RÚT GỌN PHƯƠNG TRÌNH[Toán 8]

T

trang.bui35

$1. \frac{2x}{x+3}+\frac{x}{x-3}-\frac{3x^2+3}{x^2-9}$
\Leftrightarrow $\frac{2x(x-3)+x(x+3)-3x^2-3}{(x+3)(x-3)}$
\Leftrightarrow $\frac{2x^2-6x+x^2+3x-3x^2-3}{(x+3)(x-3)}$
\Leftrightarrow $\frac{-3(x+1)}{x^2-9}$
2. $\frac{4}{x+1} + \frac{2}{1-x} - \frac{x-5}{x^2-1}$
ĐKXĐ: x khác +- 1
\Leftrightarrow $ \frac{4(x-1)-2(x+1)-x+5}{(x+1)(x-1)}$
\Leftrightarrow $\frac{4x-4-2x-2-x+5}{(x+1)(x-1)}$
\Leftrightarrow $ \frac{x-1}{(x+1)(x-1)}$
\Leftrightarrow$\frac{1}{x+1}$
 
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Q

quynhsieunhan

Câu 2:đk: $x \not= \ \pm 1$
$\frac{4}{x + 1} + \frac{2}{1 - x} - \frac{x - 5}{x^2 - 1}$
= $\frac{4(x - 1) - 2(x + 1) - (x - 5)}{x^2 - 1}$
= $\frac{4x - 4 - 2x - 2 - x + 5}{x^2 - 1}$
= $\frac{x - 1}{(x - 1)(x + 1)}$
= $\frac{1}{x + 1}$
 
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