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a) 3x^2 y - 6xy^2 = 3xy ( x - 2y)
b) 9 - ( x- y)^2 = ( 3 )^2 - ( x- y)^2
= ( 3 -x + y )( 3 + x + y )
a/ \(3x^2y-6xy^2\)\(=3xy\left(x-2y\right)\) ( đây là p2 đặt nhân tử chung )
b/9-(x -y )2 =( 3 -x +y ) ( 3 + x+y ) ( dùng hđt số 3 để giải )
c/ 3x-9xy-9y2+6y-1
=3x.(1-3y)-(1-6y+9y2)
=3x.(1-3y)-(1-3y)2
=(1-3y)[3x-(1-3y)]
=(1-3y)(3x-1+3y)
d/ x4+x2y2+y4
=x4+2x2y2+y4-x2y2
=(x2+y2)2-x2y2
=(x2-xy+y2)(x2+xy+y2)
1. = (x-2)^2 - y^2 = (x - 2 - y)(x-2+y)
2. = (x-y-x-y)(x-y+x+y) = 2(-y)2x = -4xy
Phân tích đa thức thành nhân tử:(em làm luôn đấy,ko ghi lại đề)
\(\left(x^3+y^3\right)-\left(x+y\right)+3xy\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)+3xy\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)\(=\left(x+y\right)\left[\left(x+y\right)^2-1^2\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
\(8x^3+12x^2+6x+1=0.\)
\(\Leftrightarrow\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3=0\)
\(\Leftrightarrow\left(2x+1\right)^3=0\)
\(\Leftrightarrow2x+1=0\)
\(\Leftrightarrow x=-\frac{1}{2}\)
\(2x^2+5x-3=0\Leftrightarrow\left(2x^2+6x\right)+\left(-x-3\right)=0\)
\(\Leftrightarrow2x\left(x+3\right)-\left(x+3\right)=0\Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}\)
\(x^2-2x-3=0\Leftrightarrow\left(x^2-3x\right)+\left(x-3\right)=0\)
\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}.}\)
\(\left(5x-1\right)+2\left(1-5x\right)\left(4+5x\right)+\left(5x+4\right)^2\)
\(=5x-1+2\left(4+5x-20x-25x^2\right)+25x^2+40x+16\)
\(=25x^2+45x+15+8+10x-40x-50x^2\)
\(=-25x^2+15x+23\)
\(\left(x-y\right)^3+\left(y+x\right)^3+\left(y-x\right)^3-3xy\left(x+y\right)\)
\(=\left(x-y\right)^3-\left(x-y\right)^3+\left(x+y\right)^3-3x^2y-3xy^2\)
\(=\left(x+y\right)^3-3x^2y-3xy^2\)
\(=x^3+3x^2y+3xy^2+y^3-3xy^2-3x^2y\)
\(=x^3+y^3\)
x3-x+3x2y+3xy2+y3-y
=x2(x-1)+3(x2y+xy2)+y2(y-1)
=x2(x-1)+3(x2.y+y2.x)+y2(y-1)
=x2(x-1)+3{[x(x+1)+y(y+1)]}+y2(y-1)
=x2(x-1)+3.x(x+1)+3.y(y+1)+y2(y-1)
=x2(x-1)+2x2+3.x(x+1)+3.y(y+1)+y2(y-1)+2y2-2x2-2y2
=x2(x+1)+3.x(x+1)+3.y(y+1)+y2(y+1)-2x2-2y2
=(x2+3)(x+1)+(y2+3)(y+1)-2(x2+y2)
ta có : (x*3+3x*2y+3xy*2+y*3)-(x+y)
=(x+y)*3-(x+y)
=(x+y)((X+Y)*2-1)
(x+y)(x+y+1)(x+Y-1)