Question

плз,
а) х^4 – 50х^2 + 49 = 0

б) х^4 – 5х^2 - 36 = 0

в) 4х^4 – 13х^2 + 3 = 0

г) 3х^4 + 8х^2 - 3 = 0

Answer 1

(см. объяснение)

Объяснение:

Решим уравнения методом разложения на множители:

1-ое уравнение:

${x}^{4} - 50 {x}^{2} + 49 = 0 \\ {x}^{4} - 49 {x}^{2} - {x}^{2} + 49 = 0 \\ {x}^{2}( {x}^{2} - 49) - ( {x}^{2} - 49) = 0 \\ ( {x}^{2} - 49)( {x}^{2} - 1) = 0 \\ (x + 7)(x - 7)(x + 1)(x - 1) = 0 \\ x = - 7 \\ x = 7 \\ x = - 1 \\ x = 1$

2-ое уравнение:

${x}^{4} - 5 {x}^{2} - 36 = 0 \\ {x}^{4} + 4 {x}^{2} - 9 {x}^{2} - 36 = 0 \\ {x}^{2} ( {x}^{2} + 4) - 9( {x}^{2} + 4) = 0 \\ ( {x}^{2} + 4)( {x}^{2} - 9) = 0 \\ ( {x}^{2} + 4)(x + 3)(x - 3) = 0 \\ x = - 3 \\ x = 3$

3-е уравнение:

$4 {x}^{4} - 13 {x}^{2} + 3 = 0 \\ 4 {x}^{4} - 12 {x}^{2} - {x}^{2} + 3 = 0 \\ 4 {x}^{2} ( {x}^{2} - 3) - ( {x}^{2} - 3) = 0 \\ ( {x}^{2} - 3)(4 {x}^{2} - 1) = 0 \\ (x + \sqrt{3} )(x - \sqrt{3} )(2x + 1)(2x - 1) = 0 \\ x = - \sqrt{3} \\ x = \sqrt{3} \\ x - \frac{1}{2} \\ x = \frac{1}{2}$

4-ое уравнение:

$3 {x}^{4} + 8 {x}^{2} - 3 = 0 \\ 3 {x}^{4} + 9 {x}^{2} - {x}^{2} - 3 = 0 \\ 3 {x}^{2} ( {x}^{2} + 3) - ( {x}^{2} + 3) = 0 \\ ( {x}^{2} + 3)(3 {x}^{2} - 1) = 0 \\ ( {x}^{2} + 3)( \sqrt{3} x - 1)( \sqrt{3} x + 1) = 0 \\ x = \frac{ \sqrt{3} }{3} \\ x = - \frac{ \sqrt{3} }{3}$