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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: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)
\(=\frac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}\)
\(=\frac{x\left(x+y\right)+2y\left(x+y\right)}{\left(x^2-y^2\right)\left(x+2y\right)}\)
\(=\frac{\left(x+2y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x+2y\right)}=\frac{1}{x-y}\left(đpcm\right)\)
a)\(\frac{x^3-x}{3x+3}=\frac{x.\left(x^2-1\right)}{3.\left(x+1\right)}=\frac{x.\left(x-1\right).\left(x+1\right)}{3.\left(x+1\right)}=\frac{x.\left(x+1\right)}{3}=\frac{x^2+x}{3}\)
Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)
\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)
\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)
\(=\frac{16+x}{x^2-2x}-\frac{18}{x^2-2x}\)
\(=\frac{16+x-18}{x\left(x-2\right)}\)
\(=\frac{-2+x}{x\left(x-2\right)}\)
a) \(\frac{16+x}{x^2-2x}+\frac{18}{2x-x^2}=\frac{16+x-18}{x^2-2x}=\frac{x-2}{x\left(x-2\right)}=\frac{1}{x}\)
b) \(\frac{2y}{2x^2-xy}+\frac{4x}{xy-2x^2}=\frac{2y-4x}{2x^2-xy}=\frac{-2\left(2x-y\right)}{x\left(2x-y\right)}=\frac{-2}{x}\)
c) \(\frac{4-x^2}{x-3}+\frac{2x-2x^2}{3-x}+\frac{5-4x}{x-3}=\frac{4-x^2+2x^2-2x+5-4x}{x-3}=\frac{x^2-6x+9}{x-3}=\frac{\left(x-3\right)^2}{x-3}=x-3\)
a) \(\frac{3x+6}{x^2-4}=\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{3}{x-2}\)( ĐKXĐ : x ≠ ±2 )
\(\frac{2x+6}{x^3+3x^2-9x-27}=\frac{2\left(x+3\right)}{x^2\left(x+3\right)-9\left(x+3\right)}=\frac{2\left(x+3\right)}{\left(x+3\right)\left(x^2-9\right)}=\frac{2}{\left(x-3\right)\left(x+3\right)}\)( ĐKXĐ : x ≠ ±3 )
MTC : ( x - 2 )( x - 3 )( x + 3 )
=> \(\hept{\begin{cases}\frac{3}{x-2}=\frac{3\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{3\left(x^2-9\right)}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{3x-27}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}\\\frac{2}{\left(x-3\right)\left(x+3\right)}=\frac{2\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{4x-4}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}\end{cases}}\)
b) \(\frac{x^2-4x+4}{2x^2-3x+1}=\frac{\left(x-2\right)^2}{2x^2-2x-x+1}=\frac{\left(x-2\right)^2}{2x\left(x-1\right)-\left(x-1\right)}=\frac{\left(x-2\right)^2}{\left(x-1\right)\left(2x-1\right)}\)( ĐKXĐ : \(\hept{\begin{cases}x\ne1\\x\ne\frac{1}{2}\end{cases}}\))
\(\frac{x+4}{2x-2}=\frac{x+4}{2\left(x-1\right)}\)( ĐKXĐ : x ≠ 1 )
MTC : \(2\left(x-1\right)\left(2x-1\right)\)
=> \(\hept{\begin{cases}\frac{\left(x-2\right)^2}{\left(x-1\right)\left(2x-1\right)}=\frac{2\left(x^2-4x+4\right)}{2\left(x-1\right)\left(2x-1\right)}=\frac{2x^2-8x+8}{2\left(x-1\right)\left(2x-1\right)}\\\frac{x+4}{2\left(x-1\right)}=\frac{\left(x+4\right)\left(2x-1\right)}{2\left(x-1\right)\left(2x-1\right)}=\frac{2x^2+7x-4}{2\left(x-1\right)\left(2x-1\right)}\end{cases}}\)
