Question

(x-1)(x-3)(x-5)(x-7)-20

 

Answer 1

(x- 1)(x- 3)(x- 5)(x- 7)- 20

= (x- 1)(x- 7)(x- 3)(x- 5)- 20

= (x²- 8x+ 7)(x²- 8x+ 15)- 20

Đătk x²- 8x+ 11= t

Ta có (t- 4)(t+ 4) -20

= t²- 16- 20

= t²- 36= (t- 6)(t+ 6)

Thay t= x²- 8x+ 11

=> (x²- 8x+ 11- 6)(x²- 8x +11+ 6)

= (x²- 8x+ 5)(x²- 8x+ 17)

= (x²- 8x+ 16- 11)(x²- 8x+ 17)

= ((x- 4)²- 11)(x²- 8x+ 17)

\(=\left(x-4-\sqrt{11}\right)\left(x-4+\sqrt{11}\right)\left(x^2-8x+17\right)\)

Answer 2

Đặt đa thức trên là A
<=>A=(x-1)(x-7)(x-3)(x-5)-20
<=>A=(x²-8x-7)(x²-8x+15)-20
Đặt t=x²-8x-7, ta có:

A=t(t+22)-20
   =t²+22t-20
   =t²+22t+121-141
   =(t+11)²-141
thayt=x²-8x- 7vào
 

Answer 3

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