Question

can someone help me plsssssss ><

Answer 1

Bạn xem lại đề bài

Answer 2

\(A=\dfrac{x+\sqrt{x}-6}{x-9}+\dfrac{x-7\sqrt{x}+19}{x+\sqrt{x}-12}-\dfrac{x-5\sqrt{x}}{x+4\sqrt{x}}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)+3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{x-7\sqrt{x}+19}{\sqrt{x}\left(\sqrt{x}-3\right)+4\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\sqrt{x}\left(\sqrt{x}+4\right)}\\ =\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{x-7\sqrt{x}+19}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-5}{\sqrt{x}+4}\\ =\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}+\dfrac{x-7\sqrt{x}+19}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}-\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}\\ =\dfrac{x-2\sqrt{x}+4\sqrt{x}-8+x-7\sqrt{x}+19-x+5\sqrt{x}+3\sqrt{x}-15}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}\\ =\dfrac{x+3\sqrt{x}-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\)