Question

[trần Qunag Khải]

ai làm giusp mình

Answer 1

[Đoàn Đức Hà]

a) \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15=\left(x^2+x\right)^2-5\left(x^2+x\right)+3\left(x^2+x\right)-15\)

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

b) \(x^2+2xy+y^2-x-y-12=\left(x+y\right)^2-\left(x+y\right)-12=\left(x+y\right)^2-4\left(x+y\right)+3\left(x+y\right)-12\)

\(=\left(x+y+3\right)\left(x+y-4\right)\)

c) \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)

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

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

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

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

\(=\left(x^2+7x+11\right)^2-1-24\)

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