Question

очень быстро нужно пожалуйста за теоремою Виета со сменно t
или за дискриминантом

Answer 1

$\displaystyle\bf\\3)\\\\\frac{1}{(x-1)(x+4)} -\frac{1}{x(x+3)} =\frac{1}{3} \\\\\\\frac{x(x+3)-(x-1)(x+4)}{x(x+3)(x-1)(x+4)} =\frac{1}{3} \\\\\\\frac{x^{2}+3 x-x^{2} -3x+4}{x(x+3)(x-1)(x+4)} =\frac{1}{3} \\\\\\\frac{4}{x(x+3)(x-1)(x+4)} =\frac{1}{3} \\\\\\x(x+3)(x-1)(x+4)=12\\\\\\x+1,5=m\\\\(m-1,5)\cdot(m+1,5)\cdot(m-2,5)\cdot(m+2,5)=12\\\\(m^{2} -2,25)\cdot(m^{2}-6,25)=12\\\\m^{4}-8,5m^{2} +14,0625=12$

$\displaystyle\bf\\m^{4}-8,5m^{2} +2,0625=0 \ \ \ , \ \ m^{2} =a, \ a > 0\\\\a^{2} -8,5a+2,0625=0\\\\D=(-8,5)^{2} -4\cdot 2,0625=72,25-8,25=64=8^{2}\\\\\\a_{1} =\frac{8,5-8}{2} =0,25\\\\\\a_{2} =\frac{8,5+8}{2} =8,25\\\\1) \ m^{2} =0,25\\\\m_{1,2} =\pm \ \sqrt{0,25} =\pm \ 0,5\\\\x_{1} =-0,5-1,5=-2\\\\x_{2} =0,5-1,5=-1\\\\2)m^{2} =8,25=\frac{33}{4} \\\\m_{3,4} =\pm \ \frac{\sqrt{33} }{2}$

$\displaystyle\bf\\x_{3} =-\frac{\sqrt{33} }{2} -1,5=-\frac{\sqrt{33} +3}{2} \\\\\\x_{4} =\frac{\sqrt{33} }{2} -1,5=\frac{\sqrt{33} -3}{2} \\\\\\Otvet \ : \ -2 \ ; \ -1 \ ; \ -\frac{\sqrt{33} +3}{2} \ ; \ \frac{\sqrt{33} -3}{2}$

$\displaystyle\bf\\4) \ \Big(x+\frac{1}{x} \Big)^{2} =x^{2} +2\cdot \underbrace{x\cdot \frac{1}{x} }_{1}+\frac{1}{x^{2} } =x^{2} +\frac{1}{x^{2} } +2\\\\\\x^{2} +\frac{1}{x^{2} } =\Big(x+\frac{1}{x} \Big)^{2} -2\\\\\\x+\frac{1}{x} =m\\\\\\x^{2} +\frac{1}{x^{2} } =m^{2} -2\\\\\\3\cdot(m^{2} -2)+2m=10\\\\3m^{2} -6+2m-10=0\\\\3m^{2} +2m-16=0\\\\D=2^{2} -4\cdot 3\cdot(-16)=4+192=196=14^{2} \\\\\\m_{1} =\frac{-2-14}{6} =-\frac{16}{6} =-\frac{8}{3}= -2\frac{2}{3}$

$\displaystyle\bf\\m_{2} =\frac{-2+14}{6} =\frac{12}{6} =2\\\\\\1) \ m=-2\frac{2}{3} \\\\\\x+\frac{1}{x} =-\frac{8}{3} \\\\\\x+\frac{1}{x} +\frac{8}{3}=0\\\\\\3x^{2}+8x+3=0 \ , \ x\neq 0\\\\D=8^{2} -4\cdot 3\cdot 3=64-36=28=(2\sqrt{7} )^{2} \\\\\\x_{1} =\frac{-8-2\sqrt{7} }{6} =-\frac{4+\sqrt{7} }{3} \\\\\\x_{2} =\frac{-8+2\sqrt{7} }{6} =\frac{\sqrt{7} -4}{3} \\\\\\2) \ m=2\\\\\\x+\frac{1}{x} =2\\\\\\x+\frac{1}{x} -2=0\\\\\\x^{2} -2x+1=0\\\\(x-1)^{2} =0\\\\x-1=0$

$\displaystyle\bf\\x_{3} =1\\\\\\Otvet \ : \ -\frac{4+\sqrt{7} }{3} \ ; \ \frac{\sqrt{7} -4}{3} \ ; \ 1$