\(-\left(x-1\right)\left(-x^2+2y\right)\)
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22 tháng 12 2023
Bài 2:
1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)
=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)
=>(2x-1)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
2: \(9x^3-x=0\)
=>\(x\left(9x^2-1\right)=0\)
=>x(3x-1)(3x+1)=0
=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)
=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)
=>(2x-3)(2x-3-2)=0
=>(2x-3)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
=>\(2x^2+10x-5x-25-10x+25=0\)
=>\(2x^2-5x=0\)
=>\(x\left(2x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)
Bài 1:
1: \(3x^3y^2-6xy\)
\(=3xy\cdot x^2y-3xy\cdot2\)
\(=3xy\left(x^2y-2\right)\)
2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+3y-2\right)\)
3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)
\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)
\(=(x-2y)(3x-1+5x)\)
\(=\left(x-2y\right)\left(8x-1\right)\)
4: \(x^2-y^2-6y-9\)
\(=x^2-\left(y^2+6y+9\right)\)
\(=x^2-\left(y+3\right)^2\)
\(=\left(x-y-3\right)\left(x+y+3\right)\)
5: \(\left(3x-y\right)^2-4y^2\)
\(=\left(3x-y\right)^2-\left(2y\right)^2\)
\(=\left(3x-y-2y\right)\left(3x-y+2y\right)\)
\(=\left(3x-3y\right)\left(3x+y\right)\)
\(=3\left(x-y\right)\left(3x+y\right)\)
6: \(4x^2-9y^2-4x+1\)
\(=\left(4x^2-4x+1\right)-9y^2\)
\(=\left(2x-1\right)^2-\left(3y\right)^2\)
\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)
8: \(x^2y-xy^2-2x+2y\)
\(=xy\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-2\right)\)
9: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
TA
22 tháng 8 2017
\(B=\left[\left(\frac{x}{y}-\frac{y}{x}\right):\left(x-y\right)-2.\left(\frac{1}{y}-\frac{1}{x}\right)\right]:\frac{x-y}{y}\)
\(=\left[\frac{x^2-y^2}{xy}.\frac{1}{x-y}-2.\frac{x-y}{xy}\right].\frac{y}{x-y}\)
\(=\left(\frac{\left(x-y\right)\left(x+y\right)}{xy.\left(x-y\right)}-\frac{2.\left(x-y\right)}{xy}\right).\frac{y}{x-y}\)
\(=\left(\frac{x+y}{xy}-\frac{2x-2y}{xy}\right).\frac{y}{x-y}=\frac{x+y-2x+2y}{xy}.\frac{y}{x-y}=\frac{y.\left(3y-x\right)}{xy.\left(x-y\right)}=\frac{3y-x}{x.\left(x-y\right)}\)
\(C=\left(\frac{x+y}{2x-2y}-\frac{x-y}{2x+2y}-\frac{2y^2}{y-x}\right):\frac{2y}{x-y}\)
\(=\left(\frac{x+y}{2.\left(x-y\right)}-\frac{x-y}{2.\left(x+y\right)}+\frac{2y^2}{x-y}\right).\frac{x-y}{2y}\)
\(=\frac{\left(x+y\right)^2-\left(x-y\right)^2+2.2y^2.\left(x+y\right)}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}\)
\(=\frac{\left(x+y+x-y\right)\left(x+y-x+y\right)+4y^2.\left(x+y\right)}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}\)
\(=\frac{4xy+4xy^2+4y^3}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}=\frac{4y.\left(x+xy+y^2\right).\left(x-y\right)}{4y.\left(x-y\right)\left(x+y\right)}=\frac{x+xy+y^2}{x+y}\)
\(D=3x:\left\{\frac{x^2-y^2}{x^3+y^3}.\left[\left(x-\frac{x^2+y^2}{y}\right):\left(\frac{1}{x}-\frac{1}{y}\right)\right]\right\}\)
\(=3x:\left\{\frac{\left(x+y\right)\left(x-y\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}.\left[\frac{xy-x^2-y^2}{y}:\frac{y-x}{xy}\right]\right\}\)
\(=3x:\left[\frac{x-y}{x^2-xy+y^2}.\left(\frac{xy-x^2-y^2}{y}.\frac{xy}{y-x}\right)\right]\)
\(=3x:\left(\frac{x-y}{x^2-xy+y^2}.\frac{xy.\left(x^2-xy+y^2\right)}{y.\left(x-y\right)}\right)\)
\(=3x:\frac{xy.\left(x-y\right)\left(x^2-xy+y^2\right)}{y.\left(x-y\right)\left(x^2-xy+y^2\right)}=3x:x=3\)
\(E=\frac{2}{x.\left(x+1\right)}+\frac{2}{\left(x+1\right)\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+3\right)}\)
\(=2.\left(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}\right)\)
\(=2.\frac{\left(x+2\right)\left(x+3\right)+x.\left(x+3\right)+x.\left(x+1\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=2.\frac{x^2+2x+3x+6+x^2+3x+x^2+x}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=2.\frac{3x^2+9x+6}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=2.\frac{3.\left(x^2+3x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\frac{6.\left(x^2+x+2x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6.\left[x.\left(x+1\right)+2.\left(x+1\right)\right]}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\frac{6.\left(x+1\right)\left(x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6}{x.\left(x+3\right)}\)
22 tháng 12 2018
A.(x+2y).(x+2y-1) = x^2 +4xy + 4y^2 - x - 2y
B. (x-2y).(x+2y-1) = x^2 - x - 4y^2 + 2y
C. (x-2y).(x-2y+1) = x^2 - 4xy + 4y^2 + x - 2y
D.(x+2y).(x-2y) = x^2 - 4y^2
=>....
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 12 2021
Hệ này không giải được em nhé
Phương trình dưới phải là:
\(...+6x\left(x+1\right)+2=0\) mới giải được
Khi đó pt dưới sẽ phân tích được thành:
\(2\left(x+1\right)^3+3\left(x+1\right)^2y+4y^3=0\)
Dạng pt đẳng cấp khá cơ bản
\(-\left(x-1\right)\left(-x^2+2y\right)\)
\(=\left(x-1\right)\left(x^2-2y\right)\)
\(=x^3-2xy-x^2+2y\)
\(-(x-1)(-x^2+2y)\)
\(=-\left\lbrack x(-x^2+2y)-1(-x^2+2y)\right\rbrack\)
\(=-\left\lbrack-x^3+2xy+x^2-2y\right\rbrack\)
\(=x^3-x^2-2xy+2y\)