a) A = 1/(5.6) + 1/(6.7)+...+1/(24.25)`
`= 1/5 - 1/6 + 1/6 - 1/7 +...+1/24-1/25`
`= 1/5-1/25`
`= 5/25 - 1/25`
`= 4/25`
Vậy:`A = 4/25`
`b) B = 2/(1.3)+2/(3.5)+...+2/(99.101)`
`= 1- 1/3 + 1/3 - .... +1/99-1/101`
`= 1 - 1/101`
`= 100/101`
Vậy: `B = 100/101`
\(a^3b-ab^3=ab\left(a^2-b^2\right)=ab\left(a^2-ab+ab-b^2\right)=ab\left(a-b\right)\left(a+b\right)\)
Với a hoặc b chẵn \(\Leftrightarrow ab\left(a-b\right)\left(a+b\right)⋮2\)
Với a và b lẻ \(\Leftrightarrow\left(a-b\right)⋮2\Leftrightarrow ab\left(a-b\right)\left(a+b\right)⋮2\)
Vậy \(ab\left(a-b\right)\left(a+b\right)⋮2,\forall a,b\left(1\right)\)
Với a hoặc b chia hết cho 3 thì \(ab\left(a-b\right)\left(a+b\right)⋮3\)
Với \(a=3k+1;b=3q+1\Leftrightarrow\left(a-b\right)=3\left(k-q\right)⋮3\)
\(\Leftrightarrow ab\left(a-b\right)\left(a+b\right)⋮3\)
Với \(a=3k+1;b=3q+2\Leftrightarrow\left(a+b\right)=\left(3k+1+3q+2\right)=3\left(k+q+1\right)⋮3\)
\(\Leftrightarrow ab\left(a-b\right)\left(a+b\right)⋮3\)
Mà a,b có vai trò tương đương nên \(ab\left(a-b\right)\left(a+b\right)⋮3,\forall a,b\left(2\right)\)
\(\left(1\right)\left(2\right)\Leftrightarrowđpcm\)
Ta có : a3b -ab3
=a3b -ab -ab3 +ab
=ab (a2 -1) -ab (b2 -1)
=ab (a-1)(a+1) -ab (b-1)(b+1)
Vì a (a-1)(a+1) là 3 số tự nhiên liên tiếp nên chia hết cho 6 .Tương tự b (b-1)(b+1) cũng chia hết cho 6
=> a3b -ab3 chia hết cho 6 (đpcm )
a: 12h: 0 độ
10h: 60 đọ
6h: 180 độ
5h: 150 độ
b:
a: góc nhọn: góc yMz; góc tMz
b: góc vuông: góc yMt, góc xMt
c: góc tù: góc xMz
d: góc bẹt: góc xMy
Bài 4:
\(a,\Rightarrow5⋮x\Rightarrow x\inƯ\left(5\right)=\left\{1;5\right\}\\ b,\Rightarrow x-2+7⋮x-2\\ \Rightarrow x-2\inƯ\left(7\right)=\left\{1;7\right\}\\ \Rightarrow x\in\left\{3;9\right\}\\ c,\Rightarrow3\left(x+1\right)+4⋮x+1\\ \Rightarrow x+1\inƯ\left(4\right)=\left\{1;2;4\right\}\\ \Rightarrow x\in\left\{0;1;3\right\}\\ d,\Rightarrow10x+6⋮2x-1\\ \Rightarrow5\left(2x-1\right)+11⋮2x-1\\ \Rightarrow2x-1\inƯ\left(11\right)=\left\{1;11\right\}\\ \Rightarrow x\in\left\{1;6\right\}\\ e,\Rightarrow x\left(x+3\right)+11⋮x+3\\ \Rightarrow x+3\inƯ\left(11\right)=\left\{1;11\right\}\\ \Rightarrow x=8\left(x\in N\right)\\ f,\Rightarrow x\left(x+3\right)+2\left(x+3\right)+5⋮x+3\\ \Rightarrow x+3\inƯ\left(5\right)=\left\{1;5\right\}\\ \Rightarrow x=2\left(x\in N\right)\)
3:
a: =8/24+9/24-14/24=3/24=1/8
b: =-12/56+35/56-28/56=-5/56
c: =9/36-24/36-22/36=-37/36
d: \(=\dfrac{6}{24}+\dfrac{10}{24}-\dfrac{21}{24}-\dfrac{1}{13}=\dfrac{-5}{24}-\dfrac{1}{13}=\dfrac{-89}{24\cdot13}\)
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{\dfrac{x\left(x+1\right)}{2}}=\dfrac{2019}{2021}\)
=>\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2019}{2021}\)
=>\(2\left(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2019}{2021}\)
=>\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2019}{4042}\)
=>\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2019}{4042}\)
=>\(\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{2019}{4042}=\dfrac{2021-2019}{4042}=\dfrac{2}{4042}=\dfrac{1}{2021}\)
=>x+1=2021
=>x=2020
\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2019}{2021}\)
\(\Leftrightarrow\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2019}{2021}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2019}{2021}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2019}{2021}\)
\(\Leftrightarrow1-\dfrac{2}{x+1}=\dfrac{2019}{2021}\)
\(\Leftrightarrow\dfrac{2}{x+1}=1-\dfrac{2019}{2021}\)
\(\Leftrightarrow\dfrac{2}{x+1}=\dfrac{2}{2021}\)
\(\Rightarrow x+1=2021\)
\(\Rightarrow x=2020\)