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a) Ta có: \(4\left(2-x\right)^2+xy-2y\)
\(=4\left(x-2\right)^2+y\left(x-2\right)\)
\(=\left(x-2\right)\left[4\left(x-2\right)+y\right]\)
\(=\left(x-2\right)\left(4x-8+y\right)\)
b) Ta có: \(3a^2x-3a^2y+abx-aby\)
\(=3a^2\left(x-y\right)+ab\left(x-y\right)\)
\(=\left(x-y\right)\left(3a^2+ab\right)\)
\(=a\left(x-y\right)\left(3a+b\right)\)
c) Ta có: \(x\left(x-y\right)^3-y\left(y-x\right)^2-y^2\left(x-y\right)\)
\(=x\left(x-y\right)^3-y\left(x-y\right)^2-y^2\left(x-y\right)\)
\(=\left(x-y\right)\left[x\left(x-y\right)^2-y\left(x-y\right)-y^2\right]\)
\(=\left(x-y\right)\left[x\left(x^2-2xy+y^2\right)-yx+y^2-y^2\right]\)
\(=\left(x-y\right)\left(x^3-2x^2y+xy^2-xy\right)\)
d) Ta có: \(2ax^3+6ax^2+6ax+18a\)
\(=2ax^2\left(x+3\right)+6a\left(x+3\right)\)
\(=\left(x+3\right)\left(2ax^3+6a\right)\)
\(=2a\left(x+3\right)\left(x^3+3\right)\)
e) Ta có: \(x^2y-xy^2-3x+3y\)
\(=xy\left(x-y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-3\right)\)
a) \(4x^2-6x=2x\left(2x-3\right)\)
b) \(9x^4y^3+3x^2y^4=3x^2y^3\left(3x^2+y\right)\)
c) \(3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(5x+3\right)\left(x-y\right)\)
d) \(x^3-2x^2+5x=x\left(x^2-2x+5\right)\)
e) \(5\left(x+3y\right)-15x\left(x+3y\right)=\left(5-15x\right)\left(x+3y\right)\)
\(=5\left(1-3x\right)\left(x+3y\right)\)
f) \(2x^2\left(x+1\right)-4\left(x+1\right)=\left(2x^2-4\right)\left(x+1\right)\)
\(=\left(\sqrt{2}x-2\right)\left(\sqrt{2}x+2\right)\left(x+1\right)\)
Bài 2:
a) x(x - 3)- y(3 - x)
= x(x - 3) + y(x - 3)
= (x - 3)(x + y) (1)
Thay x = \(\frac{1}{3}\); y = \(\frac{8}{3}\)vào (1)
Ta có: (\(\frac{1}{3}\)- 3)(\(\frac{1}{3}\)+ \(\frac{8}{3}\))
= \(\frac{-8}{3}\). 3
= -8
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
\(a.\: 2a^2b\left(x+y\right)-4a^3b\left(-x-y\right)\\ =\left(x+y\right)\left(2a^2b+4a^3b\right)\\ =2a^2b\left(x+y\right)\left(1+2a\right)\)
\(b.\:-3a\left(x-y\right)-a^2\left(7-x\right)\\ =a\left(3y-3x-7a+ax\right)\)
a) x(y - x)3 + y(x - y)2 + xy(x - y)
= x(y - x).(y - x)2 + y(x - y)2 + xy(x - y)
= x(y - x)(x - y)2 + y(x - y)2 + xy(x - y)
= (x - y)[x(y - x)(x - y) + y(x - y) + xy]
= (x - y)[x(y - x)(x - y) + y(x - y) + xy]
b) 3a2x - 3a2y + abx - aby
= 3a2(x - y) + ab(x - y)
= a(x - y)(3a + b)
a) x( y - x )3 - y( x - y )2 + xy( x - y )
= -x( x - y )3 - y( x - y )2 + xy( x - y )
= ( x - y )[ -x( x - y )2 - y( x - y ) + xy ]
= ( x - y )[ -x( x2 - 2xy + y2 ) - yx + y2 + xy ]
= ( x - y )( -x3 + 2x2y - xy2 - yx + y2 + xy )
= ( x - y )( -x3 + 2x2y - xy2 + y2 )
b) 3a2x - 3a2y + abx - aby
= 3a2( x - y ) + ab( x - y )
= ( x - y )( 3a2 + ab )
= ( x - y )a( 3a + b )
\(a.10x\left(x-y\right)-6y\left(y-x\right)\\ =10x\left(x-y\right)+6y\left(x-y\right)\\ =\left(10x-6y\right)\left(x-y\right)\\ =2\left(5x-3y\right)\left(x-y\right)\)
