Question

(9-4x)^2=5(4x+1) зробити рівняння​

Answer 1

$(9-4x)^{2}=5(4x+1)\\81-72x+16x^{2} =20x+5\\81-72x+16x^{2} -20x-5=0\\76-92x+16x^{2} =0\\19-23x+4x^{2} =0\\4x^{2} -23x+19=0\\4x^{2} -4x-19x+19=0\\4x*(x-1)-19(x-1)=0\\(x-1)*(4x-19)=0\\x-1=0, 4x-19=0\\x_{1}=1, x_{2}=\frac{19}{4}$