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\(x\left(y^2-z^2\right)+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)\)
\(=xy^2-xz^2+yz^2-x^2y+zx^2-zy^2\)
\(=xy^2-xz^2+yz^2-x^2y+zx^2-zy^2-xyz+xyz\)
\(=\left(yz^2-xz^2-xyz+x^2z\right)-\left(zy^2-xyz-xy^2+x^2y\right)\)
\(=z\left(yz-xz-xy+x^2\right)-y\left(zy-xz-xy+x^2\right)\)
\(=\left(z-y\right)\left(yz-xz-xy+x^2\right)\)
\(=\left(z-y\right)\left[y\left(z-x\right)-x\left(z-x\right)\right]\)
\(=\left(z-y\right)\left(y-x\right)\left(z-x\right)\)
\(\left(3x+1\right)^2-\left(3x-1\right)^2\)
\(=\left(3x+1-3x+1\right)\left(3x+1+3x-1\right)\)
\(=2\cdot6x\)
\(=12x\)
_________
\(\left(x+y\right)^2-\left(x-y\right)^2\)
\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)
\(=2x\cdot2y\)
\(=4xy\)
\(\left(x+y\right)^3+\left(x-y\right)^3\)
\(=\left(x+y+x-y\right)\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=2x\cdot\left(x^2+2xy+y^2-x^2+y^2+x^2-2xy+y^2\right)\)
\(=2x\cdot\left(x^2+3y^2\right)\)
______
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x-y\right)+z^3+3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x-y-z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3z\left(x+y\right)-3xy\right]\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yz-3xz-3yz-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2-xy-xz-yz\right)\)
\(a,\Rightarrow\left(x-3-5+2x\right)\left(x-3+5-2x\right)=0\\ \Rightarrow\left(3x-8\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{8}{3}\end{matrix}\right.\\ b,=\left(x+y\right)^2-\left(x-2y\right)^2\\ =\left(x+y-x+2y\right)\left(x+y+x-2y\right)=3y\left(2x-y\right)\\ c,=\left(x+y-x+y\right)\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\\ =2y\left(3x^2+y^2\right)\\ d,=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
b: \(=\dfrac{12\left(y-z\right)^4+3\left(y-z\right)^5}{6\left(y-z\right)^2}=2\left(y-z\right)^2+\dfrac{1}{2}\left(y-z\right)^3\)
x(y+z)^2 - y(z-x)^2 +z(x+y)^2 - x^3 + y^3 - z^3 - 4xyz
=xy^2+2xyz+xz^2-yz^2+2xyz-x^2y+x^2z+2xyz+zy^2-x^3+y^3-z^3-4xyz
=xy^2+xz^2-yz^2-x^2y+x^2z+y^2z-x^3+y^3-z^3+2xyz
=(xy^2+2xyz+xz^2)-x^3-(yz^2+2xyz+x^2y)+y^3+(x^2z+2xyz+y^2z)-z^3
=x[(y+z)^2-x^2)-y[(z+x)^2-y^2]+z[(x+y)^2-z^2]
=x(-x+y+z)(x+y+z)-y(x-y+z)(x+y+z)+z(x+y-z)(x+y+z)
=(x+y+z)[-x^2+xy+xz-xy+y^2-yz+xz+yz-z^2]
=(x+y+z)[-x(x-y-z)-y(x-y-z)+z(x-y-z)]
=(x+y+z)(x-y-z)(z-x-y)
x(y+z)^2 - y(z-x)^2 +z(x+y)^2 - x^3 + y^3 - z^3 - 4xyz
=xy^2+2xyz+xz^2-yz^2+2xyz-x^2y+x^2z+2xyz+zy^2-x^3+y^3-z^3-4xyz
=xy^2+xz^2-yz^2-x^2y+x^2z+y^2z-x^3+y^3-z^3+2xyz
=(xy^2+2xyz+xz^2)-x^3-(yz^2+2xyz+x^2y)+y^3+(x^2z+2xyz+y^2z)-z^3
=x[(y+z)^2-x^2)-y[(z+x)^2-y^2]+z[(x+y)^2-z^2]
=x(-x+y+z)(x+y+z)-y(x-y+z)(x+y+z)+z(x+y-z)(x+y+z)
=(x+y+z)[-x^2+xy+xz-xy+y^2-yz+xz+yz-z^2]
=(x+y+z)[-x(x-y-z)-y(x-y-z)+z(x-y-z)]
=(x+y+z)(x-y-z)(z-x-y)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a) \(\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)
\(=\left[\left(x-2\right)\left(x-5\right)\right]\left[\left(x-3\right)\left(x-4\right)\right]+1\)
\(=\left(x^2-7x+10\right)\left(x^2-7x+12\right)+1\)
Đặt: \(x^2-7x+11=t\)
\(\Rightarrow\hept{\begin{cases}x^2-7x+10=t-1\\x^2-7x+12=t+1\end{cases}}\)
\(\Rightarrow\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)
\(=\left(x^2-7x+10\right)\left(x^2-7x+12\right)+1\)
\(=\left(t-1\right)\left(t+1\right)+1\)
\(=t^2-1+1\)
\(=t^2\)
Vậy: \(\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)
\(=\left(x^2-7x+11\right)^2\)