a) \(A=(x^2+7x+10)(x^2+7x+12)-24\)
Đặt \(t=x^2+7x+10\)
=> \(A=t\left(t+2\right)-24\)
\(=t^2-2t-24\)
\(=(t-1)^2-25\)
\(=(t+4)(t-6)\)
\(=(x^2+7x+14)(x^2+7x+4)\)
Hinhf như còn tách típ đc đó :)))
b) \(=(x+y)^2-(x+y)-12\)
\(=(x+y)^2+3(x+y)-4(x+y)-12\)
\(=(x+y)(x+y+3)-4(x+y+3)\)
\(=(x+y-4)(x+y+3)\)
c, Chưa ra :v
a) Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24\)
\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)
\(=\left(x^2+7x\right)^2+6\left(x^2+7x\right)+16\left(x^2+7x\right)+96\)
\(=\left(x^2+7x\right)\left(x^2+7x+6\right)+16\left(x^2+7x+6\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
b) Ta có: \(x^2+2xy+y^2-x-y-12\)
\(=\left(x^2+2xy+y^2\right)-\left(x+y\right)-12\)
\(=\left(x+y\right)^2-\left(x+y\right)-12\)
\(=\left(x+y\right)^2-4\left(x+y\right)+3\left(x+y\right)-12\)
\(=\left(x+y\right)\left(x+y-4\right)+3\left(x+y-4\right)\)
\(=\left(x+y-4\right)\left(x+y+3\right)\)