(x+3)(x+4)(x+5)(x+6)-24=[(x+3)(x+6)][(x+4)(x+5)]-24

                                  =(x²+6x+3x+3.6)(x²+5x+4x+5.4)-24

                                  =(x²+9x+18)(x²+9x+20)-24

                                  =(x²+9x+18)(x²+9x+18+2)-24    (*)

đặt x²+9x+18 là t   (1)

(*) trở thành

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)           (2)

thay (2) vào (1), ta được:

(x+3)(x+4)(x+5)(x+6)-24=(x²+9x+18-4)(x²+9x+18+6)

                                   =(x²+9x+14)(x²+9x+24)

                                   =(x²+7x+2x+14)(x²+9x+24)

                                   =[(x²+7x)+(2x+14)](x²+9x+24)

                                   =x(x+7)+2(x+7)(x²+9x+24)

                                   =(x+7)(x+2)(x²+9x+24)

(mình đã cố gắng giải thật chi tiết và phân tích triệt để nhất có thể rồi. có j sai sót thì góp ý nha!)

chả hiểu cái đề -_- 

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

= (x² +5x+4) (x² +5x+6)-24

  Đặt x² +5x+4 =a

=>(x² +5x+4)(x²+5x+6)-24

= a(a+2)-24 = a² +2a-24

= a² +6a-4a-24

= a(a+6) - 4(a+6) = (a-4)(a+6)

= (x² +5x+a-4)(x² +5x+4+6) = (x² +5x)(x² +5x+10)

\(\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}=\sqrt{\left(1+\frac{1}{a}-\frac{1}{a+1}\right)^2}=\left|1+\frac{1}{a}-\frac{1}{a+1}\right|\)

Đây xem nhá

\(\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}=\sqrt{\frac{\left(a\left(a+1\right)\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}}=\sqrt{\frac{\left(a\left(a+1\right)\right)^2+a^2+2a+1+a^2}{a^2\left(a+1\right)^2}}=\sqrt{\frac{\left(a\left(a+1\right)\right)^2+2a\left(a+1\right)+1^2}{a^2\left(a+1\right)^2}}\)

\(=\sqrt{\left(\frac{a\left(a+1\right)+1}{a\left(a+1\right)}\right)^2}=\left|\frac{a\left(a+1\right)+1}{a\left(a+1\right)}\right|\)