Question

найти корни уравнения (4x-7)(9x+1)=(9x+1)²​

Answer 1

(4x-7)(9x+1)=(9x+1)²​

(4x-7)(9x+1)-(9x+1)² = 0

​(9x+1)( (4x-7)-(9x+1)) = 0

(9x + 1)(-5x - 8) = 0

x₁ = -1/9

x₂ = -8/5

Answer 2

Ответ:

$x_{1} =-1,6, x_{2} =-\frac{1}{9}$

Объяснение:

$36x^{2} +4x-7-63x=81x^{2} +18x+1$

$45x^{2}+77x+8=0$

$D=77^{2} -4*45*8=5929-1440=4489$

$\sqrt{D} =67$

$x_{1} =\frac{-77-67}{2*45}=-1,6$

$x_{2} =\frac{-77+67}{90}=-\frac{10}{90}=-\frac{1}{9}$

$(x+1,6)(x+\frac{1}{9} )=0$