Question

решить эти шесть уравнений и если можно с объяснением большое!

Answer 1

1а) x = 4/7

$\frac{x(2-x)}{2} + \frac{x(3+2x)}{4} = 1\\\\2x(2-x) + x(3+2x) = 4\\\\4x - 2x^2 + 3x + 2x^2 = 4\\\\x = \frac{4}{7}$

1б) x1 = –5, x2 = 5

$17 - 2x + \frac{x(3x+4)}{2}=54\frac{1}{2}\\\\34 - 4x + x(3x+4) = 109\\\\34 - 4x + 3x^2 + 4x = 109\\\\3x^2=75\\\\x^2 = 25\\\\x_{1,2} = \pm 5$

2а) x1 = –3, x2 = 4

$(x^2 - x + 1)(x^2 - x - 7) = 65\\\\\\y = x^2 - x + 1\\\\y(y - 8) = 65\\\\y^2 - 8y - 65 = 0\\\\D = (-8)^2 + 4\cdot65 = 324\\\\y_{1,2} = \frac{8\pm\sqrt{324}}{2}\\\\y_1=-5, \; y_2=13\\\\\\x^2 - x + 1 = -5\\\\x^2 - x + 6 = 0\\\\D = (-1)^2 - 4\cdot6 = -23 < 0\\\\\\x^2 - x + 1 = 13\\\\x^2 - x - 12 = 0\\\\D = (-1)^2 + 4\cdot12 = 49\\\\x_{1,2}=\frac{1\pm\sqrt{49}}{2}\\\\x_1=-3, \; x_2=4$

2б) x1 = –√2, x2 = √2, x3 = –4, x4 = 4

$(x^2 - 7)^2 - 4(x^2 - 7) - 45 = 0\\\\\\y = x^2 - 7\\\\y^2 - 4y - 45 = 0\\\\D = (-4)^2 + 4\cdot45 = 196\\\\y_{1,2} = \frac{4\pm\sqrt{196}}{2}\\\\y_1=-5, \; y_2 = 9\\\\\\x^2 - 7 = -5\\\\x^2 = 2\\\\x_{1,2} = \pm\sqrt{2}\\\\\\x^2 - 7 = 9\\\\x^2 = 16\\\\x_{3,4}=\pm4$

3a) x1 = –2, x2 = 2

$x^4 + 7x^2 - 44 = 0\\\\\\y = x^2\\\\y^2 + 7y - 44 = 0\\\\D=7^2+4\cdot44=225\\\\y_{1,2}=\frac{-7\pm\sqrt{225}}{2}\\\\y_1 = -11, \; y_2 = 4\\\\\\x^2 = -11 < 0 \\\\\\x^2=4\\\\x_{1,2}=\pm2$

3б) нет решений

$x^4+9x^2+8=0\\\\\\y=x^2\\\\y^2+9y+8=0\\\\D=9^2-4\cdot8=49\\\\y_{1,2}=\frac{-9\pm\sqrt{49}}{2}\\\\y_1=-8, \; y_2 = -1\\\\\\x^2 = -8\\\\\\x^2 = -1$