đặt ẩn phụ a) (x+2)(x+3)(x+4)(x+5)-24 b) (x^2+x)+4x^2+4x-12

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1. $\text{a) (x+2)(x+3)(x+4)(x+5)-24}$

   $=[(x+2)(x+5)][x+3)(x+4)]-24$

   $=(x²+5x+2x+10)(x²+4x+3x+12)-24$

   $=(x²+7x+10)(x²+7x+10+2)-24$

   $\text{Đặt x²+7x+10=t, ta có:}$

   $(x²+7x+10)(x²+7x+10+2)-24$

   $=t(t+2)=24$

   $=t²+2t-24$

   $=t²+6t-4t-24$

   $=t(t+6)-4(t+6)$

   $=(t-4)(t+6)$

   $\text{Thay t=x²+7x+10 ta có:}$

   $(t-4)(t+6)$

   $=(x²+7x+10-4)(x²+7x+10+6)$

   $=(x²+7x+6)(x²+7x+16)$

   $=(x²+x+6x+6)(x²+7x+16)$

   $=[x(x+1)+6(x+1)](x²+7x+16)$

   $=(x+6)(x+1)(x²+7x+16)$

   $\text:{b) (x²+x)+4x²+4x-12}$

   $=(x²+x)+4(x²+x)-12$

   $\text{Đặt x²+x= t, ta có:}$

   $(x²+x)+4(x²+x)-12$

   $=t+4t-12$

   $=5t-12$

   $\text{Thay t=x²+x, ta có}$

   $5t-12$

   $=5(x²+x)-12$

   `=5x²+5x-12`

    

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2. $\begin{array}{l}a) \, (x+2)(x+3)(x+4)(x+5)-24\\ = [(x+2)(x+5)][(x+3)(x+4)] – 24\\ = (x^2 + 7x + 10)(x^2 + 7x + 12) – 24\\ Đặt\,\,t = x^2 + 7 + 10\,,ta\,\, được:\\ t(t + 2) – 24\\ = t^2 + 2t – 24\\ =t^2 -4t + 6t – 24\\ = t(t – 4) + 6(t – 4)\\ = (t -4)(t + 6)\\ \text{Do đó ta được:}\\ (x^2 + 7x + 10 – 4)(x^2 + 7x + 10 + 6)\\ = (x^2 + 7x + 6)(x^2 + 7x + 16)\\ Vậy\,\,(x+2)(x+3)(x+4)(x+5)-24 = (x^2 + 7x + 6)(x^2 + 7x + 16)\\ b)\,(x^2+x)^2+4x^2+4x-12\\ = (x^2 + x)^2 + 4(x^2 + x) – 12\\ Đặt\,\,t = x^2 + x\,,ta \,\,được:\\ t^2 + 4t – 12\\ = t^2 – 2t + 6t – 12\\ = t(t -2) + 6(t – 2)\\ = (t – 2)(t + 6)\\ \text{Do đó ta được:}\\ (x^2 + x – 2)(x^2 + x + 6)\\ Vậy\,\,(x^2+x)^2+4x^2+4x-12 =(x^2 + x – 2)(x^2 + x + 6) \end{array}$

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