cho A = x+1/x+3 B= 3/x-3 – 6x/9-x^2 + x/x+3 a) Rút gọn B b) tính P= A . B 28/11/2021 Bởi Caroline cho A = x+1/x+3 B= 3/x-3 – 6x/9-x^2 + x/x+3 a) Rút gọn B b) tính P= A . B
Đáp án: a) $B = \dfrac{x+3}{x-3}$ b) $P = \dfrac{x+1}{x-3}$ Giải thích các bước giải: $\begin{array}{l}A = \dfrac{x+1}{x+3};\quad B = \dfrac{3}{x-3} -\dfrac{6x}{9-x^2} + \dfrac{x}{x+3}\qquad (x \ne \pm3)\\ a) \quad B = \dfrac{3(x+3)}{(x-3)(x+3)}+\dfrac{6x}{(x-3)(x+3)} + \dfrac{x(x-3)}{(x-3)(x+3)}\\ \to B = \dfrac{3x +9 + 6x +x^2 – 3x}{(x-3)(x+3)}\\ \to B = \dfrac{x^2 + 6x + 9}{(x-3)(x+3)}\\ \to B = \dfrac{(x+3)^2}{(x-3)(x+3)}\\ \to B = \dfrac{x+3}{x-3}\\ b)\quad P = A\cdot B\\ \to P = \dfrac{x+1}{x+3}\cdot\dfrac{x+3}{x-3}\\ \to P =\dfrac{x+1}{x-3}\end{array}$ Bình luận
Đáp án:
a) $B = \dfrac{x+3}{x-3}$
b) $P = \dfrac{x+1}{x-3}$
Giải thích các bước giải:
$\begin{array}{l}A = \dfrac{x+1}{x+3};\quad B = \dfrac{3}{x-3} -\dfrac{6x}{9-x^2} + \dfrac{x}{x+3}\qquad (x \ne \pm3)\\ a) \quad B = \dfrac{3(x+3)}{(x-3)(x+3)}+\dfrac{6x}{(x-3)(x+3)} + \dfrac{x(x-3)}{(x-3)(x+3)}\\ \to B = \dfrac{3x +9 + 6x +x^2 – 3x}{(x-3)(x+3)}\\ \to B = \dfrac{x^2 + 6x + 9}{(x-3)(x+3)}\\ \to B = \dfrac{(x+3)^2}{(x-3)(x+3)}\\ \to B = \dfrac{x+3}{x-3}\\ b)\quad P = A\cdot B\\ \to P = \dfrac{x+1}{x+3}\cdot\dfrac{x+3}{x-3}\\ \to P =\dfrac{x+1}{x-3}\end{array}$