Toán 8 Thực hiện phép tính

Bài 2:
a) $\dfrac{x^3-x}{3x+3} \ \ (DK: x \neq -1)$
$= \dfrac{x(x^2-1)}{3(x+1)} \\
= \dfrac{x(x-1)(x+1)}{3(x+1)} \\
= \dfrac{x(x-1)}{3}$
b) $\dfrac{x^2+3xy}{x^2-9y^2} \ \ (DK: x-3y \neq 0, x+3y \neq 0)$
$=\dfrac{x(x+3y)}{(x-3y)(x+3y)} \\
= \dfrac{x}{x-3y}$

Bài 3:
a) $\dfrac{x}{x-3} + \dfrac{9-6x}{x^2-3x} \ \ (DK: x \neq 0; \ x \neq 3)$
$= \dfrac{x^2+9-6x}{x(x-3)} \\
= \dfrac{(x-3)^2}{x(x-3)} \\
= \dfrac{x-3}{x}$
b) $\dfrac{6x-3}{x} : \dfrac{4x^2-1}{3x^2} \ \ \left( DK: x \neq \left\{ - \dfrac{1}{2} ; 0 ; \dfrac{1}{2} \right \} \right ) $
$= \dfrac{6x-3}{x} . \dfrac{3x^2}{4x^2-1} \\
=\dfrac{3(2x-1)}{x}. \dfrac{3x^2}{(2x-1)(2x+1)} \\
= \dfrac{9x}{2x+1}$

Lê Thị Minh Tâm said:

Bài 1: Tính:
a) (x^2-9)/(2x+6) : (3-x)/2
b) (2x)/(x-y) - (2y)/(x-y)
c) (x+15)/(x^2-9) + 2/(x+3)
d)(x+y)/(2x+2y) - (x-y)/(2x+2y) - (y^2+x^2)/(y^2-x^2)
Bài 2: Rút gọn:
a) (x^3-x)/(3x+3)
b) (x^2+3xy)/(x^2-9y^2)
Bài 3: Thực hiện phép tính:
a) x/(x-3) + (9-6x)/(x^2-3x)
b) (6x-3)/x : (4x^2-1)/(3x^2)

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Bài 1:
a) (x^2-9)/(2x+6) : (3-x)/2 (ĐK:x khác + - 3)
=[(x-3)(x+3)] / [2(x+3)] * (-2)/(x-3)
=(x-3)/ 2* (-2)/(x-3)= -1

b) (2x)/(x-y) - (2y)/(x-y) (ĐK: x khác y)
=(2x-2y)/(x-y)
=[2(x-y)] / (x-y)
=2

c) (x+15)/(x^2-9) + 2/(x+3) (ĐK:x khác + - 3)
= (x+15) / [(x-3)(x+3)] + 2(x-3) / [(x-3)(x+3)]
=(x+15+ 2x-6)/ [(x-3)(x+3)]
= (3x+9)/ [(x-3)(x+3)]
=[3(x+3)] /[(x-3)(x+3)]
=3/(x-3)

d)(x+y)/(2x+2y) - (x-y)/(2x+2y) - (y^2+x^2)/(y^2-x^2) (ĐK : x khác + - y)
=(x+y- x+y)/(2x+2y)- (y^2+x^2)/(y^2-x^2)
=y/(x+y)- (y^2+x^2)/ [(y-x)(x+y)]
=y(x-y)/[(x+y)(x-y)]+ (y^2+x^2)/ [(x-y)(x+y)]
= (xy -y^2 +y^2+x^2) / [(x-y)(x+y)]
=(xy+x^2)/ [(x-y)(x+y)]
=[x(x+y)]/ [(x-y)(x+y)]
= x /(x-y)