\(b.14x^2y-21xy^2+28x^3y^2\\ =7xy\left(x-y+xy\right)\)
\(c.x^2-4+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2+x-2\right)\\ =2x\left(x-2\right)\)
\(d.\left(x+1\right)^2-25\\ =\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)
Bài 2:
1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)
=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)
=>(2x-1)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
2: \(9x^3-x=0\)
=>\(x\left(9x^2-1\right)=0\)
=>x(3x-1)(3x+1)=0
=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)
=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)
=>(2x-3)(2x-3-2)=0
=>(2x-3)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
=>\(2x^2+10x-5x-25-10x+25=0\)
=>\(2x^2-5x=0\)
=>\(x\left(2x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)
Bài 1:
1: \(3x^3y^2-6xy\)
\(=3xy\cdot x^2y-3xy\cdot2\)
\(=3xy\left(x^2y-2\right)\)
2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+3y-2\right)\)
3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)
\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)
\(=(x-2y)(3x-1+5x)\)
\(=\left(x-2y\right)\left(8x-1\right)\)
4: \(x^2-y^2-6y-9\)
\(=x^2-\left(y^2+6y+9\right)\)
\(=x^2-\left(y+3\right)^2\)
\(=\left(x-y-3\right)\left(x+y+3\right)\)
5: \(\left(3x-y\right)^2-4y^2\)
\(=\left(3x-y\right)^2-\left(2y\right)^2\)
\(=\left(3x-y-2y\right)\left(3x-y+2y\right)\)
\(=\left(3x-3y\right)\left(3x+y\right)\)
\(=3\left(x-y\right)\left(3x+y\right)\)
6: \(4x^2-9y^2-4x+1\)
\(=\left(4x^2-4x+1\right)-9y^2\)
\(=\left(2x-1\right)^2-\left(3y\right)^2\)
\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)
8: \(x^2y-xy^2-2x+2y\)
\(=xy\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-2\right)\)
9: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
a) Xem lại đề
b) x³ - 4x²y + 4xy² - 9x
= x(x² - 4xy + 4y² - 9)
= x[(x² - 4xy + 4y² - 3²]
= x[(x - 2y)² - 3²]
= x(x - 2y - 3)(x - 2y + 3)
c) x³ - y³ + x - y
= (x³ - y³) + (x - y)
= (x - y)(x² + xy + y²) + (x - y)
= (x - y)(x² + xy + y² + 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
f) 3x² - 6xy + 3y² - 5x + 5y
= (3x² - 6xy + 3y²) - (5x - 5y)
= 3(x² - 2xy + y²) - 5(x - y)
= 3(x - y)² - 5(x - y)
= (x - y)[(3(x - y) - 5]
= (x - y)(3x - 3y - 5)
Lời giải:
a)
$5(2-x)^2+xy-2y=5(x-2)^2+y(x-2)=(x-2)[5(x-2)+y]=(x-2)(5x+y-10)$
b)
$3a^2x-3a^2y+abx-aby=3a^2(x-y)+ab(x-y)$
$=(x-y)(3a^2+ab)=a(x-y)(3a+b)$
c)
$x(x-y)^3-y(y-x)^2-y^2(x-y)=x(x-y)^3-y(x-y)^2-y^2(x-y)$
$=(x-y)[x(x-y)^2-y(x-y)-y^2]$
$=(x-y)(x^3-2x^2y+xy^2-xy)$
$=x(x-y)(x^2-2xy+y^2-y)$
d)
$2ax^3+6ax^2+6ax+18a$
$=2a(x^3+3x^2+3x+9)
$=2a[x^2(x+3)+3(x+3)]$
$=2a(x+3)(x^2+3)$
e) f) Biểu thức không phân tích được thành nhân tử. Bạn xem lại đề.