Question

МНОГО БАЛЛОВ! Решите уравнение 1/x+6+1/x+7-1/x+9-1/x+10=21/20​

Answer 1

$\displaystyle\bf\\\frac{1}{x+6} +\frac{1}{x+7} -\frac{1}{x+9} -\frac{1}{x+10} =\frac{21}{20} \\\\\\\Big(\frac{1}{x+6} -\frac{1}{x+10} \Big)+\Big(\frac{1}{x+7} -\frac{1}{x+9}\Big) =\frac{21}{20} \\\\\\\frac{x+10-x-6}{(x+6)(x+10)} +\frac{x+9-x-7}{(x+7)(x+9)} =\frac{21}{20}\\\\\\\frac{4}{x^{2} +16x+60}+\frac{2}{x^{2} +16x+63} =\frac{21}{20} \\\\\\x^{2} +16x+60=m \ \ \Rightarrow \ \ x^{2} +16x+63=m+3$

$\displaystyle\bf\\\frac{4}{m} +\frac{2}{m+3} -\frac{21}{20} =0\\\\\\\frac{80m+240+40m-21m^{2}-63m }{20m(m+3)} =0\\\\\\\frac{21m^{2}-57m -240}{m(m+3)}=0\\\\\\\left \{ {{21m^{2}-57m-240=0 } \atop {m\neq 0 \ ; \ m\neq -3}} \right. \\\\\\21m^{2} -57m-240=0\\\\D=(-57)^{2} -4\cdot 21\cdot(-240)=3249+20160=23409=153^{2}\\\\\\m_{1} =\frac{57+153}{42} =5\\\\\\m_{2} =\frac{57-153}{42}=-2\frac{2}{7}$

$\displaystyle\bf\\1)\\x^{2}+16x+60=5\\\\x^{2} +16x+55=0\\\\Teorema \ Vieta:\\\\x_{1} =-11\\\\x_{2} =-5\\\\2)\\x^{2} +16x+60=-2\frac{2}{7} \\\\x^{2} +16x+60+\frac{16}{7} =0\\\\7x^{2} +112x+420+16=0\\\\7x^{2} +112x+436=0\\\\D=112^{2} -4\cdot 7\cdot436=12544-12208=336=(4\sqrt{21} )^{2} \\\\\\x_{3} =\frac{-112-4\sqrt{21} }{14} =-\frac{56+2\sqrt{21} }{7} \\\\\\x_{4} =\frac{-112+4\sqrt{21} }{14}=\frac{2\sqrt{21}-56 }{7}$