GIẢI PT: (1/x^2-5x+6)+(!/x^2-7x+12)+(1/x^2-9x+20)+(1/x^2-11x+30)=1/8

1 bình luận về “GIẢI PT: (1/x^2-5x+6)+(!/x^2-7x+12)+(1/x^2-9x+20)+(1/x^2-11x+30)=1/8”

1. Giải đáp: –2; 10

   Lời giải và giải thích chi tiết: ĐKXĐ: $x\ne2;3;4;5;6$

   1/(x^2-5x+6)+1/(x^2-7x+12)+1/(x^2-9x+20)+1/(x^2-11x+30)=1/8$\\$<=>1/((x-2)(x-3))+1/((x-3)(x-4))+1/((x-4)(x-5))+1/((x-5)(x-6))=1/8$\\$<=>1/((x-3)(x-2))+1/((x-4)(x-3))+1/((x-5)(x-4))+1/((x-6)(x-5))=1/8$\\$<=>1/(x-3)-1/(x-2)+1/(x-4)-1/(x-3)+1/(x-5)-1/(x-4)+1/(x-6)-1/(x-5)=1/8$\\$<=>-1/(x-2)+1/(x-6)=1/8$\\$<=>1/(x-6)-1/(x-2)=1/8$\\$<=>(1(x-2)8)/((x-6)(x-2)8)-(1(x-6)8)/((x-2)(x-6)8)=(1(x-6)(x-2))/(8(x-6)(x-2))$\\$<=>(8x-16)-(8x-48)=x^2-2x-6x+12$\\$<=>8x-16-8x+48=x^2-8x+12$\\$<=>32=x^2-8x+12$\\$<=>x^2-8x+12-32$\\$<=>x^2-8x-20=0$\\$<=> x^2+2x-10x-20=0$\\$<=>(x^2+2x)-(10x+20)=0$\\$<=>x(x+2)-10(x+2)=0$\\$<=>(x+2)(x-10)=0$\\$<=>$\left[\begin{matrix} x+2=0\\ x-10=0\end{matrix}\right.$$\\$<=>$\left[\begin{matrix} x=-2\\ x=10\end{matrix}\right.$

   Vậy \text{S =} {$-2;10$}

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