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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 )y3 - ( a - y )x3 + ( x - y )a3
= ay3 + a2y2 - ax2y - a2xy - a2y2 - a3y + a2x2 + a3x - xy3 - axy2 + x3y + ax2y + axy2 + a2xy - ax3 - a2x2
= ay( y2 +ay -x2 - ax ) - a2( y2 + ay -x2 -ax ) - xy( y2 + ay - x2 -ax ) + ax( y2 + ay -x2 -ax )
= ( y2 + ay - x2 - ax )( ay - a2 - xy + ax )
= ( y2 + xy +ay -xy -ax -x2 )[ ( y -a )a - x( y-a ) ]
= [ y( y +x +a ) - x( y + x + a )]( a - x )( a - y)
= ( y + x + a)( y -x )( a - x)( y - a)
b: Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-3\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-3\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+21\)
\(=\left(x^2+5x+3\right)\left(x^2+5x+7\right)\)
t chỉ cho kết quả thôi nhá, còn nhóm nhân tử you tự xử nhá !
=(x-y)(z-x)(z-y)(x+y+z)
\(\left(x-y\right)z^3+\left(z-z\right)y^3+\left(y-z\right)x^3\)
\(=z^3\left(x-y\right)+y^3\left(z-x\right)+x^3\left(y-z\right)\)
\(=xz^3-yz^3+\left(z-x\right)y^3+\left(y-z\right)x^3\)
\(=xz^3-yz^3+y^3z-xy^3+\left(y-z\right)x^3\)
\(=xz^3-yz^3+y^3z-xy^3+y^3z-xy^3+x^3y-x^3z\)
Mk ko chắc
\(a,\)
\(x^3y-y\)
\(=y\left(x^3-1\right)\)
\(=y\left[\left(x-1\right)\left(x^2+x+1\right)\right]\)
\(=y\left(x-1\right)\left(x^2+x+1\right)\)
\(b,\)
\(x^3y+y\)
\(=y\left(x^3+1\right)\)
\(=y\left[\left(x+1\right)\left(x^2-x+1\right)\right]\)
\(=y\left(x+1\right)\left(x^2-x+1\right)\)
\(c,\)
\(\left(x-y\right)^2-x\left(y-x\right)\)
\(=\left(x-y\right)^2+x\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y+x\right)\)
\(=\left(x-y\right)\left(2x-y\right)\)
bài này hơi căng đêy
đề \(=ay^3-xy^3-ax^3+x^3y+a^3x-a^3y\)
\(=ay^3+a^2y^2-ax^2y-a^2xy-a^2y^2-a^3y+a^2x^2+a^3x-xy^3-axy^2+x^3y+ax^2y+axy^2+a^2xy-ax^3-a^2x^2\)
\(=ay\left(y^2+ay-x^2-ax\right)-a^2\left(y^2+ay-x^2-ax\right)-xy\left(y^2+ay-x^2-ax\right)+ax\left(y^2+ay-x^2-ax\right)\)
\(=\left(y^2+ay-x^2-ax\right)\left(ay-a^2-xy+ax\right)\)
\(=\left(y^2+xy+ay-xy-x^2-ax\right)\left[a\left(y-a\right)-x\left(y-a\right)\right]\)
\(=\left[y\left(y+x+a\right)-x\left(y+x+a\right)\right]\left(a-x\right)\left(y-a\right)\)
\(=\left(y+x+a\right)\left(y-x\right)\left(a-x\right)\left(y-a\right)\)
khó quá nỏ biết