Question

кто решит тому 100 балов

Answer 1

$1)\displaystyle\\\frac{x^&2}{x-3} +\frac{x}{3-x} =\frac{6}{x-3} \\\\\frac{x^2-6}{x-3} +\frac{x}{3-x} =0\\\\\frac{x^2-6}{x-3} -\frac{x}{x+3} =0\\\\\frac{x^2-x-6}{x-3} =0|\times(x-3);x\neq3\\x^2-x-6=0\\D=(-1)^2-4\times(-6)=25=5^2\\\\x_{1,2}=\frac{1\pm5}{2} =\bigg[_{-2}^{3}$

х не может быть равен 3 => х=-2

$2)\displaystyle\\y=\frac{x^2+x-12}{x+4} ;y=0=>\\\\\frac{x^2+x-12}{x+4}=0|\times(x+4);x\neq-4\\x^2+x-12=0\\D=1^2-4\times(-12)=1+48=49=7^2\\\\x_{1,2}=\frac{-1\pm7}{2} =\bigg[^3_{-4}$

х не может быть равен -4 => х=3

$3)\displaystyle\\\\\frac{2x+1}{x} +\frac{4x}{2x+1} =5\\\\\frac{(2x+1)^2+4x^2}{x(2x-1)} =\frac{5x(2x+1)}{x(2x+1)} \\\\\frac{(2x+1)^2+4x^2-5x(2x+1)}{x(2x+1)} =0|\times x(2x+1);x\neq0\:\text{and}\:x\neq-0.5\\\\(2x+1)^2+4x^2-5x(2x+1)=0\\4x^2+4x+1+4x^2-10x^2-5x=0\\-2x^2-x+1=0\\2x^2+x-1=0\\D=1^2-4\times2\times({-1})=9=3^2\\\\x_{1,2}=\frac{-1\pm3}{4} =\bigg[^{-1}_{\dfrac{1}{2} }$