(x^2-2x)(x^2+4x+3)-24. phan tich da thuc thanh nhan tu

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1. Đáp án:

   \(\left( {{x^2} + x – 3 – \sqrt {33} } \right)\left( {{x^2} + x – 3 + \sqrt {33} } \right)\)

   Giải thích các bước giải:

   \(\begin{array}{l}
   \,\,\,\,\left( {{x^2} – 2x} \right)\left( {{x^2} + 4x + 3} \right) – 24\\
   = x\left( {x – 2} \right)\left( {x + 1} \right)\left( {x + 3} \right) – 24\\
   = x\left( {x + 1} \right)\left( {x – 2} \right)\left( {x + 3} \right) – 24\\
   = \left( {{x^2} + x} \right)\left( {{x^2} + x – 6} \right) – 24\\
   = {\left( {{x^2} + x} \right)^2} – 6\left( {{x^2} – x} \right) – 24\\
   = {\left( {{x^2} + x} \right)^2} – 6\left( {{x^2} – x} \right) + 9 – 33\\
   = {\left( {{x^2} + x – 3} \right)^2} – 33\\
   = \left( {{x^2} + x – 3 – \sqrt {33} } \right)\left( {{x^2} + x – 3 + \sqrt {33} } \right)
   \end{array}\)

   Trả lời

2. $(x^2-2x).(x^2+4x+3)-24_{}$

   $=x.(x-2).(x^2+x+3x+3)-24_{}$

   $=x.(x-2).[ x.(x+1)+3.(x+1)]-24_{}$

   $=x.(x+1)(x-2)(x+3)-24_{}$

   $=(x^2+x).(x^2+x-6)-24_{}$

   $=(x^2+x)-6.(x^2-x)-24_{}$

   $=(x^2+x)^2-6.(x^2-x)+9-33_{}$

   $=(x^2+x-3)^2-33_{}$

   $=(x^2+x-3-\sqrt{33}).(x^2+x-3+\sqrt{33}_{})$

   $Good_{}$ $luck_{}$

   Trả lời