x^2 +[1-3x]/(x +4)=16-[3x-1]/(x+ 4)[5]/(2x+3)+[3-2x]/(x+2)=10 помогите решить. Created by saniya200. algebra-ru

(x+2)=10 помогите решить...

saniya200 @saniya200

July 2022 1 9 Report

x^2 +[1-3x]/(x +4)=16-[3x-1]/(x+ 4)
[5]/(2x+3)+[3-2x]/(x+2)=10 помогите решить

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$x^2+ \frac{1-3x}{x+4}=16- \frac{3x-1}{x+4}\\\\ \frac{x^2(x+4)+(1-3x)}{x+4}= \frac{16(x+4)-(3x-1)}{x+4}\\\\x+4 \neq 0\\x \neq -4\\\\x^3+4x^2+1-3x=16x+64-3x+1\\x^3+4x^2-16x-64=0\\x^2(x+4)-16(x+4)=0\\(x+4)(x^2-16)=0\\(x+4)(x+4)(x-4)=0$
x+4=0 или x-4=0
x₁=-4 x₂=4
Ответ: -4;4

$\frac{5}{2x+3} + \frac{3-2x}{x+2} =10\\\\ \frac{5(x+2)+(3-2x)(3+2x)}{(2x+3)(x+2)}=10\\\\ \left \{ {{2x+3 \neq 0} \atop {x+2 \neq 0}} \right.=\ \textgreater \ \left \{ {{x \neq -1,5} \atop {x \neq -2}} \right.\\\\5x+10+9-4x^2=10(2x+3)(x+2)\\5x-4x^2+19=10(2x^2+7x+6)\\5x-4x^2+19=20x^2+70x+60\\24x^2+65x+41=0\\D=4225-3936=289=17^2\\x_1=(-65+17)/(2*24)=-48/48=-1\\x_2=(-65-17)/(2*24)=-82/48=-41/24=-1 \frac{17}{24}$

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