Toán Lớp 8: Phân tích các đa thức sau thành nhân tử Q(x)= (4x-2)(10x+4)(5x+7)(2x+1)+17 R(x)= (3x+2)(3x-5)(x-1)(9x+10)+24x^2 K(x)= (6x+5)^2(3x+2)(x+

Question

Toán Lớp 8:  Phân tích các đa thức sau thành nhân tử Q(x)= (4x-2)(10x+4)(5x+7)(2x+1)+17 R(x)= (3x+2)(3x-5)(x-1)(9x+10)+24x^2 K(x)= (6x+5)^2(3x+2)(x+1)-35

ngoclankhue · Answer

Q(x)=(     4    x    &minus;    2     )      (     10    x    +    4     )      (     5    x    +    7     )      (     2    x    +    1     )     +    17    =    0     (4x&minus;2)(10x+4)(5x+7)(2x+1)+17=0            &hArr;     (     4    x    &minus;    2     )      (     5    x    +    7     )      (     10    x    +    4     )      (     2    x    +    1     )     +    17    =    0     &hArr;(4x&minus;2)(5x+7)(10x+4)(2x+1)+17=0            &hArr;     (     20      x      2      +    18    x    &minus;    14     )      (     20      x      2      +    18    x    +    4     )     +    17    =    0     &hArr;(20x2+18x&minus;14)(20x2+18x+4)+17=0      Đặt t=         20      x      2      +    18    x    +    4     (     t    &ge;    0     )      20x2+18x+4(t&ge;0)      ta c&oacute;:   (t-18).t +17=0         &hArr;      t      2      &minus;    18    t    +    17    =    0     &hArr;t2&minus;18t+17=0            &hArr;     (     t    &minus;    17     )      (     t    &minus;    1     )     =    0           &hArr; ( left[  begin{array}{l}t=17  t=1 end{array}  right. )         &hArr; ( left[  begin{array}{l}20x^{2}+18x+4=17  20x^{2}+18x+4=1 end{array}  right. )        &hArr;  ( left[  begin{array}{l}20x^{2}+18x-13=0  20x^{2}+18x+3=0 end{array}  right. )        &hArr; ( left[  begin{array}{l}(20x+9-&radic;341)(20x+9+&radic;341)=0  (20x+9&minus;&radic;21)(20x+9+&radic;21)=0 end{array}  right. )