Question

[Đỗ thị như quỳnh]

Phân tích thành nhân tử ;

1, \(\left(x+2\right)\cdot\left(x+3\right)\cdot\left(x+4\right)\cdot\left(x+5\right)-24\)

2, \(x\cdot\left(x+4\right)\cdot\left(x+6\right)\cdot\left(x+10\right)+128\)

3, \(\left(x^2+5x+6\right)\cdot\left(x^2-15x+56\right)-144\)

4, \(\left(x-18\right)\cdot\left(x-7\right)\cdot\left(x+35\right)\cdot\left(x+90\right)-67x^2\)

5, \(\left(x-2\right)\cdot\left(x-3\right)\cdot\left(x-4\right)\cdot\left(x-6\right)-72x^2\)

Answer 1

[Kaito Kid]

1,(x+2)(x+5)(x+3)(x+4)-24=(x²+7x+10)(x²+7x+12)-24

Đặt x²+7x+10= t ta có t(t+2)-24=t²+2t-24=(t-4)(t+6)

hay (x²+7x+6)(x²+7x+16)

2,x(x+10)(x+4)(x+6)+128=(x²+10x)(x²+10x+24)+128

Đặt x²+10x=t ta có t(t+24)+128=t²+24t+128=(t+8)(t+16)

hay (x²+10x+8)(x²+10x+16)

3,(x+2)(x-7)(x+3)(x-8)-144=(x^2-5x-14)(x²-5x-24)-144

Đặt x²-5x-14=t ta có t(t-10)-144=t²-10t-144=(t-18)(t+8)

Hay (x²-5x-32)(x²-5x-6)=(x²-5x-32)(x+1)(x-6)

Answer 2

[Thụy Lâm]

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