Câu 3:
a) A = $\frac{2x}{x²-25}$ - $\frac{5}{x-5}$ -$\frac{1}{x+5}$ (ĐK: x $\neq$ ±5)
= $\frac{2x}{(x-5)(x+5)}$ - $\frac{5}{x-5}$ -$\frac{1}{x+5}$
= $\frac{2x}{(x-5)(x+5)}$ - $\frac{5(x+5)}{x-5}$ -$\frac{x-5}{x+5}$
= $\frac{2x-5(x+5)-(x-5)}{(x-5)(x+5)}$
= $\frac{2x-5x-25-x+5}{(x-5)(x+5)}$
= $\frac{-4x-20}{(x-5)(x+5)}$
= $\frac{-4(x+5)}{(x-5)(x+5)}$
= $\frac{-4}{(x+5)}$
b) Ta có x = $\frac{4}{5}$ (thỏa mãn điều kiện)
-> Thay x = $\frac{4}{5}$ vào biểu thức A
-> A = $\frac{-4}{(x-5)}$
= $\frac{-4}{\frac{4}{5}+-5}$
= $\frac{-4}{\frac{4}{5} -\frac{25}{5}}$
= $\frac{-4}{\frac{-21}{5}}$
= $\frac{20}{21}$
Cho mik 5* vs CTLHN nhé!
