Toán Lớp 8: 2x^4-21x^3+34x^2+105x+50=0 làm theo cách hồi quy

Question

Toán Lớp 8:  2x^4-21x^3+34x^2+105x+50=0 làm theo cách hồi quy

maianhtuanh · Answer

2x^4 - 21x^3 + 34x^2 + 105x + 50 = 0   &hArr; 2x^4 - 9x^3 - 12x^3 - 10x^2 + 54x^2 - 10x^2 + 60x + 45x + 50 = 0   &hArr; 2x^4 - 9x^3 - 10x^2 - 12x^3 + 54x^2 + 60x - 10x^2 + 45x + 50 = 0   &hArr; x^2(2x^2 - 9x - 10) - 6x(2x^2 - 9x - 10) - 5(2x^2 - 9x - 10) = 0   &hArr; (2x^2 - 9x - 10)(x^2 - 6x - 5) = 0   &rArr;  ( left[  begin{array}{l}2x^2 - 9x - 10 = 0 (1)  x^2 - 6x - 5 = 0 (2) end{array}  right. )    text(Giải phương tr&igrave;nh (1))   2x^2 - 9x - 10 = 0   &hArr; x^2 - 9/2x - 5 = 0   &hArr; x^2 - 2 . x . 9/4 + 81/16 - 81/16 - 5 = 0   &hArr; (x - 9/4)^2 = 161/16   &rArr;  ( left[  begin{array}{l}x = ( sqrt{161} + 9)/4   x = (-  sqrt{161} + 9)/4 end{array}  right. )    text(Giải phương tr&igrave;nh (2))   x^2 - 6x - 5 = 0   &hArr; x^2 - 2 . x . 3 + 9 - 9 - 5 = 0   &hArr; (x - 3)^2 = 14   &rArr;  ( left[  begin{array}{l}x =  sqrt{14} + 3   x = - sqrt{14} + 3 end{array}  right. )    Vậy  S = {  sqrt{14} + 3 ; -  sqrt{14} + 3 ; ( sqrt{161} + 9)/4 ; - ( sqrt{161} + 9)/4}