Question

1) (x^2+x)\4 - (3-7x)\20 = 0.3 2) x^4-4x^2-5=0 3) 1\x-2 + 1\x+2 = 3\8 \= деление ^(цифра) = степень

Answer 1

1)
$\frac{ {x}^{2} + x }{4} - \frac{3 - 7x}{20} = 0.3 \\ \frac{5( {x}^{2} + x) }{20} - \frac{3 - 7x}{20} = \frac{6}{20} \\ \frac{5 {x}^{2} + 5x - 3 + 7x - 6 }{20} = 0 \\ \frac{5 {x}^{2} + 12x - 9 }{20} = 0 \\ 5 {x}^{2} + 12x - 9 = 0 \\ d = {b}^{2} - 4ac = 144 - 4 \times 5 \times ( - 9) = 324 \\ x1 = \frac{ - 12 + 18}{2 \times 5} = \frac{6}{10} = 0.6 \\ x2 = \frac{ - 12 - 18}{2 \times 5} = \frac{ - 30}{10} = - 3$
Ответ: -3; 0.6.

2)
${x}^{4} - 4 {x}^{2} - 5 = 0$
Пусть x^2 = t, t > 0
${t}^{2} - 4t - 5 = 0 \\ d = {b}^{2} - 4ac = 16 - 4 \times 1 \times ( - 5) = 36 \\ t1 = \frac{4 + 6}{2} = \frac{10}{2} = 5 \\ t2 = \frac{4 - 6}{2} = \frac{ - 2}{2} = - 1$
корень t2 не удовл.

Вернемся к замене:
${x}^{2} = 5 \\ x = + - \sqrt{5}$
Ответ: +- sqrt(5).

3)
$\frac{1}{x - 2} + \frac{1}{x + 2} = \frac{3}{8} \\ \frac{1}{x - 2} + \frac{1}{x + 2} - \frac{3}{8} = 0 \\ \frac{8(x + 2) + 8(x - 2) - 3(x - 2)(x + 2)}{8(x - 2)(x + 2)} = 0 \\ \frac{8x + 16 + 8x - 16 - 3( {x}^{2} - 4) }{8(x - 2)(x + 2)} = 0 \\ \frac{16x - 3 {x}^{2} + 12}{8(x - 2)(x + 2)} = 0$
ОДЗ: x не равно +-2
$- 3 {x}^{2} + 16x + 12 = 0 \\ 3 {x}^{2} - 16x - 12 = 0 \\ d = {b}^{2} - 4ac = 256 - 4 \times 3 \times ( - 12) = 400 \\ x1 = \frac{16 + 20}{2 \times 3} = \frac{36}{6} = 6 \\ x2 = \frac{16 - 20}{2 \times 3} = \frac{ - 4}{6} = - \frac{2}{3}$
Ответ: -2/3; 6.

Answer 2

1)
(х^2+х)/4-(3-7х)/20=0.3
5(х^2+х)-(3-7х)=0.3*20
5х^2+5х-3+7х=6
5х^2+12х-9=0
Д=144+4*5*9=144+180=324
х1=(-12+18)/10=6/10=0.6
х2=(-12-18)/10=-30/10=-3
2)
х^4-4х^2-5=0
х^2=t
t^2-4t-5=0
Д=16+4*5=16+20=36
t1=(4+6)/2=10/2=5
t2=(4-6)/2=-1 не подходит
х^2=5
х=±√5
3)
1/(х-2) + 1/(х+2)= 3/8
8(х+2)+8(х-2)=3(х^2-4)
16х=3х^2-12
3х^2-16х-12=0
Д=256+4*3*12=256+144=400
х1=(16+20)/6=6
х2=(16-20)/6=-4/6=-2/3
ОДЗ х≠2, х≠-2, корни подходят