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a) \(x^4+2x^3-4x-4=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)
\(=\left(x^2-2\right)\left(x^2+2x+2\right)\)
a) Ta có: \(x^4+2x^3-4x-4\)
\(=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2\)
\(=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)
\(=\left(x^2-2\right)\cdot\left(x^2+2x+2\right)\)
x2 + y2 - x2y2 + xy - x - y = (x2-x) + (y2-y) + (-x2y2 + xy) = x(x+1) + y(y+1) + xy(xy+1) = ( x+ y+ xy)( x + 1 + y + 1 + xy + 1)
\(a,=\dfrac{\left(x+1\right)\left(x+y\right)}{\left(x-y\right)\left(x+1\right)}=\dfrac{x+y}{x-y}\\ b,=\dfrac{\left(x-3\right)^2}{3x\left(x-3\right)}=\dfrac{x-3}{3x}\\ c,=\dfrac{\left(y-x\right)\left(y+x\right)}{xy\left(x-y\right)}=\dfrac{-x-y}{xy}\)
Lời giải:
a.
\(\frac{x^2+xy+x+y}{x^2-xy+x-y}=\frac{x(x+y)+(x+y)}{x(x+1)-y(x+1)}=\frac{(x+y)(x+1)}{(x+1)(x-y)}=\frac{x+y}{x-y}\)
b.
\(\frac{x^2-6x+9}{3x^2-9x}=\frac{(x-3)^2}{3x(x-3)}=\frac{x-3}{3x}\)
c.
\(\frac{y^2-x^2}{x^2y-xy^2}=\frac{(y-x)(y+x)}{-xy(y-x)}=\frac{x+y}{-xy}\)
a) (x - y)(x + y + 3). b) (x + y - 2xy)(2 + y + 2xy).
c) x 2 (x + l)( x 3 - x 2 + 2). d) (x – 1 - y)[ ( x - 1 ) 2 + ( x - 1 ) y + y 2 ].
a: \(x^2+4x+4=x^2+2\cdot x\cdot2+2^2=\left(x+2\right)^2\)
b: \(4x^2-4x+1=\left(2x\right)^2-2\cdot2x\cdot1+1^2=\left(2x-1\right)^2\)
c: \(2x-1-x^2\)
\(=-\left(x^2-2x+1\right)=-\left(x-1\right)^2\)
d: \(x^2+x+\dfrac{1}{4}=x^2+2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)
e: \(9-x^2=3^2-x^2=\left(3-x\right)\left(3+x\right)\)
g: \(\left(x+5\right)^2-4x^2=\left(x+5+2x\right)\left(x+5-2x\right)\)
\(=\left(5-x\right)\left(5+3x\right)\)
h: \(\left(x+1\right)^2-\left(2x-1\right)^2\)
\(=\left(x+1+2x-1\right)\left(x+1-2x+1\right)\)
\(=3x\left(-x+2\right)\)
i: \(=x^2y^2-4xy+4-3\)
\(=\left(xy-2\right)^2-3=\left(xy-2-\sqrt{3}\right)\left(xy-2+\sqrt{3}\right)\)
k: \(=y^2-\left(x-1\right)^2\)
\(=\left(y-x+1\right)\left(y+x-1\right)\)
l: \(=x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=\left(x+2\right)^3\)
m: \(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2-y^3=\left(2x-y\right)^3\)
\(a,Sửa:x^2+4xy-9+4y^2=\left(x+2y\right)^2-9=\left(x+2y-3\right)\left(x+2y+3\right)\\ b,=\left(x-6y\right)^2-1=\left(x-6y-1\right)\left(x-6y+1\right)\\ c,=36-\left(x-5y\right)^2=\left(6-x+5y\right)\left(6+x-5y\right)\)
\(4\left(x^2y^2+z^2t^2+2xyzt\right)-\left(x^2+y^2-z^2-t^2\right)^2\)
\(=\left(2xy-2tz\right)^2-\left(x^2+y^2-z^2-t^2\right)\)
\(=\left(2xy-2tz-x^2-y^2+z^2+t^2\right)\left(2xy-2tz+x^2+y^2-z^2-t^2\right)\)
\(=\left[-\left(x-y\right)^2+\left(z-t\right)^2\right]\left[\left(x+y\right)^2-\left(t+z\right)^2\right]\)
\(=-\left(x-y-z+t\right)\left(x-y+z-t\right)\left(x+y-t-z\right)\left(x+y+t+z\right)\)
4(x2y2+z2t2+2xyzt)−(x2+y2−z2−t2)24(x2y2+z2t2+2xyzt)−(x2+y2−z2−t2)2
=[2(xy+zt)]2−(x2+y2−z2−t2)2=[2(xy+zt)]2−(x2+y2−z2−t2)2
=(2xy+2zt)2−(x2+y2−z2−t2)2=(2xy+2zt)2−(x2+y2−z2−t2)2
=(2xy+2zt−x2−y2+z2+t2)(2xy+2zt+x2+y2−z2−t2)2
a: \(16x^5-25x^3\)
\(=x^3\left(16x^2-25\right)\)
\(=x^3\left(4x-5\right)\left(4x+5\right)\)
b: \(x^2-2xy+y^2-16\)
\(=\left(x-y\right)^2-16\)
\(=\left(x-y-4\right)\left(x-y+4\right)\)
c: \(x^2-5y-xy+5x\)
\(=x\left(x-y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+5\right)\)
d: \(x^2\left(x^2+4\right)-x^2+4\)
\(=\left(x^2+4\right)\left(x^2-1\right)\)
\(=\left(x^2+4\right)\left(x-1\right)\left(x+1\right)\)
Bài 3:
a: Ta có: C=A+B
\(=x^2-2y+xy+1+x^2+y-x^2y^2-1\)
\(=2x^2-y+xy-x^2y^2\)
b: Ta có: C+A=B
\(\Leftrightarrow C=B-A\)
\(=x^2+y-x^2y^2-1-x^2+2y-xy-1\)
\(=-x^2y^2+3y-xy-2\)
