Помогите пожалуйста решить
6.003;6.004;6.005;6.008

6.003) Делаем эту замену и получаем:z + 3/z + 4 = 0z^2 + 4z + 3 = 0(z + 1)(z + 3) = 01) z = (x^2+x-5)/x = -1x^2 + x - 5 = -xx^2 + 2x - 5 = 0D = 4 + 4*5 = 24 = (2√6)^2x1 = (-2 - 2√6)/2 = -1 - √6; x2 = -1 + √62) z = (x^2+x-5)/x = -3x^2 + x - 5 = -3xx^2 + 4x - 5 = 0(x + 5)(x - 1) = 0x3 = -5; x4 = 16.004) Умножаем все на 2Замена 2x^4 - 7 = zz - 21 = 100/zz^2 - 21z - 100 = 0(z -25)(z + 4) = 01) z = 2x^4 - 7 = -42x^4 = 3; x^4 = 1,52) z = 2x^4 - 7 = 252x^4 = 32x3 = -2; x4 = 26.005) Замена x^2 + 2x = z1/z - 1/(z+1) = 1/12Умножаем на 12z(z+1)12(z + 1) - 12z = z(z + 1)12z + 12 - 12z = z^2 + zz^2 + z - 12 = 0(z + 4)(z - 3) = 01) z = x^2 + 2x = -4x^2 + 2x + 4 = 0Решений нет2) z = x^2 + 2x = 3x^2 + 2x - 3 = 0(x + 3)(x - 1) = 0x1 = -3; x2 = 16.008. Умножаем все на (x-1)(x+1)(x-2)(x+2) = (x^2-1)(x^2-4)(x^2-4)*[ (x-3)(x+1) + (x+3)(x-1) ] = (x^2-1)*[ (x+6)(x-2) + (x-6)(x+2) ](x^2-4)*(x^2-3x+x-3+x^2+3x-x-3) = (x^2-1)*(x^2+6x-2x-12+x^2-6x+2x-12)(x^2-4)(2x^2-6) = (x^2-1)(2x^2-24)2x^4 - 8x^2 - 6x^2 + 24 = 2x^4 - 2x^2 - 24x^2 + 24-14x^2 = -26x^212x^2 = 0x = 0