a) (Mn^2-1) . (m^2n+5) b) 4(x-1/2) . (x+1/2).(4x^2+1) c) (m^2-2m+4) . (m+2)-m^3+(m+3) . (m-3)-m^2-18 d) [(t+2) . (t^2+4)] . (t-2)

0 bình luận về “a) (Mn^2-1) . (m^2n+5) b) 4(x-1/2) . (x+1/2).(4x^2+1) c) (m^2-2m+4) . (m+2)-m^3+(m+3) . (m-3)-m^2-18 d) [(t+2) . (t^2+4)] . (t-2)”

1. Đáp án:

    $a) m^3n^3+5mn^2-m^2n-5\\
   b) 16x^4-1\\
   c) -37\\
   d) t^4-16$

   Giải thích các bước giải:

    $a) (mn^2-1) . (m^2n+5)\\
   =mn^2 . (m^2n+5)-1 . (m^2n+5)\\
   =mn^2.m^2n+mn^2.5-m^2n-5\\
   =m^3n^3+5mn^2-m^2n-5\\
   b) 4(x-1/2) . (x+1/2).(4x^2+1)\\
   =4.\left [x^2-\left (\dfrac{1}{2}  \right )^2  \right ].(4x^2+1)\\
   =4.\left ( x^2-\dfrac{1}{4} \right ).(4x^2+1)\\
   =4.\dfrac{1}{4}.(4x^2-1).(4x^2+1)\\
   =(4x^2)^2-1\\
   =16x^4-1\\
   c) (m^2-2m+4) . (m+2)-m^3+(m+3) . (m-3)-m^2-18\\
   =(m+2)(m^2-2m+4)-m^3+m^2-3^2-m^2-18\\
   =m^3+2^3-m^3+m^2-27-m^2-18\\
   =(m^3-m^3)+(m^2-m^2)+(-27-18+2^3)\\
   =-37\\
   d) [(t+2) . (t^2+4)] . (t-2)\\
   =(t+2).(t-2).(t^2+4)\\
   =(t^2-4)(t^2+4)\\
   =(t^2)^2-4^2\\
   =t^4-16$

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