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a,xy(x+y)-y(x+y)^2+y^2(x-y) b,x(x+y)^2-y(x+y)^2+xy-x^2
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Những câu hỏi liên quan
MN
2
9 tháng 10 2023
a: \(=y\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(y-1\right)\)
b: \(=\left(x^2y^2-8-1\right)\left(x^2y^2-8+1\right)\)
\(=\left(x^2y^2-9\right)\left(x^2y^2-7\right)\)
\(=\left(xy-3\right)\left(xy+3\right)\left(x^2y^2-7\right)\)
c: \(=x^2-8x+x-8\)
\(=x\left(x-8\right)+\left(x-8\right)\)
\(=\left(x-8\right)\left(x+1\right)\)
T
9 tháng 10 2023
\(a,xy+y^2-x-y\)
\(=\left(xy+y^2\right)-\left(x+y\right)\)
\(=y\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(y-1\right)\)
\(---\)
\(b,\left(x^2y^2-8\right)^2-1\)
\(=\left(x^2y^2-8-1\right)\left(x^2y^2-8+1\right)\)
\(=\left[\left(xy\right)^2-9\right]\left(x^2y^2-7\right)\)
\(=\left(xy-3\right)\left(xy+3\right)\left(x^2y^2-7\right)\)
\(---\)
\(c,x^2-7x-8\)
\(=x^2+x-8x-8\)
\(=\left(x^2+x\right)-\left(8x+8\right)\)
\(=x\left(x+1\right)-8\left(x+1\right)\)
\(=\left(x+1\right)\left(x-8\right)\)
\(Toru\)
N
1
11 tháng 8 2023
b: =x^3+2x^2-x^2+4
=x^2(x+2)-(x+2)(x-2)
=(x+2)(x^2-x+2)
c: =x^3-2x^2+x^2-4
=x^2(x-2)+(x-2)(x+2)
=(x-2)(x^2+x+2)
d: =(x-y)(x+y)-7(x+y)
=(x+y)(x-y-7)
12 tháng 10 2021
Bài 2:
a: \(3x^2-3xy=3x\left(x-y\right)\)
b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)
c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)
d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)
30 tháng 7 2021
b) Ta có: \(x^3-x^2y-xy^2+y^3\)
\(=\left(x^3+y^3\right)-\left(x^2y+xy^2\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x+y\right)\left(x-y\right)^2\)
30 tháng 7 2021
a) Ta có: \(4\left(x-2\right)^2+xy-2y\)
\(=4\left(x-2\right)^2+y\left(x-2\right)\)
\(=\left(x-2\right)\left(4x-8+y\right)\)
b) Ta có: \(x\left(x-y\right)^3-y\left(y-x\right)^2-y^2\left(x-y\right)\)
\(=x\left(x-y\right)^3-y\left(x-y\right)^2-y^2\left(x-y\right)\)
\(=\left(x-y\right)\left[x\left(x-y\right)^2-y\left(x-y\right)-y^2\right]\)
TP
2
T
27 tháng 10 2023
\(a,x-xy+y-y^2\\=(x-xy)+(y-y^2)\\=x(1-y)+y(1-y)\\=(1-y)(x+y)\\---\\b,x^2-4x-y+4(?)\\---\\c,x^2-2x-3\\=x^2+x-3x-3\\=x(x+1)-3(x+1)\\=(x+1)(x-3)\)
Bạn xem lại đề câu b nhé!
