Question

sqrt1/19+3x=0.25 решить уравнение

Answer 1

$sqrt{frac{1}{19}}+3x=0,25\sqrtfrac{1}{19}}=0,25-3x|^2 \frac{1}{19}=(0,25-3x)^2|*19\1=19(0,25-3x)^2\1=19(0,0625-1,5x+9x^2)\1=1,1875-28,5x+171x^2\0=-1+1,1875-28,5x+171x^2\0=171x^2-28,5x+0,1875|:(0,1875)\0=912x^2-152x+1\912x^2-152x+1=0\D=(-152)^2-4*912*1=23104--648=19456\sqrt{D}=sqrt{19456}=sqrt{1024*19}=sqrt{1024}*sqrt{19}=32sqrt{19}\x_{1,2}=frac{152+-32sqrt{19}}{1824}=frac{8(19+-4sqrt{19})}{1824}=frac{19+-4sqrt{19}}{228}$