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\(x^2+y^2-x^2y^2+xy-x-y\)
\(\Leftrightarrow x^2\left(1-y\right)\left(1+y\right)-y\left(1-y\right)-x\left(1-y\right)\)
\(\Leftrightarrow\left(1-y\right)\left(x^2+x^2y-y-x\right)\)
\(\Leftrightarrow\left(1-y\right)\left(x+y\right)\left(x-1\right)\left(x+1\right)\)
x(y - z) + 2(z - y)
= x(y - z) - 2(y - z)
= (x - 2)(y - z)
(2x - 3y)(x - 2) - (x + 3)(3y - 2x)
= (2x - 3y)(x - 2) + (x + 2)(2x - 3y)
= (2x - 3y)(x - 2 + x + 2)
= 2x(2x - 3y)
Bài 2
\(a,x^3+2x^2+x\)
\(=x.\left(x^2+2x+1\right)\)
\(b,xy+y^2-x-y\)
\(=y.\left(x+y\right)-\left(x+y\right)\)
\(=\left(y-1\right).\left(x+y\right)\)
bài 3
\(a,3x.\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\x^2=4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=2,x=-2\end{cases}}\)
vậy x=0,x=2 hay x=-2
\(b,xy+y^2-x-y=0\)
\(y.\left(x+y\right)-\left(x+y\right)=0\)
\(\left(y-1\right).\left(x+y\right)=0\)
\(\Rightarrow\orbr{\begin{cases}y-1=0\\x+y=0\end{cases}\Rightarrow\orbr{\begin{cases}y=1\\x=-1\end{cases}}}\)
vậy x=-1, y=1
\(x^2+5x-6\)
\(\Leftrightarrow x^2-x+6x-6\)
\(\Leftrightarrow x\left(x-1\right)+6\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+6\right)\)
Chúc bạn học tốt
\(x^3-x+3x^2y+xy^2+y^3-y\)
\(=\left(x^3+3x^2y+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
Trả lời :
Ta có :
\(x^2+2xy+7x+7y+y^2+10\)
\(=\left(x^2+2xy+y^2\right)+\left(7x+7y\right)+10\)
\(=\left(x+y\right)^2+7\left(x+y\right)+10\)
\(=\left(x+y\right)\left(x+y+2\right)+5\left(x+y+2\right)\)
\(=\left(x+y+2\right)\left(x+y+5\right)\)
Hok tốt
a) \(x^2+2xy+7x+7y+y^2+10\)
\(=\left(x^2+2xy+y^2\right)+\left(7x+7y\right)+10\)
\(=\left(x+y\right)^2+7\left(x+y\right)+10\)
\(=\left(x+y\right)^2+2\left(x+y\right)+5\left(x+y\right)+10\)
\(=\left(x+y+2\right)\left(x+y+5\right).\)
b) \(x^2y+xy^2+x+y=2010\)
\(\Leftrightarrow xy\left(x+y\right)+\left(x+y\right)=2010\)
\(\Leftrightarrow11\left(x+y\right)+1\left(x+y\right)=2010\)
\(\Leftrightarrow12\left(x+y\right)=2010\)
\(\Leftrightarrow x+y=\frac{335}{2}\)
\(\Leftrightarrow\left(x+y\right)^2=\frac{112225}{4}\)
\(\Leftrightarrow x^2+2xy+y^2=\frac{112225}{4}\)
\(\Leftrightarrow x^2+y^2+22=\frac{112225}{4}\)
\(\Leftrightarrow x^2+y^2=\frac{112137}{4}.\)
Vậy \(x^2+y^2=\frac{112137}{4}.\)
\(x^2-3x+xy-3y\)
\(=x\left(x+y\right)-3\left(x+y\right)\)
\(=\left(x+y\right)\left(x-3\right)\)
\(x^2-2xy+y^2-4=\left(x-y\right)^2-2^2=\left(x-y-2\right)\left(x-y+2\right)\)
\(x^2+x-y^2+y=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x+y\right)\left(x-y+1\right)\)
\(y\left(x-y\right)^2+xy\left(x-y\right)\)
\(=\left(xy-y^2\right)\left(x-y\right)+xy\left(x-y\right)\)
\(=\left(xy-y^2+xy\right)\left(x-y\right)\)
\(=\left(2xy-y^2\right)\left(x-y\right)\)
y ( x - y)2 + xy ( x-y) = (x - y) [(x-y) y +xy]
= (x-y) ( 2xy -y2)
\(x^3-x^2y-xy^2+y^3\)
\(=x^2\left(x-y\right)-y^2\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)^2.\left(x+y\right)\)
Chúc bn học giỏi nhoa!!!
\(=x^2\left(x-y\right)-y^2\left(x-y\right)\)( do khi ta đóng ngoặc 2 hạng tử cuôi phải đổi dấu trước dấu trừ);
\(=\left(x^2-y^2\right)\left(x-y\right)=\left(x-y\right)\left(x+y\right)\left(x-y\right)\)
\(=\left(x-y\right)^2\left(x+y\right)\)
ủng hộ nha bạn...
\(y-x^2y-xy^2-y^2\)
\(=\left(y-x^2y\right)-\left(xy^2+y^2\right)\)
\(=y\left(1-x^2\right)-y^2\left(x+1\right)\)
\(=y\left(1-x\right)\left(x+1\right)-y^2\left(x+1\right)\)
\(=\left(x+1\right)\left[y\left(1-x\right)-y^2\right]\)
\(=\left(x+1\right)\left(y-xy-y^2\right)\)
\(=\left(x+1\right)\left[y-\left(xy+y^2\right)\right]\)
\(=y\left(x+1\right)\left(1-x-y\right)\)