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a/ (\(x^3y^2\)-\(\frac{1}{2}x^3y\) + \(2xy\) - \(2x^2y^3\) + \(xy^2\) - \(4y^2\) =
Ta có: \(\frac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}\)
\(=\frac{x^2y+xy^2+xy^2+y^3}{2x^2+2xy-xy-y^2}\)
\(=\frac{xy\left(x+y\right)+y^2\left(x+y\right)}{2x\left(x+y\right)-y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)\left(xy+y^2\right)}{\left(2x-y\right)\left(x+y\right)}=\frac{xy+y^2}{2x-y}\left(đpcm\right)\)
Ta có: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)
\(=\frac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}\)
\(=\frac{x\left(x+y\right)+2y\left(x+y\right)}{\left(x^2-y^2\right)\left(x+2y\right)}\)
\(=\frac{\left(x+2y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x+2y\right)}=\frac{1}{x-y}\left(đpcm\right)\)
\(C1:=3+1-3y\)
\(=4-3y\)
\(C2:\)
\(a.=3x\left(2y-1\right)\)
\(b.=\left(x-y\right)\left(x+y\right)+4\left(x+y\right)\)
\(=\left(x-y+4\right)\left(x+y\right)\)
\(C3:\)
\(a.6x^2+2x+12x-6x^2=7\)
\(14x=7\)
\(x=\frac{1}{2}\)
\(b.\frac{1}{5}x-2x^2+2x^2+5x=-\frac{13}{2}\)
\(\frac{26}{5}x=-\frac{13}{2}\)
\(x=-\frac{13}{2}\times\frac{5}{26}\)
\(x=-\frac{5}{4}\)
Bạn Moon làm kiểu gì vậy ?
1) \(\left(3x^2y^2+x^2y^2\right):\left(x^2y^2\right)-3y\)
\(=\left[\left(x^2y^2\right)\left(3+1\right)\right]:\left(x^2y^2\right)-3y\)
\(=4-3y\)
2) a, \(6xy-3x=\left(3x\right)\left(2y-1\right)\)
b, \(x^2-y^2+4x+4y=\left(x+y\right)\left(x-y\right)+4\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+4\right)\)
3) a, \(2x\left(3x+1\right)+\left(4-2x\right)3x=7\)
\(< =>6x^2+2x+12x-6x^2=7\)
\(< =>14x=7< =>x=\frac{7}{14}\)
b, \(\frac{1}{2}x\left(\frac{2}{5}-4x\right)+\left(2x+5\right)x=-6\frac{1}{2}\)
\(< =>\frac{x}{2}.\frac{2}{5}-\frac{x}{2}.4x+2x^2+5x=-\frac{13}{2}\)
\(< =>\frac{x}{5}-2x^2+2x^2+5x=-\frac{13}{2}\)
\(< =>\frac{26x}{5}=\frac{-13}{2}\)
\(< =>26x.2=\left(-13\right).5\)
\(< =>52x=-65< =>x=-\frac{65}{52}=-\frac{5}{4}\)
\(\frac{2\left(x+y\right)\left(x-y\right)}{x}-\frac{-2y^2}{x}\)
\(=\frac{2\left(x^2-y^2\right)+2y^2}{x}\)
\(=\frac{2x^2-2y^2+2y^2}{x}\)
\(=\frac{2x^2}{x}\)
\(=2x\)
Làmmmm
1/ \(\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\)(ĐKXĐ:x\(\ne0\), x\(\ne\frac{1}{2}\))
= \(\frac{\left(1-2x\right)\left(2x-1\right)}{2x\left(2x-1\right)}+\frac{4x^2}{\left(2x-1\right)2x}-\frac{1}{2x\left(2x-1\right)}\)
\(=\frac{2x-1-4x^2+2x+4x^2-1}{2x\left(2x-1\right)}\)
\(=\frac{4x-2}{2x\left(2x-1\right)}=\frac{2\left(2x-1\right)}{2x\left(2x-1\right)}=\frac{1}{x}\)
KL:..............
2/\(\frac{x^2+2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{1-x}\)(ĐKXĐ : x\(\ne1\))
\(=\frac{x^2+2}{x^3-1}+\frac{2x-2}{x^3-1}-\frac{x^2+x+1}{x^3-1}\)
\(=\frac{x^2+2+2x-2-x^2-x-1}{x^3-1}=\frac{x-1}{x^3-1}=\frac{1}{x^2+x+1}\)
Kl:....................
\(\frac{x^3-y^3}{2y}.\left[\frac{2y}{4-2y-2x+xy}+\frac{2xy+4y}{\left(x-y\right)\left(x^2-4\right)}\right]\)
\(=\frac{x^3-y^3}{2y}.\left[\frac{2y}{\left(x-2\right)\left(y-2\right)}+\frac{2y\left(x+2\right)}{\left(x-y\right)\left(x-2\right)\left(x+2\right)}\right]\)
\(=\left(x^3-y^3\right).\left[\frac{x-y}{\left(x-y\right)\left(x-2\right)\left(y-2\right)}+\frac{y-2}{\left(x-y\right)\left(x-2\right)\left(y-2\right)}\right]\)
\(=\frac{\left(x-y\right)\left(x^2+xy+y^2\right)\left(x-2\right)}{\left(x-y\right)\left(x-2\right)\left(y-2\right)}\)
\(=\frac{x^2+xy+y^2}{y-2}\)