Question

ПОМОГИТЕ ПОЖАЛУЙСТА

Answer 1

а)
$\frac{ 4{m}^{2} - 5 }{m - 3} = \frac{2m + 1}{m - 3} \\ \frac{ 4{m}^{2} - 5 }{m - 3} - \frac{2m + 1}{m - 3} = 0 \\ \frac{4 {m}^{2} - 5 - 2m - 1}{m - 3} = 0 \\ \frac{4 {m}^{2} - 2m - 6}{m - 3} = 0 \\ \frac{2 {m}^{2} - m - 3}{m - 3} = 0 \\ 2 {m}^{2} - m - 3 = 0$
m - 3 не равно 0
m не равно 3
$d = {b}^{2} - 4ac = {( - 1)}^{2} - 4 \times 2 \times ( - 3) = 25 \\ m1 = \frac{1 + 5}{2 \times 2} = \frac{6}{4} = \frac{3}{2} \\ m2 = \frac{1 - 5}{2 \times 2} = - \frac{4}{4} = - 1$
Ответ: -1; 3/2.

б)
$\frac{3x + 20}{x - 4} = \frac{16 - 3 {x}^{2} }{4 - x} \\ \frac{3x + 20}{x - 4} = \frac{16 - 3 {x}^{2} }{ - (x - 4)} \\ \frac{3x + 20}{x - 4} = - \frac{1 6 - 3 {x}^{2} }{x - 4} \\ \frac{3x + 20}{x - 4} + \frac{16 - 3 {x}^{2} }{x - 4} = 0 \\ \frac{3x + 20 + 16 - 3 {x}^{2} }{x - 4} = 0 \\ \frac{ - 3 {x}^{2} + 3x + 36 }{x - 4} = 0 \\ \frac{ {x}^{2} - x - 12 }{x - 4} = 0 \\ {x}^{2} - x - 12 = 0$
x - 4 равно 0
х не равно 4
$d = {b}^{2} - 4ac = {( - 1)}^{2} - 4 \times 1 \times ( - 12) = 49 \\ x1 = \frac{1 + 7}{2} = 4 \\ x2 = \frac{1 - 7}{2} = - 3$
Корень x = 4 не удовл.
Ответ: -3.

в)
$\frac{ {k}^{2} + k }{4k + 32} = \frac{6 - k}{k + 8} \\ \frac{ {k}^{2} + k}{4(k + 8)} = \frac{6 - k}{k + 8} \\ \frac{ {k}^{2} + k}{4(k + 8)} - \frac{6 - k}{k + 8} = 0 \\ \frac{ {k}^{2} + k}{4(k + 8)} - \frac{4(6 - k)}{4(k + 8)} = 0 \\ \frac{ {k}^{2} + k - 4(6 - k)}{4(k + 8)} = 0 \\ \frac{ {k}^{2} + k - 24 + 4k}{4(k + 8)} = 0 \\ \frac{ {k}^{2} + 5k - 24}{4(k + 8)} = 0 \\ {k}^{2} + 5k - 24 = 0$
k + 8 не равно 0
k не равно -8
$d = {b}^{2} - 4ac = {5}^{2} - 4 \times ( - 24) = 121 \\ k1 = \frac{ - 5 + 11}{2 \times 1 } = \frac{6}{2} = 3 \\ k2 = \frac{ - 5 - 11}{2} = - 8$
Корень k = -8 не удовл.
Отвеь: 3.

Answer 2

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