Toán Lớp 8: Cho biểu thức: P= [x/(x+2)-(x-3)/(x+2)+(3x+2)/((x+2).(x-2))]. (x^2 +x-6)/(4x-16)
Hãy rút gọn P
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Toán Lớp 8: Cho biểu thức: P= [x/(x+2)-(x-3)/(x+2)+(3x+2)/((x+2).(x-2))]. (x^2 +x-6)/(4x-16)
Hãy rút gọn P
Comments ( 2 )
P=[x/(x+2)-(x-3)/(x+2)+(3x+2)/((x+2).(x-2))].(x^2+x-6)/(4x-16)
=[(x(x-2))/((x+2)(x-2))-((x-3)(x-2))/((x+2)(x-2))+(3x+2)/((x+2).(x-2))].(x^2+x-6)/(4x-16)
=[(x^2-2x-(x^2-2x-3x+6)+3x+2)/((x+2)(x-2))].(x^2-2x+3x-6)/(4x-16)
=(x^2-2x-x^2+5x-6+3x+2)/((x+2)(x-2)).(x(x-2)+3(x-2))/(4x-16)
=(6x-4)/((x+2)(x-2)).((x-2)(x+3)/(4x-16)
=(2(3x-2))/((x+2)(x-2)).((x-2)(x+3)/(4(x-4))
=((3x-2)(x+3))/((x+2)2(x-4))
=((3x-2)(x+3))/(2(x+2)(x-4))
Vậy P=((3x-2)(x+3))/(2(x+2)(x-4)).
$P=\left(\dfrac{x}{x+2}-\dfrac{x-3}{x+2}+\dfrac{3x+2}{\left(x+2\right)\left(x-2\right)}\right)\dfrac{x^2+x-6}{4x-16}$ $;(x≠2)$
$=\left(\dfrac{3}{x+2}+\dfrac{3x+2}{\left(x+2\right)\left(x-2\right)}\right)\dfrac{x^2+x-6}{4x-16}$
$=\dfrac{\left(\dfrac{3}{x+2}+\dfrac{3x+2}{\left(x+2\right)\left(x-2\right)}\right)\left(x^2+x-6\right)}{4x-16}$
$=\dfrac{\dfrac{\left(6x-4\right)\left(x+3\right)}{x+2}}{4x-16}$ $=\dfrac{\left(6x-4\right)\left(x+3\right)}{\left(x+2\right)\left(4x-16\right)}$
$=\dfrac{2\left(3x-2\right)\left(x+3\right)}{\left(x+2\right)\left(4x-16\right)}$
$=\dfrac{2\left(3x-2\right)\left(x+3\right)}{\left(x+2\right)\cdot \:4\left(x-4\right)}$
$=\dfrac{2\left(3x-2\right)\left(x+3\right)}{\left(x+2\right)\cdot \:2\cdot \:2\left(x-4\right)}$
$=\dfrac{\left(3x-2\right)\left(x+3\right)}{\left(x+2\right)\cdot \:2\left(x-4\right)}$
$=\dfrac{\left(3x-2\right)\left(x+3\right)}{2\left(x+2\right)\left(x-4\right)}$