Question

[Ngô Phương Thị Thảo]

phân tích đa thức thành nhân tử

1.(x+2)(x+3)(x+4)(x+5)-24

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

3.x(x-1)(x-1)(x+2)-3

4.(x-7)(x-5)(x-4)(x-2)-72

5.(x^2+6x+8)(x^2-8x+15)-24

6.(x^2-6x+5)(x^2-10x+21)-20

7.(x^2+5x+6)(x^2-15x+56)-144

#Mn làm giúp mik vs#

#thank kiu#

Answer 1

[olm (admin@gmail.com)]

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

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt \(x^2+7x=t\)

\(\Rightarrow BT=\left(t+10\right)\left(t+12\right)-24\)

\(=t^2+22x+96=\left(t+11\right)^2-25\ge-25\)

Vậy GTNN của bt là - 25\(\Leftrightarrow x^2+7x+11=0\)

\(\Delta=7^2-4.11=5\)

\(\orbr{\begin{cases}x_1=\frac{-22+\sqrt{5}}{2}\\x_2=\frac{-22-\sqrt{5}}{2}\end{cases}}\)

Answer 2

[olm (admin@gmail.com)]

2) \(\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20\)

\(=\left(x-1\right)\left(x-7\right)\left(x-3\right)\left(x-5\right)-20\)

\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20\)

Đặt \(x^2-8x=t\)

\(\RightarrowĐT=\left(t+7\right)\left(t+15\right)-20\)

\(=t^2+22t+85=\left(t+11\right)^2-36\ge-36\)

Vậy GTNN của bt là - 36\(\Leftrightarrow x^2-8x+11=0\)

\(\Delta=\left(-8\right)^2-4.11=20\)

\(\orbr{\begin{cases}x_1=\frac{-22-\sqrt{20}}{2}\\x_2=\frac{-22+\sqrt{20}}{2}\end{cases}}\)