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Question

Алгебра

Dima

4 год назад

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ОТВЕТЫ

Milana

Nov 30, 2020

$\frac{x-7}{y-5} +\frac{2}{x+5} =\frac{20}{25-x^{2} } \\x\neq 5\\x\neq -5\\\frac{x-7}{y-5} +\frac{2}{x+5} -\frac{20}{25-x^{2} }=0\\\frac{x-7}{y-5} +\frac{2}{x+5} -\frac{20}{-(5-x)(5+x)} }=0\\\frac{x-7}{y-5} +\frac{2}{x+5} +\frac{20}{(5-x)(5+x)} }=0\\\frac{(x-7)(x+5)+2(x-5)+20}{(x-5)(x+5)} =0\\\frac{x^2-7x+5x-35+2x-10+20}{(x-5)(x+5)} =0\\\frac{x^2-25}{(x-5)(x+5)}=0\\\frac{(x-5)(x+5)}{(x-5)(x+5)} =0\\1=0\\$

x∈∅

$\frac{x-4}{x-2} +\frac{x+6}{x+2} =\frac{8}{4-x^2} \\x\neq 2\\x\neq -2\\\frac{x-4}{x-2} +\frac{x+6}{x+2}+\frac{8}{(x-2)(2+x)} =0\\\frac{(x-4)(2+x)+(2+x)(x+6)+8}{(x-2)(2+x)} =0\\\frac{2x-8+x^2-4x+2x+12+x^2+6x+8}{(x-2)(2+x)} =0\\\frac{2x^2+12+6x}{(x-2)(2+x)} =0\\2x^2+12+6x=0 |:2\\x^2+3x+6=0\\D=3^2-4*1*6=9-24=-15\\$

x∈R

$\frac{x+1}{x-2} +\frac{x-8}{x+3} =\frac{15}{(x-2)(x+3)} \\x\neq 2\\x\neq -3\\\frac{(x+3)(x+1)+(x-2)(x-8)-15}{(x-2)(x+3)} =0\\\frac{2x^2-6x+4}{(x-2)(x+3)} =0\\\frac{2(x-1)(x-2)}{(x-2)(x+3)} =0\\\frac{2(x-1)}{x+3} =0\\2(x-1)=0 |:2\\x-1=0\\x=1$

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