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CM
12 tháng 10 2018
Đáp án là A
Ta có: x + 3 = (x + 1) + 2
Vì (x + 3) ⋮ (x + 1), (x + 1) ⋮ (x + 1) ⇒ 2 ⋮ (x + 1)
Do đó, x + 1 = ±1 hoặc x + 1 = ±2
Nếu x + 1 = ±1 thì x = 0 hoặc x = -2
Nếu x + 1 = ±2 thì x = 1 hoặc x = -3
Vậy x ∈ {-3; -2; 0; 1}
NH
2
3 tháng 2 2016
a) (3x -2)^5 = -243
(3x -2)^5 = (-3)^5
3x-2=-3
3x=-1
x=-1/3
b) ( x-7)^( x+1) - ( x-7)^(x+11) = 0
( x-7)^( x+1) - ( x-7)^(x+1)*x^10 = 0
( x-7)^( x+1) (1-x^10) = 0
Do x phải là số nguyên tố nên x=7
HT
2
NN
2
LN
0
CT
3
MH
4 tháng 2 2016
-210 = (-1) + (-2) + (-3) + ... + (-x + 1) + (-x)
=> -210 = -(1 + 2 + 3 + ...+ x - 1 + x)
=> 210 = 1 + 2 + 3 + x - 1 + x
=> 210 = \(\frac{\left(x+1\right).x}{2}\)
=> 420 = (x + 1).x
=> 21 . 20 = (x + 1) . x
=> (20 + 1) . 20 = (x + 1) . x
=> x = 20
25 tháng 1 2017
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NM
2
ND
Nguyễn Đức Trí
VIP
28 tháng 6 2023
(x-1)(x-3)(x-4)>0
Trường hợp 1 :
x-1>0; x-3>0; x-4>0
Nên x>1; x>3; x>4
Vậy x>4 (hay x∈ Z/x ∈ { 5;6;7...})
Trường hợp 2 :
x-1>0; x-3<0; x-4<0
Nên x>1; x<3; x<4
Vậy 1<x<3 (hay x∈ Z/x ∈ { 2 })
NM
2
PP
29 tháng 6 2023
(x-1)(x-3)(x-4)>0
Trường hợp 1 :
x-1>0; x-3>0; x-4>0
Nên x>1; x>3; x>4
Vậy x>4 (hay x∈ Z/x ∈ { 5;6;7...})
Trường hợp 2 :
x-1>0; x-3<0; x-4<0
Nên x>1; x<3; x<4
Vậy 1<x<3 (hay x∈ Z/x ∈ { 2 })
\(\dfrac{1}{5.8}\) + \(\dfrac{1}{8.11}\) +...+ \(\dfrac{1}{x.\left(x+3\right)}\) = \(\dfrac{12}{255}\) (đk \(x\ne\) -3; 0)
\(\dfrac{1}{3}\).(\(\dfrac{3}{5.8}\) + \(\dfrac{3}{8.11}\) + ... + \(\dfrac{3}{x.\left(x+3\right)}\)) = \(\dfrac{12}{255}\)
\(\dfrac{1}{3}\).(\(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\)) = \(\dfrac{12}{255}\)
\(\dfrac{1}{3}\) .(\(\dfrac{1}{5}\) - \(\dfrac{1}{x+3}\)) = \(\dfrac{12}{255}\)
\(\dfrac{1}{5}\) - \(\dfrac{1}{x+3}\) = \(\dfrac{12}{255}\).3
\(\dfrac{1}{5}\) - \(\dfrac{1}{x+3}\) = \(\dfrac{12}{85}\)
\(\dfrac{1}{x+3}\) = \(\dfrac{1}{5}\) - \(\dfrac{12}{85}\)
\(\dfrac{1}{x+3}\) = \(\dfrac{1}{17}\)
\(x\) + 3 = 17
\(x\) = 17 - 3
\(x\) = 14
Vậy \(x\) = 14