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a) 6 . x - 15 = 15
6 . x = 15 + 15
6 . x = 30
x = 30 : 6
x = 5
b) 2 . x + 3 = 9
2 . x = 9 - 3
2 . x = 6
x = 6 : 2
x = 3
c) 17 - 6 : x = 15
6 : x = 17 - 15
6 : x = 2
x = 6 : 2
x = 3
d) x : 3 + 12 = 14
x : 3 = 14 - 12
x : 3 = 2
x = 2 . 3
x = 6
a) 6 . x -15 = 15
⇒ 6 .x = 15 + (-15)
⇒ 6 .x = 0
⇔ x = 0
Vậy x = 0.
x-34-[(15+x)-(23-x)]=15-21
x-34-[(15+x)-(23-x)]= -6
x-34-15-x-23+x = -6
(x-x+x) + (-34-15-23)= -6
x + -72 = -6
x = -6-(-72)
x =66
\(\dfrac{15}{19}\times\dfrac{38}{5}< x< \dfrac{67}{15}-\dfrac{56}{16}\\ \dfrac{5\times3\times19\times2}{19\times5}< x< \dfrac{67\times16-56\times15}{15\times16}\\ 3\times2< x< \dfrac{232}{240}\\ 6< x< \dfrac{232}{240}\) (*)
Mà : \(\dfrac{232}{240}< \dfrac{240}{240}=1\\ \)
Nên (*) là vô lí ( Do `6>232/240` )
Vậy \(x\in\varnothing\)
Bài 2:
a: \(\Leftrightarrow x-1\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
=>\(x\in\left\{2;0;3;-1;4;-2;7;-5\right\}\)
b: \(\Leftrightarrow x+3\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)
=>\(x\in\left\{-2;-4;0;-6;2;-8;12;-18\right\}\)
c: \(\Leftrightarrow x+3\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)
=>\(x\in\left\{-2;-4;-1;-5;0;-6;1;-7;3;-9;9;-15\right\}\)
d: =>x+1+15 chia hết cho x+1
=>\(x+1\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)
=>\(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)
15 . ( x - 15 ) = 15
x - 15 = 15 : 15
x - 15 = 1
x = 1 + 15
x = 16
15 . ( x - 15 ) = 15
x - 15 =15 : 15
x - 15 = 1
x = 1 + 15
x = 16
a) 78 - x = 90
x = 78 - 90
x = -12
b) 15 + 2x = 5¹⁰ : 5⁸
15 + 2x = 5²
15 + 2x = 25
2x = 25 - 15
2x = 10
x = 10 : 2
x = 5
c) 48 : x + 17 = -33
48 : x + 17 = -33 - 17
48 : x = -50
x = 48 : (-50)
x = -24/25
d) Do x ⋮ 15 và x ⋮ 20
⇒ x ∈ BC(15; 20)
Ta có:
15 = 3.5
20 = 2².5
⇒ BCNN(15; 20) = 2².3.5 = 60
⇒ x ∈ BC(15; 20) = {0; 60; 120; ...}
Mà 50 < x < 70
⇒ x = 60
\(x^{15}\) = \(x\)
\(x^{15}\) - \(x\) = 0
\(x\) \(\times\)( \(x^{14}\) - 1) = 0
\(\left[{}\begin{matrix}x=0\\x^{14}-1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x^{14}=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy \(x\) \(\in\) { -1; 0; 1}