Đáp án:
$a)$Biểu thức không xác định $\forall \ x$
$b)$Biểu thức xác định $\forall \ x$
$c)x \le \dfrac{9}{7}\\ d) \Leftrightarrow \left[\begin{array}{l} x \ge \dfrac{4+\sqrt{7}}{3} \\ x \le \dfrac{4-\sqrt{7}}{3}\end{array} \right.$
Giải thích các bước giải:
$a)\sqrt{-x^2+10x-30}\\ =\sqrt{-x^2+10x-25-5}\\ =\sqrt{\underbrace{-(x-5)^2-5}_{<0 \ \forall \ x}}$
$\Rightarrow $Biểu thức không xác định $\forall \ x$
$b)\dfrac{1}{\sqrt{4x^2-6x+15}}\\ =\dfrac{1}{\sqrt{(2x)^2-2.2x.\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+\dfrac{51}{4}}}\\ =\dfrac{1}{\sqrt{\underbrace{\left(2x-\dfrac{3}{2}\right)^2+\dfrac{51}{4}}_{>0 \ \forall \ x}}}$
$\Rightarrow$ Biểu thức xác định $\forall \ x$
$c)C=\sqrt{(x^2-x+1)(9-7x)}\\ =\sqrt{\left(x^2-x+\dfrac{1}{4}+\dfrac{3}{4}\right)(9-7x)}\\ =\sqrt{\underbrace{\left(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right)}_{>0 \ \forall \ x}(9-7x)}$
$C$ xác định
$\Leftrightarrow 9-7x \ge 0\\ \Leftrightarrow x \le \dfrac{9}{7}\\ d)\sqrt{3x^2-8x+3}\\\ =\sqrt{(\sqrt{3}x)^2-2.\sqrt{3}x.\dfrac{4}{\sqrt{3}}+\left(\dfrac{4}{\sqrt{3}}\right)^2-\dfrac{7}{3}}\\ =\sqrt{\left(\sqrt{3}x-\dfrac{4}{\sqrt{3}}\right)^2-\left(\sqrt{\dfrac{7}{3}}\right)^2}\\ $$ =\sqrt{\left(\sqrt{3}x-\dfrac{4}{\sqrt{3}}-\sqrt{\dfrac{7}{3}}\right)\left(\sqrt{3}x-\dfrac{4}{\sqrt{3}}+\sqrt{\dfrac{7}{3}}\right)}\\ $$ =\sqrt{\left(\sqrt{3}x-\dfrac{4+\sqrt{7}}{\sqrt{3}}\right)\left(\sqrt{3}x-\dfrac{4-\sqrt{7}}{\sqrt{3}}\right)}\\ =\dfrac{1}{3}\sqrt{\left(x-\dfrac{4+\sqrt{7}}{3}\right)\left(x-\dfrac{4-\sqrt{7}}{3}\right)}\\ ĐKXĐ: \left(x-\dfrac{4+\sqrt{7}}{3}\right)\left(x-\dfrac{4-\sqrt{7}}{3}\right) \ge 0\\ \Leftrightarrow \left[\begin{array}{l} x \ge \dfrac{4+\sqrt{7}}{3} \\ x \le \dfrac{4-\sqrt{7}}{3}\end{array} \right.$
