Toán Lớp 8: Tính a) x-8/x-3 + 1/x-3 – 4/3-x b) x+3/ x^2-3x . x/x^2-9
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Toán Lớp 8: Tính a) x-8/x-3 + 1/x-3 – 4/3-x b) x+3/ x^2-3x . x/x^2-9
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Answer
a, {x – 8}/{x – 3} + 1/{x – 3} – 4/{3 – x}
= {x – 8}/{x – 3} + 1/{x – 3} + 4/{x – 3}
= {x – 8 + 1 + 4}/{x – 3}
= {x + (-8 + 1 + 4)}/{x – 3}
= {x – 3}/{x – 3}
= 1
______________________________
b, {x + 3}/{x^2 – 3x} . x/{x^2 – 9}
= {x + 3}/{x . (x – 3)} . x/{x^2 – 3^2}
= {x + 3}/{x . (x – 3)} . x/{(x – 3) . (x + 3)}
= {(x + 3) . x}/{x . (x – 3) . (x – 3) . (x + 3)}
= 1/{(x – 3) . (x – 3)}
= 1/{(x – 3)^2}
$a)$ $\dfrac{x-8}{x-3}+\dfrac{1}{x-3}-\dfrac{4}{3-x}$
$=\dfrac{x-7}{x-3}-\dfrac{4}{3-x}$
$=\dfrac{x-7}{x-3}-\dfrac{4}{-\left(x-3\right)}$
$=\dfrac{x-7}{x-3}-\dfrac{-4}{x-3}$
$=\dfrac{x-7-\left(-4\right)}{x-3}$
$=\dfrac{x-3}{x-3}$
$=1$
$b)$ $\dfrac{x+3\:}{\:x^2-3x\:}\cdot \dfrac{x}{x^2-9}$
$=\dfrac{\left(x+3\right)x}{\left(x^2-3x\right)\left(x^2-9\right)}$
$=\dfrac{\left(x+3\right)x}{x\left(x-3\right)\left(x^2-9\right)}$
$=\dfrac{x+3}{\left(x-3\right)\left(x^2-9\right)}$
$=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)\left(x-3\right)}$
$=\dfrac{x+3}{\left(x-3\right)^2\left(x+3\right)}$
$=\dfrac{1}{\left(x-3\right)^2}$