(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

thay x^2+7x+11=y

=> (y-1)(y+1)-24=y^2-1^2-24=y^2-25=(y-5)(y+5)

= (x^2+7x+11-5)(x^2+7x+11+5)=(x^2+7x+6)(x^2+7x+16)=(x^2+x+6x+6)(x^2+7x+16)=[x(x+1)+6(x+1)]((x^2+7x+16)=(x+1)(x+6)(x^2+7x+16)

(x + 2)(x + 3)(x + 5)(x + 7) - 24

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

=(x² + 7x + 10)(x² + 7x +12) - 24

Đặt x² + 7x + 11 = t ; ta có:

(t - 1)(t + 1) - 24

= t² - 1² - 24

= t² - 25

= (t - 5)(t + 5)

Thay t = x² + 7x + 11 ta được:

(x² + 7x + 11 - 5)(x² + 7x +11 + 5)

= (x² + 7x + 6)(x² + 7x + 16)

= (x + 1)(x + 6)(x² + 7x + 16)

Chúc bn học tốt

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

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

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

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

Đặt \(t=x^2+7x+10\), khi đó (*) trở thành:

\(t\left(t+2\right)-24\)

\(=t^2+2t-24\)

\(=\left(t-4\right)\left(t+6\right)\)

Thay \(t=x^2+7x+10\) vào, ta được:

\(\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

Vậy ...

Help meee plzzz

em ko lam dc anh ehh

em chua gap bai nao nhu the nay. noi dung hon la chua den lop lam haha

SORRY BUT I CAN'T HELP

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

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

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

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)

(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 + \(\sqrt{40}\))

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

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

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

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

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

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)

a) \(x^2+7x+12\)

\(=x^2+3x+4x+12\)

\(=x\left(x+3\right)+4\left(x+3\right)\)

\(=\left(x+3\right)\left(x+4\right)\)

b) \(3x^2-5x+2\)

\(=3x^2-3x-2x+2\)

\(=3x\left(x-1\right)-2\left(x-1\right)\)

\(=\left(x-1\right)\left(3x-2\right)\)

a) x² + 7x + 12 = x² + 3x + 4x + 12 = x(x + 3) + 4(x + 3) = (x + 4)(x + 3)

b) 3x² - 5x + 2 = 3x² - 3x - 2x + 2 = 3x(x - 1) - 2(x - 1) = (3x - 2)(x - 1)

c) x² + 9x - 10 = x² + 10x - x - 10 = x(x + 10) - (x + 10) = (x - 1)(x + 10)

d) x² - 7x - 8 = x² - 8x + x - 8 = x(x - 8) + (x - 8) = (x + 1)(x - 8)

e) 2x² + 3x - 5 = 2x² + 5x - 2x - 5 = x(2x + 5) - (2x + 5) = (x  - 1)(2x + 5)

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: \(x^4+4x^3+5x^2+x^2+4x+5\)

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

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