(2x+1)^-2 - 3(2x+1) ^-1 = 4 - Все ответы

$(2x + 1)^{-2} + 3*(2x +1)^{-1} = 4 \\ \\ \frac{1}{(2x+1)^2} + 3*\frac{1}{2x+1} =4 \\ 2x + 1 \neq 0 ; 2x = - 1 ; x \neq - 0,5 \\ \\ \frac{1}{(2x+1)^2} + \frac{3}{2x+1} =4 | * (2x + 1)^2 \\ \\ 1 + 3(2x+1) = 4*(2x+1)^2 \\ 1 + 6x + 3 = 4(4x^2 + 4x + 1) \\ 6x + 4 = 16x^2 + 16x + 4 \\ 16x^2+16x +4 - 6x - 4 = 0 \\ 16x^2 + 10x = 0 \\ 2x(8x + 5) = 0 \\ \\ 2x = 0 \\ x_{1} =0 \\ \\ 8x + 5 = 0 \\ 8x = - 5 \\ x = - \frac{5}{8} \\ x_{2} = - 0.625 \\ \\$

проверим:
$(2*0 + 1)^{-2} + 3*(2*0 +1)^{-1} = 1^{-2} + 3 * 1^{-1} = 1 + 3 = 4 \\ \\ (2*(-0.625) + 1)^{-2} + 3*(2*( - 0.625) +1)^{-1} = \\ = ( - 1.25 + 1)^{-2} +3*(- 1.25 + 1)^{-1} = \frac{1}{(-0.25)^2} + 3 * \frac{1}{ -0.25} = \\ = \frac{1}{0.0625} + 3 * ( - 4) = 16 - 12 = 4$

ответ: х₁ = 0 ; х₂ = - 0,625.