c) \(\frac{6a}{a-b}\)( ĐKXĐ : a ≠ b ) ; \(\frac{2b}{b-a}=\frac{-2b}{a-b}\)( ĐKXĐ : a ≠ b) ; \(\frac{5}{a^2-b^2}=\frac{5}{\left(a-b\right)\left(a+b\right)}\)( ĐKXĐ : a ≠ ±b )
MTC : \(\left(a-b\right)\left(a+b\right)\)
=> \(\frac{6a}{a-b}=\frac{6a\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}=\frac{6a^2+6ab}{\left(a-b\right)\left(a+b\right)}\)
\(\frac{-2b}{a-b}=\frac{-2b\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}=\frac{-2ab-2b^2}{\left(a-b\right)\left(a+b\right)}\)
\(\frac{5}{a^2-b^2}=\frac{5}{\left(a-b\right)\left(a+b\right)}\)
d) \(\frac{x}{x^2+11x+30}=\frac{x}{x^2+5x+6x+30}=\frac{x}{x\left(x+5\right)+6\left(x+5\right)}=\frac{x}{\left(x+5\right)\left(x+6\right)}\)( ĐKXĐ : x ≠ -5 ; x ≠ -6 )
\(\frac{5}{x^2+9x+20}=\frac{5}{x^2+4x+5x+20}=\frac{5}{x\left(x+4\right)+5\left(x+4\right)}=\frac{5}{\left(x+4\right)\left(x+5\right)}\)( ĐKXĐ : x ≠ -4 ; x ≠ -5 )
MTC : \(\left(x+4\right)\left(x+5\right)\left(x+6\right)\)
=> \(\hept{\begin{cases}\frac{x}{\left(x+5\right)\left(x+6\right)}=\frac{x\left(x+4\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}=\frac{x^2+4x}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}\\\frac{5}{\left(x+4\right)\left(x+5\right)}=\frac{5\left(x+6\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}=\frac{5x+30}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}\end{cases}}\)
Sai chỗ nào bạn bỏ qua nhé
\(A=\left(\frac{4x}{x^2-4}+\frac{2x-4}{x+2}\right).\frac{x+2}{2x}+\frac{2}{2-x}\\=\left(\frac{4x}{x^2-4}+\frac{\left(2x-4\right)\left(x-2\right)}{x^2-4}\right)\frac{x+2}{2x}+\frac{2}{2-x}=\left(\frac{4x}{x^2-4}+\frac{2x^2-4x-4x+8}{x^2-4}\right) \frac{x+2}{2x}+\frac{2}{2-x}\)
\(=\left(\frac{4x+2x^2-8x+8}{x^2-4}\right).\frac{x+2}{2x}+\frac{2}{2-x}\\ =\frac{2x\left(x+2\right)-8\left(x-1\right)}{x^2-4}.\frac{x+2}{2x}+\frac{2}{2-x}\)
phân thức 1 :3x(x^2-4)/(2x+4)(x^2-4)=? ( tính ra )
phan thức 2 ;(2x+4)(x+3)/(2x+4)(x^2-4)=?(tính ra )
kết quả tính dc là phrp1 qd của 2 phân thức đó
\(2x+4=2\left(x+2\right)\)
\(x^2-4=x^2-2^2=\left(x+2\right)\left(x-2\right)\)
\(\Rightarrow MTC=2\left(x+2\right)\left(x-2\right)\)
\(\frac{3x}{2x+4}=\frac{3x}{2\left(x+2\right)}=\frac{3x\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}\)
\(\frac{x-3}{x^2-4}=\frac{x-3}{\left(x+2\right)\left(x-2\right)}=\frac{2\left(x-3\right)}{2\left(x+2\right)\left(x-2\right)}\)
a,\(\frac{2x^2+4x}{x+2}\)=\(\frac{2x\left(x+2\right)}{x+2}\)\(=2x\)
b, \(\frac{3x}{2x+4}\)=\(\frac{3x^2-6x}{2\left(x+2\right)\left(x-2\right)}\)
\(\frac{x+3}{x^2+4}\)=\(\frac{2x+6}{2\left(x-2\right)\left(x+2\right)}\)
tick mình nhé!!