Phân tích các đa thức sau thành nhân tử(sử dụng phương pháp nhóm các số hạng)
a)\(5x^2-5xy+7y-7x\)
b)\(x^2-y^2+2x+1\)
c)\(3x^2+6xy+3y^2-3z^2\)
d)\(ab\left(x^2+y^2\right)+xy\left(a^2+b^2\right)\)
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x2 - x - y2 - y
= (x - y)(x + y) - (x + y)
= (x + y)(x - y - 1)
***
9x2 + y2 - 16z2 + 6xy
= (3x + y)2 - (4z)2
= (3x + y - 4z)(3x + y + 4z)
***
a3 - a2x - ay + xy
= a2(a - x) - y(a - x)
= (a - x)(a2 - y)
***
2x2 - 8y2 + 3x + 6y
= 2(x2 - 4y2) + 3(x + 2y)
= 2(x - 2y)(x + 2y) + 3(x + 2y)
= (x + 2y)(2x - 4y + 3)
***
xy(x + y) + yz(y + z) + xz(x + z) + 2xyz
= xy(x + y + z) + yz(x + y + z) + xz(x + z)
= y(x + y + z)(x + z) + xz(x + z)
= (x + z)(xy + y2 + yz + xz)
= (x + z)[y(x + y) + z(x + y)]
= (x + z)(x + y)(y + z)
a) \(x^3y^3+125=\left(xy\right)^3+5^3=\left(xy+5\right)\left(x^2y^2-5xy+25\right)\)
b) \(8x^3+y^3-6xy\left(2x+y\right)=\left(8x^3+y^3\right)-6xy\left(2x+y\right)=[\left(2x\right)^3+y^3]-6xy\left(2x+y\right)\)
\(=\left(2x+y\right)\left(4x^2-2xy+y^2\right)-6xy\left(2x+y\right)=\left(2x+y\right)\left(4x^2-2xy+y^2-6xy\right)\)
\(=\left(2x+y\right)\left(4x^2-8xy+y^2\right)\)
c) \(\left(3x+2\right)^2-2\left(x-1\right)\left(3x+2\right)+\left(x-1\right)^2\)
\(=[\left(3x+2\right)-\left(x-1\right)]^2=\left(3x+2-x+1\right)^2=\left(2x+3\right)^2=\left(2x+3\right)\left(2x+3\right)\)
a: \(x^4+25x^2+20x-4\)
\(=x^4-5x^3+2x^2+5x^3-25x^2+10x-2x^2+10x-4\)
\(=x^2\left(x^2-5x+2\right)+5x\left(x^2-5x+2\right)-2\left(x^2-5x+2\right)\)
\(=\left(x^2-5x+2\right)\left(x^2+5x-2\right)\)
b: \(=x^4-6x^2-x^2+9\)
\(=\left(x^2-3\right)^2-x^2\)
\(=\left(x^2-x-3\right)\left(x^2+x-3\right)\)
c: \(=abx^2+aby^2-a^2xy-b^2xy\)
\(=\left(abx^2-b^2xy\right)+\left(aby^2-a^2xy\right)\)
\(=xb\left(ax-by\right)+ay\left(by-ax\right)\)
\(=\left(ax-by\right)\cdot\left(xb-ay\right)\)
\(a.10x\left(x-y\right)-6y\left(y-x\right)\\ =10x\left(x-y\right)+6y\left(x-y\right)\\ =\left(10x-6y\right)\left(x-y\right)\\ =2\left(5x-3y\right)\left(x-y\right)\)
\(b.14x^2y-21xy^2+28x^3y^2\\ =7xy\left(x-y+xy\right)\)
\(c.x^2-4+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2+x-2\right)\\ =2x\left(x-2\right)\)
\(d.\left(x+1\right)^2-25\\ =\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)
Trả lời:
1) sửa đề: \(x^4+x^3-4x-4=x^3\left(x+1\right)-4\left(x+1\right)=\left(x+1\right)\left(x^3-4\right)\)
2) \(x^2-\left(a+b\right)x+ab=x^2-ax-bx+ab=\left(x^2-ax\right)-\left(bx-ab\right)\)
\(=x\left(x-a\right)-b\left(x-a\right)=\left(x-a\right)\left(a-b\right)\)
3) \(5xy^3-2xyz-15y^2+6z=\left(5xy^3-15y^2\right)-\left(2xyz-6z\right)\)
\(=5y^2\left(xy-3\right)-2z\left(xy-3\right)=\left(xy-3\right)\left(5y^2-2z\right)\)
B1: a)\(xy\left(3x-2y\right)-2xy^2=3x^2y-2y^2x-2xy^2=3x^2y-4xy^2\)
b) \(\left(x^2+4x+4\right):\left(x+2\right)=\left(x+2\right)^2:\left(x+2\right)=\left(x+2\right)\)
\(\dfrac{2\left(x-1\right)}{x^2}.\dfrac{x}{\left(x-1\right)}=\dfrac{2\left(x-1\right)x}{x^2\left(x-1\right)}=\dfrac{2}{x}\)
B2:
a)\(2x^2-4x+2=2\left(x^2-2x+1\right)=2\left(x-1\right)^2\)
b)\(x^2-y^2+3x-3y=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)
Mấy bài này là mấy bài rất rất rất cơ bản, học sinh TB cũng phải tự làm được, mấy bài kiểu này đừng nên đăng lên hỏi nha:vv
a) 5x2 - 5xy + 7y - 7x = ( 5x2 - 5xy ) - ( 7x - 7y ) = 5x( x - y ) - 7( x - y ) = ( x - y )( 5x - 7 )
b) x2 - y2 + 2x + 1 = ( x2 + 2x + 1 ) - y2 = ( x + 1 )2 - y2 = ( x - y + 1 )( x + y + 1 )
c) 3x2 + 6xy + 3y2 - 3z2 = 3( x2 + 2xy + y2 - z2 ) = 3[ ( x2 + 2xy + y2 ) - z2 ] = 3[ ( x + y )2 - z2 ] = 3( x + y - z )( x + y + z )
d) ab( x2 + y2 ) + xy( a2 + b2 ) = abx2 + aby2 + a2xy + b2xy
= ( a2xy + abx2 ) + ( aby2 + b2xy )
= ax( ay + bx ) + by( ay + bx )
= ( ay + bx )( ax + by )