(2x-9)3=27 32 :(3x-2)=23 6 chia hết (x-1)
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a: \(\left(15-x\right)+\left(x-12\right)=7-\left(x-5\right)\)
=>7-x+5=15-x+x-12
=>12-x=3
hay x=9
b: \(\Leftrightarrow x-\left\{57-\left[42-23-x\right]\right\}=13-\left\{47+25-32+x\right\}\)
\(\Leftrightarrow x-\left\{57-19+x\right\}=13-\left\{40+x\right\}\)
=>x-38-x=13-40-x
=>-27-x=-38
=>x+27=38
hay x=11
e: \(x^2+3x+9⋮x+3\)
\(\Leftrightarrow x\left(x+3\right)+9⋮x+3\)
\(\Leftrightarrow x+3\in\left\{1;-1;9;-9;3;-3\right\}\)
hay \(x\in\left\{-2;-4;6;-12;0;-6\right\}\)
\(1\)) \(5-\left(10-x\right)=7\)
\(10-x=5-7\)
\(10-x=-2\)
\(x=10-\left(-2\right)\)
\(x=12\)
\(2\)) \(-32-\left(x-5\right)=0\)
\(x-5=-32-0\)
\(x-5=-32\)
\(x=-32+5\)
\(x=-27\)
x + 4 chia hết cho x
4 chia hết cho x
x thuộc U(4) = {-4;-2;-1;1;2;4}
3x+ 7 chia hết cho x
7 chia hết cho x
x thuộc U(7) = {-7;-1;1;7}
8 + 6 chia hết cho x + 1
14 chia hết cho x + 1
x + 1 thuộc U(14) = {-14;-7;-2;-1;1;2;7;14}
Vậy x thuộc {-15 ; -8 ; -3 ; -2 ; 0 ; 1 ; 6 ; 13}
x + 4 chia hết cho x
4 chia hết cho x
x thuộc U(4) = {-4;-2;-1;1;2;4}
3x+ 7 chia hết cho x
7 chia hết cho x
x thuộc U(7) = {-7;-1;1;7}
8 + 6 chia hết cho x + 1
14 chia hết cho x + 1
x + 1 thuộc U(14) = {-14;-7;-2;-1;1;2;7;14}
Vậy x thuộc {-15 ; -8 ; -3 ; -2 ; 0 ; 1 ; 6 ; 13}
a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)
(\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)
- \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)
\(x\) = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))
\(x=\) - \(\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)
3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)
3\(x\) - 3,7 = - \(\dfrac{19}{2}\)
3\(x\) = - \(\dfrac{19}{2}\) + 3,7
3\(x\) = - \(\dfrac{29}{5}\)
\(x\) = - \(\dfrac{29}{5}\) : 3
\(x\) = - \(\dfrac{29}{15}\)
Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\)
a: \(\left(2x-9\right)^3=27\)
=>2x-9=3
hay x=6
b: \(32:\left(3x-2\right)=2^3\)
=>3x-2=32:8=4
=>3x=6
hay x=2
c: \(6⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
hay \(x\in\left\{2;0;3;-1;4;-2;7;-5\right\}\)