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\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
a) \(x^3y^3+x^2y^2+4\)
\(=x^3y^3-x^2y^2+2x^2y^2-2xy+2xy+4\)
\(=\left(x^3y^3-x^2y^2+2xy\right)+\left(2x^2y^2-2xy+4\right)\)
\(=xy\left(x^2y^2-xy+2\right)+2\left(x^2y^2-xy+2\right)\)
\(=\left(xy+2\right)\left(x^2y^2-xy+2\right)\)
b) \(x^3+3x^2y-9xy^2+5y^3\)
\(=x^3+5x^2y-2x^2y-10xy^2+xy^2+5y^3\)
\(=\left(5y^3-10xy^2+5x^2y\right)+\left(xy^2-2x^2y+x^3\right)\)
\(=5y\left(y^2-2xy+x^2\right)+x\left(y^2-2xy+x^2\right)\)
\(=\left(5y+x\right)\left(y^2-2xy+x^2\right)\)
\(=\left(5y+x\right)\left(y-x\right)^2\)
Bài làm :
\(\text{a)}9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)
\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)
\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)
\(=\left(-x-3y-5\right)\left(7x+9y-1\right)\)
\(\text{b)}3x^4y^2+3x^3y^2+3xy^2+3y^2\)
\(=\left(3x^4y^2+3xy^2\right)+\left(3x^3y^2+3y^2\right)\)
\(=3xy^2\left(x^3+1\right)+3y^2\left(x^3+1\right)\)
\(=\left(3xy^2+3y^2\right)\left(x^3+1\right)\)
\(=3y^2\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=3y^2\left(x+1\right)^2\left(x^2-x+1\right)\)
\(\text{c)}\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1-3xy\right)\)
\(=\left(x+y-1\right)\left(x^2+x+y^2+y+1-xy\right)\)
\(d ) x^3+3x^2+3x+1-27z^3\)
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)
a) xy+3x-7y-21
=x(y+3)-7(x+3)
=(x-7)(y+3)
b)2xy-15-6x-5y
=2x(y-3)-5(-3+y)
=(2x-5)(y-3)
c)2x^2y+2xy^2-2x-2y
=2x(xy-1)+2y(xy-1)
=(2x+2y)(xy-1)
x(x+3)-5x(x-5)-5(x+3)
=(x-5)(x+3)-5x(x-5)
=(x-5)(x+3-5x)
Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn
Trả lời:
a, x4 + 3x3 + x2 + 3x
= ( x4 + 3x3 ) + ( x2 + 3x )
= x3 ( x + 3 ) + x ( x + 3 )
= ( x3 + x ) ( x + 3 )
= x ( x2 + 1 ) ( x + 3 )
b, Sửa đề: x4 - x2 + 8x - 8
= ( x4 - x2 ) + ( 8x - 8 )
= x2 ( x2 - 1 ) + 8 ( x - 1 )
= x2 ( x - 1 ) ( x + 1 ) + 8 ( x - 1 )
= ( x - 1 ) [ x2 ( x + 1 ) + 8 ]
= ( x - 1 ) ( x3 + x2 + 8 )
a. 6x3-x2-486x+81
= 6x3-54x2+53x2-477x-9x+81
= 6x2.(x-9)+53x.(x-9)-9.(x-9)
= (x-9).(6x2+53x-9)
= (x-9)(6x2+54x-x-9)
=(x-9)[6x.(x+9)-(x+9)]=(x-9)(x+9)(6x-1)
b. x3-5x2+3x+9
= x3+x2-6x2-6x+9x+9
=x2.(x+1)-6x.(x+1)+9.(x+1)
=(x+1)(x2-6x+9)=(x+1)(x-3)2
c. x3+3x2+6x+4
= x3+x2+2x2+2x+4x+4
= x2.(x+1)+2x.(x+1)+4.(x+1)
= (x+1)(x2+2x+4)
d.
x^2 - 7xy + 10y^2
= (x^2 - 2xy) - (5xy - 10y^2)
= x(x - 2y) - 5y( x - 2y)
= (x - 5y)(x - 2y)
`a)3x-3a+yx-ya`
`=3(x-a)+y(x-a)`
`=(x-a)(y+3)`
`b)x^2-9-4(x+3)`
`=(x-3)(x+3)-4(x+3)`
`=(x+3)(x-3-4)`
`=(x+3)(x-7)`
a) \(=3x+yx-\left(3a+ya\right)\) \(=x\left(3+y\right)-a\left(3+y\right)\) \(=\left(3+y\right)\left(x-a\right)\)
b) \(=\left(x-3\right)\left(x+3\right)-4\left(x+3\right)\) \(=\left(x+3\right)\left(x-3-4\right)\) \(=\left(x+3\right)\left(x-7\right)\)