\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)
Đặt \(x^2+5x+4=t\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)
\(=t.\left(t+2\right)-15\)
\(=t^2+2t+1-16\)
\(=\left(t+1\right)^2-4^2\)
\(=\left(t-3\right)\left(t+5\right)\)
\(=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)
Ta có :
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)
\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-15\)
\(=\left[x\left(x+4\right)+1\left(x+4\right)\right]\left[x\left(x+3\right)+2\left(x+3\right)\right]-15\)
\(=\left(x^2+4x+x+4\right)\left(x^2+3x+2x+6\right)-15\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)
\(=\left(x^2+5x+4\right)\left[\left(x^2+5x+4\right)+2\right]-15\)(1)
Đặt \(x^2+5x+4=y\)thì (1) trở thành :
\(y\left(y+2\right)-15\)
\(=y^2+2y-15\)
\(=y^2+5y-3y-15\)
\(=\left(y^2+5y\right)-\left(3y+15\right)\)
\(=y\left(y+5\right)-3\left(y+5\right)\)
\(=\left(y-3\right)\left(y+5\right)\)(2)
Thay \(y=x^2+5x+4\)thì (2) trở thành:
\(\left(x^2+5x+4-3\right)\left(x^2+5x+4+5\right)\)
\(=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)
HT nha bạn
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