Answer:
a) 2x² - x + 3 = x(2x + 1) - 2x
<=> 2x² - x + 3 = 2x² + x - 2x
<=> 3 - x = - x
<=> 3 = 0
=> no solution
b) x/2 - x/3 - x/4 = 1/12
[tex] \frac{6x}{12} - \frac{4x}{12} - \frac{3x}{12} = \frac{1}{12} \\ \\ \frac{ - x}{12} = \frac{1}{12} \\ \\ = > - x = 1 \\ \\ \\ < = > x = 1[/tex]
=> the euqation has the solution x = 1
c) |x - 5| = 2|x|
=> x - 5 = 2x
or x - 5 = -2x
<=> x = -5
or 3x = 5
<=> x = -5
x = -5or x = 5/3
d) defined conditions: 2 - x 》 0
<=> x 《 2
we have:
[tex] \sqrt{x + 4} = 2 - x \\ < = > x + 4 = (2 - x) {}^{2} \\ < = > x + 4 = {x}^{2} - 4x + 4 \\ < = > {x}^{2} - 5x = 0 \\ < = > x(x - 5) = 0 \\ [/tex]
<=> x = 0 (because x 《 2)
e)
[tex] \sqrt{5} = \frac{1}{25 {}^{x} } \\ = > 5 = \frac{1}{25 {}^{2x} } \\ < = > 5.25 {}^{2x} = 1 \\ < = > 5.5 {}^{4x} = 1 \\ < = > 5 {}^{4x + 1} = 1 \\ = > 4x + 1 = 0[/tex]
<=> x = -1/4
