toán 8

N

nguyenbahiep1

[laTEX]a^2 = \frac{3-5a}{10} \\ \\ A = \frac{2a-1}{3a-1}+\frac{5-a}{3a+1} = \frac{3a^2+15a-6}{9a^2-1} \\ \\ A = \frac{3.\frac{3-5a}{10}+15a-6}{9.\frac{3-5a}{10}-1} \\ \\ A = \frac{\frac{135a-51}{10}}{\frac{17-45a}{10}} \\ \\ A = \frac{3(45a-17)}{-(45a-17)} = - 3[/laTEX]
 
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