Question

[Minh tú Trần]

Quy đồng mẫu thức các phân thức sau 

a) \(\frac{3x+6}{x^2-4}\); \(\frac{2x+6}{x^3+3x^2-9x-27}\)

b) \(\frac{x^2-4x+4}{2x^2-3x+1}\); \(\frac{x+4}{2x-2}\)

c) \(\frac{6a}{a-b}\); \(\frac{2b}{b-a}\); \(\frac{5}{a^2-b^2}\)

d) \(\frac{x}{x^2+11x+30}\); \(\frac{5}{x^2+9x+20}\)

Answer 1

[l҉o҉n҉g҉ d҉z҉]

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 : \(\left(x+4\right)\left(x+5\right)\left(x+6\right)\)

=> \(\hept{\begin{cases}\frac{x}{\left(x+5\right)\left(x+6\right)}=\frac{x\left(x+4\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}=\frac{x^2+4x}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}\\\frac{5}{\left(x+4\right)\left(x+5\right)}=\frac{5\left(x+6\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}=\frac{5x+30}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}\end{cases}}\)

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