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PA

d. \(\left(x^2+y^2-z^2\right)^2-4x^2y^2\)

e. \(\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\)

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TL
4 tháng 8 2016 lúc 11:19

d)\(\left(x^2+y^2-z^2\right)^2-4x^2y^2\)

\(=\left(x^2+y^2-z^2+2xy\right)\left(x^2+y^2-z^2-2xy\right)\)

\(=\left[\left(x^2+2xy+y^2\right)-z^2\right]\left[\left(x^2-2xy+y^2\right)-z^2\right]\)

\(=\left[\left(x+y\right)^2-z^2\right]\left[\left(x-y\right)^2-z^2\right]\)

\(=\left(x+y-z\right)\left(x+y+z\right)\left(x-y-z\right)\left(x-y+z\right)\)

e)Đặt \(x^2+3x=a\)

Có: \(\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\)

\(=\left(a+1\right)\left(a-3\right)-5\)

\(=a^2-3a+a-3-5\)

\(=a^2-2a-8\)

\(=a^2+2x-4x-8\)

\(=a\left(a+2\right)-4\left(a+2\right)\)

\(=\left(a+2\right)\left(a-4\right)\)

\(=\left(x^2+3x+2\right)\left(x^2+3x-4\right)\)

\(=\left(x^2+x+2x+2\right)\left(x^2-x+4x-4\right)\)

\(=\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x-1\right)+4\left(x-1\right)\right]\)

\(=\left(x+1\right)\left(x+2\right)\left(x-1\right)\left(x+4\right)\)

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NJ
4 tháng 8 2016 lúc 13:51

\(d,\left(x^2+y^2-z^2\right)^2-4x^2y^2\)
\(=\left(x^2+y^2-z^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2+y^2-z^2-2xy\right)\left(x^2+y^2-z^2+2xy\right)\)
\(=\left[\left(x^2-2xy+y^2\right)-z^2\right]\left[\left(x^2+2xy+y^2\right)-z^z\right]\)
\(=\left[\left(x-y\right)^2-z^2\right]\left[\left(x+y\right)^2-z^2\right]\)
\(=\left(x-y-z\right)\left(x-y+z\right)\left(x+y-z\right)\left(x+y+z\right)\)
\(e,\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\left(1\right)\)
\(\text{Đặt }x^2+3x+\frac{1-3}{2}=t\)
\(\text{hay }x^2+3x-2=t\left(2\right)\)
\(\left(1\right)\Leftrightarrow\left(t+3\right)\left(t-1\right)-5\)
\(\Rightarrow t^2-t+3t-3-5\)
\(=t^2+2t-8\)
\(=t^2-2t+4t-8\)
\(=t\left(t-2\right)+4\left(t-2\right)\)
\(=\left(t-2\right)\left(t+4\right)\left(3\right)\)
\(\text{Thay (2) vào (3),ta được:}\)
\(\left(x^2+3x-2-2\right)\left(x^2+3x-2+4\right)\)
\(=\left(x^2+3x-4\right)\left(x^2+3x+2\right)\)

\(=\left(x^2-x+4x-4\right)\left(x^2+x+2x+2\right)\)
\(=\left[x\left(x-1\right)+4\left(x-1\right)\right]\left[x\left(x+1\right)+2\left(x+1\right)\right]\)

\(=\left(x-1\right)\left(x+4\right)\left(x+1\right)\left(x+2\right)\)

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