Toán Lớp 8: A = x-3/x-2 B = x/x-2 – 1/x+2 – x^2/x^2 -4
C = B/A = ?
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Toán Lớp 8: A = x-3/x-2 B = x/x-2 – 1/x+2 – x^2/x^2 -4
C = B/A = ?
Comments ( 2 )
ĐK : x$\neq ±2$
=$\dfrac{x(x+2)}{ x²-4}$ $-\dfrac{x-2}{x^{2}-4}$ $-\dfrac{x² }{x^{2}-4 }$
= $\dfrac{x²+2x-x+2-x²}{ x^{2}-4 }$
=$\dfrac{x+2}{ x^{2}-4}$
=$\dfrac{1}{x-2}$
TA có : $C=\dfrac{B}{A}$ ( x$\neq ±2$)
hay $\dfrac{1}{x-2}$ :$\dfrac{x-3}{x-2}$
=$\dfrac{1}{x-2}$ $.\dfrac{x-2}{x-3}$
=$\dfrac{1}{x-3}$
Vậy…
B= x/(x-2)-1/(x+2)-x^2/(x^2-4)
<=>B=(x(x+2))/((x-2)(x+2))-(x-2)/((x-2)(x+2))-x^2/((x-2)(x+2))
<=>B=(x(x+2)-x+2-x^2)/((x-2)(x+2))
<=>B=(x^2+2x-x+2-x^2)/((x-2)(x+2))
<=>B=(x+2)/((x-2)(x+2))
<=>B=1/(x-2)
Mà C=B/A
->C=((1/(x-3)/(x-3)/(x-2)))
<=>C=1/(x-2):(x-3)/(x-2)
<=>C=1/(x-2).(x-2)/(x-3)
<=>C=1/(x-3)
Vậy : C=1/(x-3)