Mn giúp e với ạ
Quy đồng mẫu thức các phân thức sau:
x+5/x^2+6x+9; x+2/2x+6
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\(\dfrac{3x}{3x^3-27x}=\dfrac{3x}{3x\left(x^2-9\right)}=\dfrac{1}{\left(x-3\right)\left(x+3\right)}=\dfrac{x+2}{\left(x-3\right)\left(x+3\right)\left(x+2\right)}\\ \dfrac{x+1}{\left(x+2\right)\left(x+3\right)}=\dfrac{\left(x+1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x+2\right)}\)
Lời giải:
$\frac{4x^2-3x+8}{x^3-1}$
$\frac{2x}{x^2+x+1}=\frac{2x(x-1)}{(x-1)(x^2+x+1)}=\frac{2x^2-2x}{x^3-1}$
$\frac{6}{1-x}=\frac{-6(x^2+x+1)}{(x-1)(x^2+x+1)}=\frac{-6x^2-6x-6}{x^3-1}$
\(a,\dfrac{7x-1}{2x^2+6x}=\dfrac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}=\dfrac{7x^2-22x+3}{2x\left(x-3\right)\left(x+3\right)}\\ \dfrac{5-3x}{x^2-9}=\dfrac{2x\left(5-3x\right)}{2x\left(x-3\right)\left(x+3\right)}=\dfrac{10x-6x^2}{2x\left(x-3\right)\left(x+3\right)}\)
Ta có: 2 x 2 + 6 x = 2 x x + 3 ; x 2 - 9 = x + 3 x - 3
Mẫu thức chung: 2x(x + 3)(x – 3)
MT1: 2 x 2 + 6 x = 2 x ( x + 3 ) ; MT2: x 2 - 9 = ( x - 3 ) ( x + 3 )
MTC: 2x(x - 3)(x + 3)
NTP1: x – 3; NTP2: 2x
Quy đồng:
a) MTC: \(12x^3y^3\)
\(\dfrac{3}{4x^3y^2}=\dfrac{3\cdot3y}{4x^3y^2\cdot3y}=\dfrac{9y}{12x^3y^3}\)
\(\dfrac{2}{3xy^3}=\dfrac{2\cdot4x^2}{3xy^3\cdot4x^2}=\dfrac{8x^2}{12x^3y^3}\)
b) MTC: \(x\left(x-3\right)^2\)
\(\dfrac{5}{x^2-6x+9}=\dfrac{5}{\left(x-3\right)^2}=\dfrac{5x}{x\left(x-3\right)^2}\)
\(\dfrac{3}{x^2-3x}=\dfrac{3}{x\left(x-3\right)}=\dfrac{3\left(x-3\right)}{x\left(x-3\right)^2}=\dfrac{3x-9}{x\left(x-3\right)^2}\)