Ta có:

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

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

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

= (x² + 7x + 10)(x² + 7x + 12) - 24

Đặt x² + 7x + 10 = k 

=> k(k + 2) - 24 = k² + 2k - 24 = k² + 6x - 4x - 24 

                            = k(k + 6)  - 4(k  + 6)

                          = (k - 4)(k + 6)

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

= (x² + 7x + 10 - 4)(x² + 7x + 10 + 6)

= (x² + 7x + 6)(x² + 7x + 16)

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

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

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)(1)

Đặt \(x^2+7x+11=t\)thay vào (1) ta được:
\(\left(t-1\right)\left(t+1\right)-24\)

\(=t^2-1-24\)

\(=t^2-25\)

\(=\left(t-5\right)\left(t+5\right)\)Thay \(t=x^2+7x+11\)ta được:

\(\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)

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

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

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

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

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

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

Dat \(x^2+7x+10=y\)

ta co :\(y\left(y+2\right)-24=y^2+2y-24\)

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

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

tra lai bien ta co

\(\left(x^2+7x+10-4\right)\left(x^2+7x+12+6\right)\)

\(\left(x^2+7x+6\right)\left(x^2+7x+18\right)=\left(x^2+6x+x+6\right)\left(x^2+7x+18\right)\)

\(\left(x+1\right)\left(x+6\right)\left(x^2+7x+18\right)\)

\((x+5)^2+4(x+5)(x-5)+4(x^2-10x+25)=0\\\Rightarrow(x+5)^2+4(x+5)(x-5)+4(x^2-2\cdot x\cdot5+5^2)=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+4(x-5)^2=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+[2(x-5)]^2=0\\\Rightarrow[(x+5)+2(x-5)]^2=0\\\Rightarrow(x+5+2x-10)^2=0\\\Rightarrow(3x-5)^2=0\\\Rightarrow3x-5=0\\\Rightarrow3x=5\\\Rightarrow x=\frac53\\\text{#}Toru\)

Sao đề là phân tích mà lại "= 0" vậy bạn?

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

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

Đặt \(x^2+7x+10=y\)

\(\Rightarrow\)\(A=y.\left(y+2\right)-24\)

\(A=y^2+2y+1-25\)

\(A=\left(y+1\right)^2-5^2\)

\(A=\left(y+1-5\right)\left(y+1+5\right)\)

\(A=\left(y-4\right)\left(y+6\right)\)

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

\(A=\left[\left(x^2+x\right)+\left(6x+6\right)\right].\left(x^2+7x+16\right)\)

\(A=\left[x.\left(x+1\right)+6.\left(x+1\right)\right].\left(x^2+7x+16\right)\)

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

Đặt \(B=\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)

\(B=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)

Đặt \(12x^2+11x-1=a\)

\(\Rightarrow B=a.\left(a+3\right)-4\)

\(B=a^2+3a-4\)

\(B=\left(a^2-a\right)+\left(4a-4\right)\)

\(B=a.\left(a-1\right)+4.\left(a-1\right)\)

\(B=\left(a-1\right)\left(a+4\right)\)

\(\Rightarrow B=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)

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

= x⁴ + 10x³ + 35x² + 50x + 24 - 24

= x⁴ + 10x³ + 35x² + 50x

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

= ( x² + 5x + 4 ) .( x² + 5x + 6 ) - 24

Đặt t = x² + 5x + 5 

=> ( t - 1 ). ( t + 1 ) - 24

= t² - 1 - 24 

= t² - 25

= ( t - 5 ). ( t + 5 )

= ( x² + 5x + 5 - 5 ) . ( x² + 5x + 5 + 5 )

= ( x² + 5x ) . ( x² + 5x + 10 )

= x. ( x + 5 ) . ( x² + 5x + 10 )

`(x+3)^4+(x+5)^4-2`

`={[(x+3)^2]^2-1^2}+{[(x+5)^2]^2 -1^2}`

`=[(x+3)^2-1^2][(x+3)^2+1]+[(x+5)^2-1^2][(x+5)^2+1]`

`=(x+3-1)(x+3+1)[(x+3)^2+1]+(x+5-1)(x+5+1)[(x+5)^2+1]`

`=(x+2)(x+4)[(x+3)^2+1]+(x+4)(x+6)[(x+5)^2+1]`

`=(x+4){(x+2)[(x+3)^2+1]+(x+6)[(x+5)^2+1]}`

`=(x+4)(2x^3+24x^2+108x+176)`

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

\(=\left[\left(x+3\right)^4-1\right]+\left[\left(x+5\right)^4-1\right]\)

