Bài2: Phân tích thành nhân tử
a) xy+y2-x-y
b) (xy-8)2-1
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a) x2-xy+5y-25
= x(2-y)+ 5(y-2)
= x(2-y)-5(2-y)
= (x-5)(2-y)
a: \(=y\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(y-1\right)\)
b: \(=\left(x^2y^2-8-1\right)\left(x^2y^2-8+1\right)\)
\(=\left(x^2y^2-9\right)\left(x^2y^2-7\right)\)
\(=\left(xy-3\right)\left(xy+3\right)\left(x^2y^2-7\right)\)
c: \(=x^2-8x+x-8\)
\(=x\left(x-8\right)+\left(x-8\right)\)
\(=\left(x-8\right)\left(x+1\right)\)
\(a,xy+y^2-x-y\)
\(=\left(xy+y^2\right)-\left(x+y\right)\)
\(=y\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(y-1\right)\)
\(---\)
\(b,\left(x^2y^2-8\right)^2-1\)
\(=\left(x^2y^2-8-1\right)\left(x^2y^2-8+1\right)\)
\(=\left[\left(xy\right)^2-9\right]\left(x^2y^2-7\right)\)
\(=\left(xy-3\right)\left(xy+3\right)\left(x^2y^2-7\right)\)
\(---\)
\(c,x^2-7x-8\)
\(=x^2+x-8x-8\)
\(=\left(x^2+x\right)-\left(8x+8\right)\)
\(=x\left(x+1\right)-8\left(x+1\right)\)
\(=\left(x+1\right)\left(x-8\right)\)
\(Toru\)
a) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1=\left(x^2+3x\right)\left(x^2+3x+2\right)+1=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1=\left(x^2+3x+1\right)^2\)
b) \(\left(1+x^2\right)\left(1+y^2\right)+4xy+2\left(x+y\right)\left(1+xy\right)=25\Leftrightarrow1+x^2+y^2+x^2y^2+4xy+2\left(x+y\right)\left(1+xy\right)-25=0\Leftrightarrow\left(x+y\right)^2+2\left(x+y\right)\left(1+xy\right)+\left(1+xy\right)^2-25=0\Leftrightarrow\left(x+y+1+xy\right)^2-25=0\Leftrightarrow\left(x+y+xy-24\right)\left(x+y+xy+26\right)=0\)
a: Ta có: \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1\)
\(=\left(x^2+3x+1\right)^2\)
\(a,=\dfrac{\left(x+1\right)\left(x+y\right)}{\left(x-y\right)\left(x+1\right)}=\dfrac{x+y}{x-y}\\ b,=\dfrac{\left(x-3\right)^2}{3x\left(x-3\right)}=\dfrac{x-3}{3x}\\ c,=\dfrac{\left(y-x\right)\left(y+x\right)}{xy\left(x-y\right)}=\dfrac{-x-y}{xy}\)
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) \(3x\left(2x-y\right)+5y\left(y-2x\right)\)
\(=3x\left(2x-y\right)-5y\left(2x-y\right)\)
\(=\left(3x-5y\right)\left(2x-y\right)\)
b) \(\left(x-5\right)^2-9\left(x+y\right)^2\)
\(=\left(x-5\right)^2-3^2\left(x+y\right)^2\)
\(=\left(x-5\right)^2-\left(3x+3y\right)^2\)
\(=\left(x-5+3x+3y\right)\left(x-5-3x-3y\right)\)
\(=\left(4x+3y-5\right)\left(-2x-3y-5\right)\)
a: \(3x\left(2x-y\right)+5y\left(y-2x\right)=\left(2x-y\right)\left(3x-5y\right)\)
e: \(x^2-10x+24=\left(x-4\right)\left(x-6\right)\)
b) Ta có: \(x^3-x^2y-xy^2+y^3\)
\(=\left(x^3+y^3\right)-\left(x^2y+xy^2\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x+y\right)\left(x-y\right)^2\)
Lời giải:
a. $xy(x+y)-y(x+y)^2+y^2(x-y)$
$=y(x+y)[x-(x+y)]+y^2(x-y)$
$=y(x+y)(-y)+y^2(x-y)$
$=-y^2(x+y)+y^2(x-y)$
$=y^2(x-y)-y^2(x+y)=y^2[(x-y)-(x+y)]$
$=y^2(-2y)=-2y^3$
b.
$x(x+y)^2-y(x+y)^2+xy-x^2$
$=[x(x+y)^2-y(x+y)^2]-(x^2-xy)$
$=(x+y)^2(x-y)-x(x-y)$
$=(x-y)[(x+y)^2-x]=(x-y)(x^2+2xy+y^2-x)$
a: \(xy\left(x+y\right)-y\left(x+y\right)^2+y^2\left(x-y\right)\)
\(=\left(x+y\right)\left[xy-y\left(x+y\right)\right]+y^2\left(x-y\right)\)
\(=\left(x+y\right)\left(xy-xy-y^2\right)+y^2\left(x-y\right)\)
\(=y^2\left(-x-y\right)+y^2\left(x-y\right)\)
\(=y^2\left(-x-y+x-y\right)=-2y\cdot y^2=-2y^3\)
b: \(x\left(x+y\right)^2-y\left(x+y\right)^2+xy-x^2\)
\(=\left(x+y\right)^2\left(x-y\right)+x\left(y-x\right)\)
\(=\left(x+y\right)^2\cdot\left(x-y\right)-x\left(x-y\right)\)
\(=\left(x-y\right)\left[\left(x+y\right)^2-x\right]\)
a) \(=\left(2x-1\right)^2\)
b) \(=x\left(y^2-x^2+2x-1\right)=x\left[y^2-\left(x-1\right)^2\right]=x\left(y-x+1\right)\left(y+x-1\right)\)
a: =2(x-2)+y(x-2)
=(x-2)(2+y)
b: \(=\left(x+y\right)^2-4=\left(x+y+2\right)\left(x+y-2\right)\)
c: =(x-7)(x+2)
a.
2x - 4 + xy - 2y
= 2(x-2) +y(x-2)
= (x-2)(y+2)
c.
x^2 - 5x - 14
= x^2 + 2x - 7x - 14
= x(x+2) - 7(x+2)
= (x-7)(x+2)
b: \(\left(xy-8\right)^2-1\)
\(=\left(xy-8\right)^2-1^2\)
\(=\left(xy-8-1\right)\left(xy-8+1\right)\)
=(xy-7)(xy-9)
a: \(xy+y^2-x-y\)
\(=y\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(y-1\right)\)
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