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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
\(x^{15}\) = \(x\)
\(x^{15}\) - \(x\) = 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}