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

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

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)

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

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\\ =\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\\ =\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt \(x^2+7x+11=y\)

\(\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\\ =\left(y+1\right)\left(y-1\right)-24\\ =y^2-1-24\\ =y^2-25\\ =\left(y-5\right)\left(y+5\right)\\ =\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\\ =\left(x^2+7x+6\right)\left(x^2+7x+16\right)\\ =\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\\ =\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\\ =\left(x^2+7x+11\right)^2-1-24\\ =\left(x^2+7x+11\right)^2-25\\ =\left(x^2+7x+16\right)\left(x^2+7x+6\right)\\ =\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)

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

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

Đặt \(t=x^2+7x+10\) ta có:

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

\(=t^2-4t+6t-24\)\(=t\left(t-4\right)+6\left(t-4\right)\)

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

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

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

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

Đặt x^2+7x+10=a

a(a+2)-24

=a^2+2a-24

=(a-4)(a+6)

=(x^2+7x+6)(x^2+7x+16)

=(x+1)(x+6)(x^2+7x+16)

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

= (x²+7x+10)(x²+7x+12)-24

đặt y=x²+7x+10

ta có biểu thức:

y.(y+2)-24

= y²+2y-24

= y²+6y-4y-24

= y(y+6)-4(y+6)

= (y+6)(y-4)

= (x²+7x+10+6)(x²+7x+10-4)

= (x²+7x+16)(x²+7x+6)

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

= (x² + 3x + 2x + 6)(x² + 5x + 4x + 20) - 24

= (x² + 5x + 6)(x² + 9x + 20) - 24

= x⁴ + 9x³ + 20x² + 5x³ + 45x² + 100x + 6x² + 54x + 120 - 24

= x⁴ + 14x³ + 71x² + 100x + 96

(x + 2)(x + 3)(x + 4)(x + 5) – 24 =  (x + 2)(x + 5)(x + 4)(x + 3) – 24 

                                                            = ( x² + 7x +10) ( x² + 7x +12) – 24  đặt x² + 7x +10 = a

                                                            = a(a + 2 ) – 24 = a² + 2a – 24  = a² + 2a +1 – 25

                                                            = (a – 4 )( a + 6) 

Thay vào ta có: (x + 2)(x + 3)(x + 4)(x + 5) – 24  =  ( x² + 7x +6) ( x² + 7x +16)

                                                                              = (x + 6)(x + 1)( x² + 7x +16)

Đặt A = (x^2+5x+4)(x^2+5x+6)-24 và  x^2+5x+5=a 

Do đó A= (a-1)(a+1)-24

= a^2- 25

= a^2-5^2

=(a-5)(a+5)

= ( x^2+5x+5-5)( x^2+5x+5+5)

= ( x^2+5x)(x^2+5x+10)

Đặt A = (x^2+5x+4)(x^2+5x+6)-24 và  x^2+5x+5=a 

Do đó A= (a-1)(a+1)-24

= a^2- 25

= a^2-5^2

=(a-5)(a+5)

= ( x^2+5x+5-5)( x^2+5x+5+5)

= ( x^2+5x)(x^2+5x+10)

<=>[(x+2)(x+5)][(x+3)(x+4)]-24=\(\left(x^2+7x+10\right)\left(\left(x^2+7x+12\right)\right)-24\)(1)

đặt x^2+7x+11=t

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

<=> \(\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

(x+2)(x+3)(x+4)(x+5)=24

(x+x+x+x)(2+3+4+5)=24

(x.4)14=24

x.4=24:14

x.4=2

x=2:4

X=1/2

Phân ch đa thức a mình lớp 6 chưa học

(x+2)(x+5)(x+4)(x+3)-24=(x^2+7x+10)(x^2+7x+12)-24

đặt:x^2+7x+10=t  thi x^2+7x+12=t+2

=>t(t+2)-24=t^2+2t-25=t^2+2t+1-25=(t+1)^2-5^2=(t-4)(t+6)

thay t vao suy ra: (x^2+7x+6)(x^2+7x+16)

\(\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

thang: Em sai một chỗ rất nhỏ :) tách \(t^2+2t-24\) mới đúng nhé :)

ta có: A= (x+2)*(x+3)*(x+4)*(x+5)=\(\left(\left(x+2\right)\cdot\left(x+5\right)\right)\cdot\left(\left(x+3\right)\cdot\left(x+4\right)\right)=\left(x^2+7x+10\right)\cdot\left(x^2+7x+12\right)\)

đặt t=x^2+7x+11

=> A=(t-1)*(t+1)-24=t^2-25=(t-5)*(t+5)

=> A=(x^2+7x+6)*(x^2+7x+16)

Ta có : \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

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

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

- Đặt \(x^2+5x+5=a\)

\(=\left(a-1\right)\left(a+1\right)-24=a^2-1-24=a^2-25\)

\(=\left(a-5\right)\left(a+5\right)\)

Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24-24\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)\)

\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)

\(=x\left(x+5\right)\left(x^2+5x+10\right)\)