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6x . 6 = 2016
6x = 2016 : 6
6x = 336
=> x \(\in\varnothing\)
42x+3 : 4 = 256
42x+3 = 256 x 4
42x+3 = 1024
42x+3 = 45
2x + 3 = 5
2x = 5 - 3
2x = 2
x = 2 : 2
x = 1
[ x - 2 ]2 = 16
[ x - 2 ]2 = 42
x - 2 = 4
x = 4 + 2
x = 6
[ 2x - 1 ]3 = 27
[ 2x - 1 ]3 = 33
2x - 1 = 3
2x = 3 + 1
2x = 4
x = 4 : 2
x = 2
[ 2x - 1 ]100 = [ 2x - 1 ]100
=> x \(\in N\)
\(a,|2x-2019|=1\)
\(\Leftrightarrow\orbr{\begin{cases}2x-2019=1\\2x-2019=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=2020\\2x=2018\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1010\\x=1009\end{cases}}\)
Vậy ............
\(b,\left(2-x\right)^5=-32\)
\(\Leftrightarrow\left(2-x\right)^5=\left(-2\right)^5\)
\(\Leftrightarrow2-x=-2\)
\(\Leftrightarrow x=4\)
Vậy ..........
a; -2\(x\) - 3.(\(x-17\)) = 34 - 2.( - \(x\) + 25)
- 2\(x\) - 3\(x\) + 51 = 34 + 2\(x\) - 50
2\(x\) + 2\(x\) + 3\(x\) = - 34 + 50 + 51
7\(x\) = 67
\(x\) = 67 : 7
\(x\) = \(\dfrac{67}{7}\)
Vậy \(x\) = \(\dfrac{67}{7}\)
b; 17\(x\) + 3.(- 16\(x\) - 37) = 2\(x\) + 43 - 4\(x\)
17\(x\) - 48\(x\) - 111 = 2\(x\) - 4\(x\) + 43
- 31\(x\) - 2\(x\) + 4\(x\) = 111 + 43
- \(x\) x (31 + 2 - 4) = 154
- \(x\) x (33 - 4) = 154
- \(x\) x 29 = 154
- \(x\) = 154 : (-29)
\(x\) = - \(\dfrac{154}{29}\)
Vậy \(x=-\dfrac{154}{29}\)
\(\left(2x-1\right)^2-\dfrac{1}{4}=2\)
\(\left(2x-1\right)^2=2+\dfrac{1}{4}=\dfrac{9}{4}=\left(\dfrac{3}{2}\right)^2\)
2x - 1 = \(\dfrac{3}{2}\)
2x = \(\dfrac{3}{2}+1=\dfrac{5}{2}\)
x = \(\dfrac{5}{2}:2=\dfrac{5}{4}\)
(2x-1)^2 = 2,25
<=> 2x-1 = 1,5
<=> 2x = 2,5
<=> x = 1,25