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a) x^2 - 2xy + y^2 - xz + yz
= (x^2 - 2xy + y^2 ) - (xz + yz)
= (x - y)^2 - z(x + y)
= (x - y)(x - x + y)
a.
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1+3z\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+9z^2\right]\)
\(=\left(x+3z+1\right)\left(x^2+2x+1+3zx+3z+9z^2\right)\)
b.
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
c.
\(=x^4-1+4x^2-4\)
\(=\left(x^2-1\right)\left(x^2+1\right)+4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+5\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
a) Ta có: \(x^3+3x^2+3x+1-27z^3\)
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)
b) Ta có: \(x^2-2xy+y^2-zx+yz\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
c) Ta có: \(x^4+4x^2-5\)
\(=x^4+4x^2+4-9\)
\(=\left(x^2+2\right)^2-3^2\)
\(=\left(x^2-1\right)\left(x^2+5\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
Câu 1 : x2-y2+2yz-z2=-(y2-2yz+z2-x2) Câu 2: x2-2xy+y2-xz+yz=(x2-2xy+y2)-xz+yz
=-(y-z)2 -x2 =(x-y)2-z(x-y)
=-(y-z-x)(y-z+x) =(x-y)(x-y-z)
c1:
x3+3x2+3x+1-27z3
=(x+1)3-(3z)3
=(x+1-3z)[(x+1)2-(x+1)3z+9z2)
=(x+1-3z)(x2+2x+1-3xz-3z+9x2)
c2:
x2-2xy+y2-xz+yz
=(x-y)2-z(x-y)
=(x-y)(x-y-z)
\(1,\\ a,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ b,=a^2\left(a-x\right)-y\left(a-x\right)=\left(a^2-y\right)\left(a-x\right)\\ c,=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\\ d,=x\left(x-2y\right)+t\left(x-2y\right)=\left(x+t\right)\left(x-2y\right)\\ 2,\\ \Rightarrow x^2-4x+4-x^2+9=6\\ \Rightarrow-4x=-7\Rightarrow x=\dfrac{7}{4}\\ 3,\\ a,x^2+2x+2=\left(x+1\right)^2+1\ge1>0\\ b,-x^2+4x-5=-\left(x-2\right)^2-1\le-1< 0\)
c) x2 + 2xy + y2 – xz – yz = (x + y)2 – z(x + y) = (x + y)(x + y – z)
\(x^2-2xy+y^2-xz+yz\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
\(x^2-2xy+y^2-xz+yz=\left(x^2-2xy+y^2\right)-\left(xz-yz\right)=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)