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4 tháng 9 2022
a: \(=3y^2-5x^2y^3-2y^2+3x^2y^3=y^2-2x^2y^3\)
b: \(=6x-y+2x^2+3y^2-2x^2+x=7x-y+3y^2\)
c: \(=x-y+4y^2-6xy+\dfrac{10x^2}{y}\)
23 tháng 9 2018
\(a.\left(9x^2y^3-15x^4y^4\right):3x^2y-\left(2-3x^2y\right)y^2\)
\(=3y^2-5x^2y^3-2y^2+3x^2y^3\)
\(=y^2-2x^2y^3\)
\(b.\left(6x^2-xy\right):x+\left(2x^3y+3xy^2\right):xy-\left(2x-1\right)x\)
\(=6x-y+2x^2+3y-2+x\)
\(=2x^2+7x+2y-2\)
\(c.\left(x^2-xy\right):x+\left(6x^2y^5-9x^3y^4+15x^4y^3\right):\dfrac{3}{2}x^2y^3\)
\(=x-y+4y^2-6xy+10x^2\)
4 tháng 9 2022
a: \(=\dfrac{27a^6b^3\cdot a^2b^6}{a^8b^8}=27b\)
b: \(=3y^2-5x^2y^3-2y^2+3x^2y^3\)
\(=y^2-2x^2y^3\)
c: \(=6x-y+2x^2+3y-2x^2+x\)
\(=7x+2y\)
d: \(=x-y+2y^2-6xy+\dfrac{10x^2}{y}\)
BA
21 tháng 10 2019
\(a,x^2-5x-xy+5y\)
\(=x\cdot\left(x-y\right)-5\cdot\left(x-y\right)\)
\(=\left(x-y\right)\cdot\left(x-5\right)\)
\(b,x^3+6x^2+9x\)
\(=x\cdot\left(x^2+6x+9\right)\)
\(=x\cdot\left(x+3\right)^2\)
\(c,x^2+x-2\)
\(=x^2-x+2x-2\)
\(=x\cdot\left(x-1\right)+2\cdot\left(x-1\right)\)
\(=\left(x-1\right)\cdot\left(x+2\right)\)
\(d,4x^2-\left(x^2+1\right)\)
\(=\left(2x-x^2-1\right)\cdot\left(2x+x^2+1\right)\)
\(=\left(2x-x^2-1\right)\cdot\left(x+1\right)^2\)
AH
Akai Haruma
Giáo viên
6 tháng 8 2017
1. \(x^3-x^2+x-1=(x^3-x^2)+(x-1)\)
\(=x^2(x-1)+(x-1)=(x^2+1)(x-1)\)
2. \(6x^2y-2xy^2+3x-y=2xy(3x-y)+(3x-y)\)
\(=(3x-y)(2xy+1)\)
3. \(4x^2+1\) thì còn cái gì để phân tích hả bạn? Hay ý bạn là \(4x^4+1\)?
\(4x^4+1=(2x^2)^2+1=(2x^2)^2+1+4x^2-4x^2\)
\(=(2x^2+1)^2-(2x)^2=(2x^2+1-2x)(2x^2+1+2x)\)
4. \(x^2-9x+8=(x^2-x)-(8x-8)\)
\(=x(x-1)-8(x-1)=(x-1)(x-8)\)
5. \(x^3-2x^2y+3xy^2=x(x^2-2xy+3y^2)\)
6. \(x^2-6x+y-y^2\) (sai đề)
7. \(x^2-xy-2x+2y=(x^2-xy)-(2x-2y)\)
\(=x(x-y)-2(x-y)=(x-y)(x-2)\)
A
24 tháng 7 2019
1/ \(\left(x+3\right)^2-\left(2x-5\right)\left(x+3\right).\)
\(=x^2+6x+9-\left(2x^2+6x-5x-15\right)\)
\(=x^2+6x+9-2x^2-6x+5x+15\)
\(=-x^2+5x+24\)
\(=-\left(x^2-5x-24\right)\)
\(=-\left(x^2-8x+3x-24\right)\)
\(=-\left[x\left(x-8\right)+3\left(x-8\right)\right]\)
\(=-\left(x-8\right)\left(x+3\right)\)
2/ \(x^2-xy+x-y\)
\(=\left(x^2+x\right)-\left(xy+y\right)\)
\(=x\left(x+1\right)-y\left(x+1\right)\)
\(=\left(x-y\right)\left(x+1\right)\)
3/ \(x^3+6x^2+9x\)
\(=x\left(x^2+6x+9\right)\)
\(=x\left(x+3\right)^2\)
B
24 tháng 7 2019
1,Bạn tự lm
\(2,x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x-1\right)\left(x+1\right)\)
\(3,x^3+6x^2+9x=x\left(x^2+6x+9\right)=x\left(x+3\right)^2\)
\(4,x^2y+xy+x+1=xy\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(xy+1\right)\)
\(5,ax+by+ay+bx=\left(ax+ay\right)+\left(by+bx\right)=a\left(x+y\right)+b\left(y+x\right)=\left(x+y\right)\left(a+b\right)\)
\(6,x^2-\left(a+b\right)x+ab=x^2-ax-bx+ab=\left(x^2-bx\right)-\left(ax-ab\right)=x\left(x-b\right)-a\left(x-b\right)=\left(x-b\right)\left(x-a\right)\)
24 tháng 5 2017
a)\(\frac{x^2+y^2-1+2xy}{x^2-y^2+1+2x}\)
\(\Leftrightarrow\frac{\left(x+y\right)^2-1}{\left(x+1\right)^2-y^2}\)
\(\Leftrightarrow\frac{\left(x+y+1\right)\left(x+y-1\right)}{\left(x+1-y\right)\left(x+1+y\right)}\)
\(\Leftrightarrow\frac{x+y-1}{x-y+1}\)
b)\(\frac{3x^3-6x^2y+xy^2-2y^3}{9x^5-18x^4y-xy^4+2y^5}\)
\(\Leftrightarrow\frac{3x^2\left(x-2y\right)+y^2\left(x-2y\right)}{9x^4\left(x-2y\right)-y^4\left(x-2y\right)}\)
\(\Leftrightarrow\frac{\left(3x^2+y^2\right)\left(x-2y\right)}{\left(9x^4-y^4\right)\left(x-2y\right)}\)
\(\Leftrightarrow\frac{3x^2+y^2}{\left(3x^2-y^2\right)\left(3x^2+y^2\right)}\)
\(\Leftrightarrow\frac{1}{3x^2-y^2}\)
\(x^3+6x^2+9x-xy^2\)
\(=x\left(x^2+6x+9-y^2\right)\)
\(=x\left[\left(x+3\right)^2-y^2\right]\)
\(=x\left(x+3-y\right)\left(x+3+y\right)\)
\(x^3+6x^2+9x-xy^2\)
\(=x\left(x^2+6x+9-y^2\right)\)
\(=x\left[\left(x^2+6x+9\right)-y^2\right]\)
\(=x\left[\left(x+3\right)^2-y^2\right]\)
\(=x\left(x+3+y\right)\left(x+3-y\right)\)