Toán Lớp 8: a, $\frac{9x^2-7}{3x-2}$ – $\frac{6-9x^2}{2-3x}$ + $\frac{5-6x}{3x-2}$
b, $\frac{x-3}{x-2}$ – $\frac{1-x}{x+2}$ – $\frac{8x-x^2+8}{4-x^2}$
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Toán Lớp 8: a, $\frac{9x^2-7}{3x-2}$ – $\frac{6-9x^2}{2-3x}$ + $\frac{5-6x}{3x-2}$
b, $\frac{x-3}{x-2}$ – $\frac{1-x}{x+2}$ – $\frac{8x-x^2+8}{4-x^2}$
Comments ( 2 )
Giải đáp:
a) -2 với x\ne2/3
b) (x+2)/(x-2) với x\ne±2
Lời giải và giải thích chi tiết:
a) (9x^2-7)/(3x-2)-(6-9x^2)/(2-3x)+(5-6x)/(3x-2) (đkxđ: x\ne2/3)
=(9x^2-7)/(3x-2)+(6-9x^2)/(3x-2)+(5-6x)/(3x-2)
=(9x^2-7+6-9x^2+5-6x)/(3x-2)
=(-6x-4)/(3x+2)
=-(2(3x+2))/(3x+2)
=-2
b) (x-3)/(x-2)-(1-x)/(x+2)-(8x-x^2+8)/(4-x^2) (đkxđ: x\ne±2)
=(x-3)/(x-2)+(x-1)/(x+2)+(8x-x^2+8)/(x^2-2^2)
=(x-3)/(x-2)+(x-1)/(x+2)+(8x-x^2+8)/((x-2)(x+2))
=((x-3)(x+2)+(x-1)(x-2)+8x-x^2+8)/((x-2)(x+2))
=(x^2+2x-3x-6+x^2-2x-x+2+8x-x^2+8)/((x-2)(x+2))
=(x^2+4x+4)/((x-2)(x+2))
=(x^2+2*x*2+2^2)/((x-2)(x+2))
=(x+2)^2/((x-2)(x+2))
=(x+2)/(x-2)
$a,=\dfrac{9x^2-7+6-9x^2+5-6x}{3x-2}(x\neq\dfrac{2}{3})\\ =\dfrac{-6x+4}{3x-2}=\dfrac{-2(3x-2)}{3x-2}=-2\\ b,=\dfrac{(x-3)(x+2)+(x-1)(x-2)+8x-x^2+8}{(x-2)(x+2)}(x\neq -2;2)\\ =\dfrac{x^2-x-6+x^2-3x+2+8x-x^2+8}{(x-2)(x+2)}\\ =\dfrac{x^2+4x+4}{(x-2)(x+2)}=\dfrac{(x+2)^2}{(x-2)(x+2)}\\ =\dfrac{x+2}{x-2}$