Toán Lớp 8: Tóm x :(1/x^2+3x+2)+(1/x^2+5x+6)+(1/x^2+7x+12)+(1/x^2+9x+20)+1(/x^2+11x+30)=-5/36

Question

Toán Lớp 8:  Tóm x :(1/x^2+3x+2)+(1/x^2+5x+6)+(1/x^2+7x+12)+(1/x^2+9x+20)+1(/x^2+11x+30)=-5/36

thucquyenlan · Answer

$ $ ĐKXĐ : x ne -1, x ne -2,x ne -3,x ne -4, x ne -5, x ne -6 1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)+1/(x^2+9x+20) +1/(x^2+11x+30)=(-5)/36 <=>1/(x^2+2x+x+2)+1/(x^2+2x+3x+6)+1/(x^2+4x+3x+12)+1/(x^2+4x+5x+20)+1/(x^2+5x+6x+30)=(-5)/36 <=>1/(x(x+2)+(x+2)) + 1/(x(x+2)+3(x+2)) + 1/(x(x+4)+3(x+4)) + 1/(x(x+4)+5(x+4)) + 1/(x(x+5)+6(x+5)) = (-5)/36 <=> 1/((x+1)(x+2)) + 1/((x+2)(x+3)) + 1/((x+3)(x+4))+1/((x+4)(x+5))+1/((x+5)(x+6))=(-5)/36 <=> 1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)+1/(x+4)-1/(x+5)+1/(x+5)-1/(x+6)=(-5)/36 <=>1/(x+1) - 1/(x+6)=(-5)/36 <=> (x+6)/((x+1)(x+6)) - (x+1)/((x+1)(x+6))=(-5)/36 <=>(x+6-x-1)/((x+1)(x+6)) = (-5)/36 <=> 5/((x+1)(x+6)) = 5/(-36) <=>(x+1)(x+6)+36=0 <=>x^2+6x+x+6+36=0 <=>x^2+7x+42=0 <=>x^2+2.x. 7/2 + (7/2)^2 =(-119)/4 <=>(x+7/2)^2=(-119)/4 (Vô lí) => Không có x thỏa mãn Vậy không có x thỏa mãn