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

= [(x + 1)(x + 4)][(x + 2)(x + 3)] - 24

= (x² + 4x + x +4)(x² + 3x + 2x + 12) - 24

= (x² + 5x + 4)(x² + 5x + 12) - 24

Đặt t = x² + 5x + 8

Ta có: x² + 5x + 4 = x² + 5x + 8 - 4 (1)

x² + 5x + 12 = x² + 5x + 8 + 4 (2)

Thay t = x² + 5x + 8 vào (1) và (2), ta có:

⇒ (t - 4)(t + 4) - 24

= t² - 16 - 24

= t² - 40

= (t - \(\sqrt{40}\))(t + \(\sqrt{40}\))

= (x² + 5x + 8 - \(\sqrt{40}\))(x² + 5x + 8 + \(\sqrt{40}\))

a) x³ + 4x² - 29x + 24                                                           

= x³ - 3x² + 7x² - 21x - 8x + 24

= x²(x-3) + 7x(x-3) - 8(x-3)

= (x-3)(x²+7x-8)

=(x-3)(x²+8x-x-8)

= (x-3)[(x²+8x)-(x+8)]

= (x-3)[x(x+8)-(x+8)]

= (x-3)(x+8)(x-1)

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-8\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)\(=\left(x^2+7x+11-1\right)\left(x^2+7x+11+1\right)-8\)

\(=\left(x^2+7x+11\right)^2-9\)

\(=\left(x^2+7x+11-3\right)\left(x^2+7x+11+3\right)=\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)

a) =x³-2x²+6x²-12x -12x +24

= x²(x-2)+6x(x-2)-12(x-2)

= (x-2)(x²+6x-12)

mk giải đc câu a thôi, bn zô jup mk lại vs

\(a,x^3+4x^2-24x+24\)

\(=x^3+6x^2-12x-2x^2-12x+24\)

\(=\left(x^3-2x^2\right)+\left(6x^2-12x\right)-\left(12x-24\right)\)

\(=x^2\left(x-2\right)+6x\left(x-2\right)-12\left(x-2\right)\)

\(=\left(x-2\right)\left(x^2+6x-12\right)\)

b)

x⁸+x⁴+1

= x⁸+x⁷+x⁶-x⁷-x⁶-x⁵+x⁵+x⁴+x³-x³-x²-x+x²+x+1

= x⁶(x²+x+1)-x⁵(x²+x+1) +x³(x²+x+1)-x(x²+x+1)+(x²+x+1)

= (x²+x+1)(x⁶-x⁵+x³-x+1)

= (x²+x+1)(x⁶-x⁵+x⁴-x⁴+x³-x²+x²-x+1)

= (x²+x+1)[x⁴(x²-x+1) - x²(x²-x+1) + (x²-x+1)]

= (x²+x+1)(x²-x+1)(x⁴-x²+1)

Giải giùm em \(\left(x^2+4x+8\right)^2+3x^3+14x^2+24x\) nha

\(=\left(a-1\right)\left(a+4\right)\left(a+3\right)\left(a-2\right)-24=\left(a-2\right)\left(a+4\right)\left(a-1\right)\left(a+3\right)-24\)\(=\left(a^2+2a-8\right)\left(a^2+2a-3\right)-24.dat:a^2+2a-8=h\)\(\Rightarrow\left(a^2+2a-8\right)\left(a^2+2a-3\right)-24=h\left(h+5\right)-24=h^2+5h-24=\left(h-3\right)\left(h+8\right)\)\(=\left(a^2+2a-11\right)a\left(a+2\right)\)

\(-\left(x+2\right)+3\left(x^2-4\right)\)

\(=3\left(x-2\right)\left(x+2\right)-\left(x+2\right)\)

\(=\left(x+2\right)\left[3\left(x-2\right)-1\right]=\left(x+2\right)\left(3x-7\right)\)

\(\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)+1\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)

\(=\left(x^2+5x+4\right)^2+2\left(x^2+5x+4\right)+1\)

\(=\left(x^2+5x+4+1\right)^2\)

\(=\left(x^2+5x+5\right)^2\)

Ta có tổng quát: \(\left(ax^2+bx+c\right)\)\(\left(mx^2+nx+p\right)\)\(\circledast\)

-Nhân ra ta được: \(amx^4+\left(an+bm\right)x^3+\left(ap+bn+cm\right)x^2+\left(bp+cn\right)x+cp\)

-Áp dụng phương pháp hệ số bất định, ta có:

am=1

an+bm=4 (1)

ap+bn+cm=6 (2)

bp+cn=4 (3)

cp=5

-Xét a=m=1 và c=1, p=5

thay vào (1), ta được: n+b=4 (4)

thay vào (3), ta được: n+5b=4 (5)

từ (4),(5)\(\Rightarrow\)n=4 và b=0

giờ thay tất cả vào phương trình (3), ta được: 5+0+1=6 (T/M)

\(\Rightarrow\)Thay vào\(\circledast\), ta được: \(\left(x^2+1\right)\left(x^2+4x+5\right)\)

Cách 2: Ta tách \(6x^2\) thành \(5x^2+x^2\)

ta được: \(x^4+4x^3+5x^2+x^2+4x+5\)

\(\Leftrightarrow x^2\left(x^2+4x+5\right)+\left(x^2+4x+5\right)\)

\(\Leftrightarrow\left(x^2+1\right)\left(x^2+4x+5\right)\)