Đáp án:
$20)\\ a)A=3x-2\sqrt{x}\\ b)B=4\sqrt{x-1}\\ c)C=x\\ d)D=\sqrt{a}$
Giải thích các bước giải:
$20)\\ a)A=2\sqrt{x}+3\sqrt{x^2}-\sqrt{16x} \ \ \ \ x \ge 0\\ =2\sqrt{x}+3|x|-4\sqrt{x}\\ =3x-2\sqrt{x}\\ b)B=\sqrt{25x-25}-\sqrt{9x-9}+\sqrt{4x-4} \ \ \ \ x \ge 1\\ =\sqrt{25(x-1)}-\sqrt{9(x-1)}+\sqrt{4(x-1)} \\ =5\sqrt{x-1}-3\sqrt{x-1}+2\sqrt{x-1} \\ =4\sqrt{x-1}\\ c)C=\dfrac{x^2+x}{x+1} \ \ \ \ x \ne -1\\ =\dfrac{x(x+1)}{x+1} \\ =x\\ d)D=\dfrac{a+\sqrt{a}}{\sqrt{a}+1} \ \ \ \ a \ge 0\\ =\dfrac{\sqrt{a}(\sqrt{a}+1)}{\sqrt{a}+1}\\ =\sqrt{a}$