(x+2).(x+3).(x+4).(x+5)−24
=(x2+7x+10).(x2+7x+12)−24
=(x2+7x+10).(x2+7x+10+2)−24
Đặt x2+7x+10=t, ta có
t.(t+2)−24
=t2+2t−24
=t2+2t+1−25
=(t−1)2−25
=(t−1−5)(t−1+5)
=(t−6)(t+4)
=(x2+7x+10−6)(x2+7x+10+4)
(x2+7x+4)(x2+7x+14)
P/s tham khảo nha
\(\left(x+2\right).\left(x+3\right).\left(x+4\right).\left(x+5\right)-24\)
\(\Leftrightarrow\left(x^2+7x+10\right).\left(x^2+7x+12\right)-24\)
\(\Leftrightarrow\left(x^2+7x+10\right).\left(x^2+7x+10+2\right)-24\)
Đặt \(x^2+7x+10=t\), ta có
\(t.\left(t+2\right)-24\)
\(\Leftrightarrow t^2+2t-24\)
\(\Leftrightarrow t^2+2t+1-25\)
\(\Leftrightarrow\left(t-1\right)^2-25\)
\(\Leftrightarrow\left(t-1-5\right)\left(t-1+5\right)\)
\(\Leftrightarrow\left(t-6\right)\left(t+4\right)\)
\(\Rightarrow\left(x^2+7x+10-6\right)\left(x^2+7x+10+4\right)\)
\(\Leftrightarrow\left(x^2+7x+4\right)\left(x^2+7x+14\right)\)
P/s tham khảo nha
\(x^4+4=\left[\left(x^2\right)^2+2.x^2.2+2^2\right]-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
@ Trần Hoàng Việt @ là dấu bằng chứ không phải tương đương nhé em!!! Em nhớ xem và sửa lại:))