12 tháng 10 2021
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
17 tháng 12 2023
a: \(2x^2+3xy-14y^2\)
\(=2x^2+7xy-4xy-14y^2\)
\(=\left(2x^2+7xy\right)-\left(4xy+14y^2\right)\)
\(=x\left(2x+7y\right)-2y\left(2x+7y\right)\)
\(=\left(2x+7y\right)\left(x-2y\right)\)
b: \(\left(x-7\right)\left(x-5\right)\left(x-3\right)\left(x-1\right)+7\)
\(=\left(x-7\right)\left(x-1\right)\left(x-5\right)\left(x-3\right)+7\)
\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)+7\)
\(=\left(x^2-8x\right)^2+15\left(x^2-8x\right)+7\left(x^2-8x\right)+105+7\)
\(=\left(x^2-8x\right)^2+22\left(x^2-8x\right)+112\)
\(=\left(x^2-8x\right)^2+8\left(x^2-8x\right)+14\left(x^2-8x\right)+112\)
\(=\left(x^2-8x\right)\left(x^2-8x+8\right)+14\left(x^2-8x+8\right)\)
\(=\left(x^2-8x+8\right)\left(x^2-8x+14\right)\)
c: \(\left(x-3\right)^2+\left(x-3\right)\left(3x-1\right)-2\left(3x-1\right)^2\)
\(=\left(x-3\right)^2+2\left(x-3\right)\left(3x-1\right)-\left(x-3\right)\left(3x-1\right)-2\left(3x-1\right)^2\)
\(=\left(x-3\right)\left[\left(x-3\right)+2\left(3x-1\right)\right]-\left(3x-1\right)\left[\left(x-3\right)+2\left(3x-1\right)\right]\)
\(=\left(x-3+6x-2\right)\left(x-3-3x+1\right)\)
\(=\left(7x-5\right)\left(-2x-2\right)\)
\(=-2\left(x+1\right)\left(7x-5\right)\)
d: \(xy\left(x-y\right)+yz\left(y-z\right)+zx\left(z-x\right)\)
\(=x^2y-xy^2+y^2z-yz^2+zx\left(z-x\right)\)
\(=\left(x^2y-yz^2\right)-\left(xy^2-y^2z\right)+xz\left(z-x\right)\)
\(=y\left(x^2-z^2\right)-y^2\left(x-z\right)-xz\left(x-z\right)\)
\(=y\cdot\left(x-z\right)\left(x+z\right)-\left(x-z\right)\left(y^2+xz\right)\)
\(=\left(x-z\right)\left(xy+zy-y^2-xz\right)\)
\(=\left(x-z\right)\left[\left(xy-y^2\right)+\left(zy-zx\right)\right]\)
\(=\left(x-z\right)\left[y\cdot\left(x-y\right)-z\left(x-y\right)\right]\)
\(=\left(x-z\right)\left(x-y\right)\left(y-z\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 10 2021
a.
\(2x^3-x^2y+x^2+y^2-2xy-y=0\)
\(\Leftrightarrow x^2\left(2x-y+1\right)-y\left(2x-y+1\right)=0\)
\(\Leftrightarrow\left(x^2-y\right)\left(2x-y+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-y=0\\2x-y+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=x^2\\y=2x+1\end{matrix}\right.\)
Thế vào pt đầu:
\(\left[{}\begin{matrix}x^3+x-2=0\\x\left(2x+1\right)+x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x^2+x+2\right)=0\\x^2+x-1=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 10 2021
b.
\(x^2-2xy+x=-y\)
Thế vào \(y^2\) ở pt dưới:
\(x^2\left(x^2-4y+3\right)+\left(x^2-2xy+x\right)^2=0\)
\(\Leftrightarrow x^2\left(x^2-4y+3\right)+x^2\left(x-2y+1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\Rightarrow y=0\\x^2-4y+3+\left(x-2y+1\right)^2=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x^2-4xy+2x+4y^2-8y+4=0\)
\(\Leftrightarrow2\left(x^2-2xy+x\right)+4y^2-8y+4=0\)
\(\Leftrightarrow-2y+4y^2-8y+4=0\)
\(\Leftrightarrow...\)
Lời giải:
a. $xy(x+y)-y(x+y)^2+y^2(x-y)$
$=y(x+y)[x-(x+y)]+y^2(x-y)$
$=y(x+y)(-y)+y^2(x-y)$
$=-y^2(x+y)+y^2(x-y)$
$=y^2(x-y)-y^2(x+y)=y^2[(x-y)-(x+y)]$
$=y^2(-2y)=-2y^3$
b.
$x(x+y)^2-y(x+y)^2+xy-x^2$
$=[x(x+y)^2-y(x+y)^2]-(x^2-xy)$
$=(x+y)^2(x-y)-x(x-y)$
$=(x-y)[(x+y)^2-x]=(x-y)(x^2+2xy+y^2-x)$
a: \(xy\left(x+y\right)-y\left(x+y\right)^2+y^2\left(x-y\right)\)
\(=\left(x+y\right)\left[xy-y\left(x+y\right)\right]+y^2\left(x-y\right)\)
\(=\left(x+y\right)\left(xy-xy-y^2\right)+y^2\left(x-y\right)\)
\(=y^2\left(-x-y\right)+y^2\left(x-y\right)\)
\(=y^2\left(-x-y+x-y\right)=-2y\cdot y^2=-2y^3\)
b: \(x\left(x+y\right)^2-y\left(x+y\right)^2+xy-x^2\)
\(=\left(x+y\right)^2\left(x-y\right)+x\left(y-x\right)\)
\(=\left(x+y\right)^2\cdot\left(x-y\right)-x\left(x-y\right)\)
\(=\left(x-y\right)\left[\left(x+y\right)^2-x\right]\)