\(=\left[\left(x^2+6x+9-1\right)\left(x^2+6x+9+1\right)\right]+\left[\left(x^2+10x+25-1\right)\left(x^2+10x+25+1\right)\right]\)

\(=\left(x^2+6x+8\right)\left(x^2+6x+10\right)+\left(x^2+10x+24\right)\left(x^2+10x+26\right)\)

\(=\left(x+2\right)\left(x+4\right)\left(x^2+6x+10\right)+\left(x+4\right)\left(x+6\right)\left(x^2+10x+26\right)\)

\(=\left(x+4\right)\left[\left(x+2\right)\left(x^2+6x+10\right)+\left(x+6\right)\left(x^2+10x+26\right)\right]\)

\(=\left(x+4\right)\left(x^3+6x^2+10x+2x^2+12x+20+x^3+10x^2+26x+6x^2+60x+156\right)\)

\(=\left(x+4\right)\left(2x^3+24x^2+108x+176\right)\)

\(=2\left(x+4\right)\left(x^3+12x^2+54x+88\right)\)

\(5x(2x+3)+6x+9\\=5x(2x+3)+3(2x+3)\\=(2x+3)(5x+3)\)

a: \(5x\left(2x+3\right)+6x+9\)

\(=5x\left(2x+3\right)+\left(6x+9\right)\)

\(=5x\left(2x+3\right)+3\left(2x+3\right)\)

\(=\left(2x+3\right)\left(5x+3\right)\)

b: \(3x\left(x+4\right)+48\left(x+4\right)+5\left(x+4\right)\)

\(=\left(x+4\right)\left(3x+48+5\right)\)

=(x+4)(3x+53)

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

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

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

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

\(=\left(x^2+7x+11\right)^2-25\)

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

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

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

Đặt \(y=x^2+7x+10\), khi đó biểu thức trở thành:

\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(x^2+7x+10+1-5)(x^2+7x+10+1+5)\\=(x^2+7x+6)(x^2+7x+16)\\=(x^2+x+6x+6)(x^2+7x+16)\\=[x(x+1)+6(x+1)](x^2+7x+16)\\=(x+1)(x+6)(x^2+7x+16)\\Toru\)

(x + 2)(x + 3)(x + 4)(x + 5) - 24

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

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

= (x² + 7x + 10)(x² + 7x + 12) - 24 (1)

Đặt t = x² + 7x + 10

(1) = t.(t + 2) - 24

= t² + 2t - 24

= t² - 4t + 6t - 24

= (t² - 4t) + (6t - 24)

= t(t - 4) + 6(t - 4)

= (t - 4)(t + 6)

= (x² + 7x + 10 - 4)(x² + 7x + 10 + 6)

= (x² + 7x + 6)(x² + 7x + 16)

= (x² + x + 6x + 6)(x² + 7x + 16)

= [(x² + x) + (6x + 6)](x² + 7x + 16)

= [x(x + 1) + 6(x + 1)](x² + 7x + 16)

= (x + 1)(x + 6)(x² + 7x + 16)

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

= [(x - 5)² + 2(2x - 1)(x - 5) + (2x - 1)²) - [2(x - 3)]²

= (x - 5 + 2x - 1)² - (2x - 6)²

= (3x - 6)² - (2x - 6)²

= (3x - 6 - 2x + 6)(3x - 6 + 2x - 6) = x(5x - 12)

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

= [ ( x - 5 )² + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )² ] - 2²( x - 3 )²

= ( x - 5 + 2x - 1 )² - ( 2x - 6 )²

= ( 3x - 6 )² - ( 2x - 6 )²

= ( 3x - 6 - 2x + 6 )( 3x - 6 + 2x - 6 )

= x( 5x - 12 )

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

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

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

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

\(=\left[\left(3x-6\right)-\left(2x-6\right)\right].\left[\left(3x-6\right)+\left(2x-6\right)\right]\)

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

\(=\left(5x-12\right)x\)