Решите уравнение (x+3)^2-16=(1-2x)^2 

[latex]( x+3)^{2}-16= (1-2x)^{2} [/latex] [latex] x^{2} +6x+9-16=1-4x+4 x^{2} [/latex] [latex] x^{2} +6x-7=4 x^{2} -4x+1[/latex] [latex] x^{2} -4 x^{2} +6x+4x-7-1=0[/latex] [latex]-3 x^{2} +10x-8=0[/latex] [latex]D= 10^{2}-4*(-3)*(-8)=100-96=4 [/latex] [latex] x_{1} = \frac{-10- \sqrt{4} }{2*(-3)}= \frac{-10-2}{-6}= \frac{-12}{-6}=2 [/latex] [latex] x_{2}= \frac{-10+2}{-6}= \frac{-8}{-6}=4/3[/latex] Ответ:(2;4/3)

x^2+6x+9-16=1-4x+4x^2 -3x^2+10x-8=0 k=5 дискриминант=25-24=1 x1,2=(5+-1)/6 x1=2 x2=4/3