1the ki co may giay

What?!?!?

\(A=1+5^2+5^3+...+5^{2015}+5^{2016}\)

\(5A=5+5^3+5^4+...+5^{2016}+5^{2017}\)

\(4A=\left(5+5^3+5^4+...+5^{2016}+5^{2017}\right)-\left(1+5^2+5^3+...+5^{2015}+5^{2016}\right)\)

\(=5+5^{2017}-\left(1+5^2\right)\)

\(=4+5^{2017}-5^2\)

\(A=\frac{4+5^{2017}-5^2}{4}\)

Ta có : 5A = 5 + 5^3 + 5^4 + ... + 5^2016 + 5^2017

  =>    5A - A = ( 5 + 5^3 + 5^4 + ... + 5^2016 + 5^2017 ) - ( 1 + 5^2 + 5^3 + ... + 5^2015 + 5^2016 )

  =>         4A = 4 + 5^2 + 5^2017

  =>           A = ( 4 + 5^2 + 5^2017 )/4

1a,
(x/2)*3+x*4=231
=>(3/2+4)x=231
=>(11/2)x=231
=>x=231/(11/2)
=>x=231(*2/11)
=>x=42
Vậy x=42

&&&&

1a,
(x/2)*3+x*4=231
=>(3/2+4)x=231
=>(11/2)x=231
=>x=231/(11/2)
=>x=231(*2/11)
=>x=42
Vậy x=42

^HT^
^^^^^^

\(f,=\left(5^2+3\right):7=28:7=4\\ g,=7^2-9+8\cdot25=49-9+200=240\\ h,=600+72+18=690\\ i,=5^2+5-20=10\\ j,=45-28+83=100\)

\(2A=\frac{4}{1.5}+\frac{6}{5.11}+\frac{8}{11.19}+\frac{10}{19.29}+\frac{12}{29.41}\)

\(=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{19}+...+\frac{1}{29}-\frac{1}{41}=1-\frac{1}{41}=\frac{40}{41}\)

\(\Rightarrow A=\frac{20}{21}\)

\(3B=\frac{3}{1.4}+\frac{6}{4.10}+\frac{9}{10.19}+\frac{12}{19.31}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{10}+\frac{1}{10}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}\)

\(=1-\frac{1}{31}=\frac{30}{31}\)

\(\Rightarrow B=\frac{10}{31}=\frac{20}{62}<\frac{20}{41}\)

Do đó $A>B$

Ta có: \(A=\dfrac{2}{1.5}+\dfrac{3}{5.11}+\dfrac{4}{11.19}+\dfrac{5}{19.29}+\dfrac{6}{29.41}\)

\(2A=1-\dfrac{1}{5}+\dfrac{1}{5}+...+\dfrac{1}{29}-\dfrac{1}{41}\)

\(2A=1-\dfrac{1}{41}=\dfrac{40}{41}\)

\(A=\dfrac{20}{41}\)

Lại có: \(B=\dfrac{1}{1.4}+\dfrac{2}{4.10}+\dfrac{3}{10.19}+\dfrac{4}{19.31}\)

\(3B=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}\)

\(3B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+...+\dfrac{1}{19}-\dfrac{1}{31}\)

\(3B=1-\dfrac{1}{31}=\dfrac{30}{31}\)

\(B=\dfrac{10}{31}\)

Vì \(\dfrac{20}{41}>\dfrac{10}{31}\) nên...

đk : \(n\ne-\dfrac{1}{3}\)

gọi d là ƯCLN(18n+3,21n+7)

ta có 18n+3chia hết cho d

          21n+7 chia hết cho d

⇔21n+7-18n-3 chia hết cho d

⇔126n+42-126n-21 chia hết cho d 

21 chia hết cho d

⇒d∈Ư(21)=1;3;7;21

n ≠ 3k-1;3k-3;3k-7;3k-21

\(\frac{n-3}{n+2}\inℤ\Leftrightarrow n-3⋮n+2\)

=> n + 2 - 5 ⋮ n + 2

     n + 2 ⋮ n + 2

=> 5 ⋮ n + 2

=> n + 2 thuộc {-1; 5; 1; -5}

=> n thuộc {-3; 3; -1; -7}

vậy_

Bài giải : 

n−3n+2 ∈ Z ⇔n−3 ⋮ n+2

=> n + 2 - 5 ⋮ n + 2

     n + 2 ⋮ n + 2

=> 5 ⋮ n + 2

=> n + 2 € {-1; 5; 1; -5}

=> n € {-3; 3; -1; -7}

Vậy n € { -3 ; 3 ; -1 ; -7 }

\(\dfrac{x}{1.4}\)+\(\dfrac{x}{4.7}+\dfrac{x}{7.10}+\dfrac{x}{10.13}+\dfrac{x}{13.16}=\dfrac{5}{2}\) \(x.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right)=\dfrac{5}{2}\) \(x.\left(\dfrac{1}{1}-\dfrac{1}{16}\right)=\dfrac{5}{2}\) \(x.\dfrac{15}{16}=\dfrac{5}{2}\) \(x=\dfrac{5}{2}:\dfrac{15}{16}\) \(x=\dfrac{80}{30}=\dfrac{8}{3}\) DAY LA BAI LAM CUA MK NHO TICK CHO MK NHA CAM ON BAN TRUOC

Nếu đề là tìm n để phím chia hết thì làm như sau
 n^2 +3n -7 : n-3
n(n+3)-7: n-3
 vì n(n+3) chia hết cho n+3 nên để n^2 +3n -7 chia hết cho n+3 thì -7 chia hết cho n+3
=> n+3 thuộc Ư(7)={1,7,-1,-7}
n+3=1 => n= -2
n+3=7 => n= 4
n+3 = -1 => n=-4
n+3=7 => n =-10
 

b, n^2 +5 : n+1 
n^2 -1+6 : n+1
(n-1)(n+1) + 6: n+1         ( n^2 -1 =(n+1)(n-1) là dùng hằng đẳng thức lớp 8 sẽ học)
vì (n-1)(n+1) chia hết cho n+1 nên để n^2 +5 chia hết n+1 thì 6 phải chia hết cho n+1
=> n+1 thuộc Ư(6)={1,2,3,6,-1,-2,-3,-6}
n+1 =1 =>n=0
n+1=2=>n=1
n+1=3=>n=2
n+1=6=>n=5
n+1=-1=>n=-2
n+1=-2=>n=-3
n+1=-3=>n=-4
n+1=-6=>n=-7