(x+1)(x+2)(x+3)(x+4)-15
= (x+1)(x+4)(x+2)(x+3)-15
= (x2+5x+4)(x2+5x+6)-15
Đặt x2+5x+5= y
\(\Rightarrow\left(y+1\right)\left(y-1\right)-15\)
\(\Rightarrow y^2-1-15\)
\(\Rightarrow y^2-16\)
\(\Rightarrow\left(y-4\right)\left(y+4\right)\)
\(\Rightarrow\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)
Ta có: (x+1)(x+4)(x+2)(x+3)-15=(x^2+5x+4)(x^2+5x+6)-15
Đặt x^2+5x+4=a, ta có:
a(a+2)-15=a^2+2a-15=a^2+5a-3a-15=a(a+5)-3(a+5)=(a-3)(a+5)
\(\Rightarrow\)(x^2+5x+1)(x^2+5x+9)
Bài dễ
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\\ \\ =\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-15\\ =\left(x^2+4x+x+4\right)\left(x^2+2x+3x+6\right)-15\\ \\ =\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\text{ }\text{ }\text{ }\left(1\right)\)
Đặt \(x^2+5x+5=y\) \(\left(\text{*}\right)\)
Thay \(\left(\text{*}\right)\) vào \(\left(1\right)\), ta được:
\(\left(1\right)=\left(y-1\right)\left(y+1\right)-15\\ \\ =y^2-1-15\\ \\ =y^2-16\\ \\ =\left(y+4\right)\left(y-4\right)\text{ }\text{ }\text{ }\left(2\right)\)
Thay \(\left(\text{*}\right)\) vào \(\left(2\right)\), ta lại được:
\(\left(2\right)=\left(x^2+5x+5+4\right)\left(x^2+5x+5-4\right)\\ \\=\left(x^2+5x+9\right)\left(x^2+5x+1\right)\)
nhờ mn là giúp mình với ạ , minh đang cần gấp :( 
