Question

Решите пж (4x-5)^2=4x*(4x-7)

Answer 1

$(4x-5)^2=4x(4x-7) \ 16x^2-40x+25=16x^2-28x \ 12x=25 \ x= frac{25}{12} =2 frac{1}{12}$

Answer 2

16x²-40x+25=16x²-28x
-40x+25=-28x
-40x+28x=-25
40x-28x=25
12x=25
x=25/12=2,1/12