Question

рациональные уравнения 9 класс​

Answer 1

Ответ:

Объяснение:

$\displaystyle\\1)\frac{x+4}{x+1}-\frac{10}{x^2-1}=\frac{10}{3} ;ODZ:x\neq  -1;x\neq 1\\\\\\\frac{x+4}{x+1}-\frac{10}{(x-1)(x+1)}=\frac{10}{3}\\\\\\\frac{3(x+4)(x-1)-30-10(x^2-1)}{3(x+1)(x-1)}=0\\\\\\3(x^2+4x-x-4)-30-10x^2+10=0\\\\3x^2+9x-12-20-10x^2=0\\\\=-7x^2+9x-32=0\\\\D=81-4*7*32=81-896=-815<0$

Ответ: ∅

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$\displaystyle\\2)\frac{x}{x+4}}+\frac{5}{x-4} =\frac{32}{x^2-16} ;ODZ:x\neq  4;x\neq -4\\\\\\\frac{x(x-4)+5(x+4)-32}{x^2-16} =0\\\\\\x^2-4x+5x+20-32=0\\\\x^2+x-12=0\\\\D=1+4*12=49=7^2\\\\x_1=(-1-7)/2=-4;ne\ udowl.odz\\\\x_2=(-1+7)/2=3\\\\Otvet:x=3$

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$\displaystyle\\3)\frac{x-3}{x+2} +\frac{x-3}{2-x} =\frac{20}{x^2-4};ODZ:x\neq  -2;x\neq 2\\\\\\\frac{(x-3)(x-2)-(x-3)(x+2)-20}{x^2-4}=0\\\\\\x^2-3x-2x+6-x^2+3x-2x+6-20=0\\\\-4x-8=0\\\\x=-2;ne\ udowl.odz$

Ответ: ∅

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$\displaystyle\\4)\frac{x+5}{x-2}+\frac{x-4}{x+1}=\frac{21}{(x-2)(x+1)} ;ODZ:x\neq   2;x\neq -1\\\\\\(x+5)(x+1)+(x-4)(x-2)-21=0\\\\x^2+5x+x+5+x^2-4x-2x+8-21=0\\\\2x^2-8=0\\\\x^2=4\\\\x_1=2;ne\ udowl.odz\\\\x_2=-2\\\\Otvet:x=-2$