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\(\dfrac{xy}{2}-x+\dfrac{x^2}{4}=x\left(\dfrac{y}{2}-1+\dfrac{x}{4}\right)\)
a: \(=-x^2y\cdot x+x^2y\cdot y=x^2y\left(-x+y\right)\)
b: \(=-xy^2\cdot x^2-xy^2\cdot z=-xy^2\left(x^2+z\right)\)
c: x^2y^3-xy^2
=xy^2*xy-xy^2
=xy^2(xy-1)
d: -x^3z-z
=z(-x^3-1)
=-z(x+1)(x^2-x+1)
e: =x(x-y)+(x-y)
=(x-y)(x+1)
n: =x^2(x-1)-(x-1)
=(x-1)(x^2-1)
=(x-1)^2(x+1)
2(x-y)2 -y(x-y)2 +xy2-x2y= 2(x-y)2-y(x-y)2+(xy^2-x^2y)=2(x-y)2-y(x-y)2+xy(x-y)=(x-y)\(\left[2\left(x-y\right)-y\left(x-y\right)+xy\right]\)=(x-y)(2x-2y-xy+y2+xy)=(x-y)(2x-2y+y2)
\(2\left(x-y\right)^2-y\left(x-y\right)^2+xy^2-x^2y\)
\(=\left(x-y\right)^2\left(2-y\right)+xy\left(y-x\right)\)
\(=\left(x-y\right)^2\cdot\left(2-y\right)-xy\left(x-y\right)\)
\(=\left(x-y\right)\left[\left(x-y\right)\left(2-y\right)-xy\right]\)
\(x\left(x-y\right)^2-y\left(x-y\right)^2+xy^2-x^2y\)
\(=\left(x-y\right)^2\left(x-y\right)-xy\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-2xy+y^2-xy\right)\)
\(=\left(x-y\right)\left(x^2-3xy+y^2\right)\)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
Phân tích thành nhân tử
\(=\left(my+nx\right)\left(ny+mx\right)\)
mn(x2 +y2) +xy(m2 +n2)= mnx2 +mny2 +xym2 +xyn2
=mx(nx + my) +ny( my +nx)
=(mx+ny)(nx+my)
Bài 4:
Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
mk ko bt 123