Question

Решить уравнения, алгебра 8 класс. Решите , очень надо! Надо решить уравнения методом замены переменной ​

Answer 1

$1)\dfrac{x^{2} }{(3x+1)^{2} } -\dfrac{6x}{3x+1}+5=0\\\\\dfrac{x^{2}-6x\cdot(3x+1)+5\cdot(9x^{2} +6x+1) }{(3x+1)^{2} }=0\\\\\dfrac{x^{2}-18x^{2}-6x+45x^{2}+30x+5}{(3x+1)^{2} }=0\\\\\dfrac{28x^{2}+24x+5 }{(3x+1)^{2} }=0\\\\\left \{ {{28x^{2}+24x+5=0 } \atop {3x+1\neq0 }} \right.\\\\\left \{ {{28x^{2}+24x+5=0 } \atop {x\neq-\frac{1}{3} }} \right. \\\\28x^{2}+24x+5=0\\\\D=24^{2}-4\cdot 28\cdot 5=576-560=16=4^{2}\\\\x_{1}=\frac{-24-4}{56}=-\frac{28}{56} =-0,5$

$x_{2} =\frac{-24+4}{56} =-\frac{5}{14}$

$2)\dfrac{x-5}{x+3}+\dfrac{x+3}{x-5} =-2\dfrac{1}{2}\\\\\dfrac{x-5}{x+3}+\dfrac{x+3}{x-5} +\dfrac{5}{2}=0\\\\\dfrac{2\cdot(x-5)\cdot(x-5)+2\cdot (x+3)\cdot(x+3)+5\cdot(x+3)(x-5)}{2(x+3)(x-5)}=0 \\\\\dfrac{2x^{2}-20x+50+2x^{2}+12x+18+5x^{2}-10x-75}{2(x+3)(x-5)} =0\\\\\dfrac{9x^{2}-18x-7 }{2(x+3)(x-5)}=0 \\\\\left \{ {{9x^{2}-18x-7=0 } \atop {x\neq -3 \ ; \ x\neq 5}} \right. \\\\9x^{2}-18x-7=0\\\\D=(-18)^{2} -4\cdot9\cdot(-7)=324+252=576=24^{2}\\\\x_{1} =\frac{18-24}{18}=-\frac{1}{3}$

$x_{2}=\frac{18+24}{18} =2\frac{1}{3}$