Trắc nghiệm Bài tập lý thuyết Biến đổi đơn giản biểu thức chứa căn thức bậc hai có đáp án

$\sqrt{54}=3\sqrt{6}$

$3\sqrt{5}=\sqrt{15}$

$\sqrt{18{\mathrm{a}}^2}=3\mathrm{a}\sqrt{2}$

$-\frac{2}{3}.\sqrt{\mathrm{xy}}=\sqrt{\frac{4}{9}.\mathrm{xy}}$ với x, y > 0

$2\mathrm{y}\sqrt{2}$

$-2\mathrm{y}\sqrt{2}$

$4\mathrm{y}\sqrt{2}$

$-4\mathrm{y}\sqrt{2}$

$\mathrm{x}.\sqrt{\frac{11}{\mathrm{x}}}=11\mathrm{x}$

$\mathrm{x}.\sqrt{\frac{11}{\mathrm{x}}}=\frac{11}{\mathrm{x}}$

$\mathrm{x}.\sqrt{\frac{11}{2}}=\sqrt{11\mathrm{x}}$ với x > 0

$\mathrm{x}.\sqrt{\frac{11}{\mathrm{x}}}=-\sqrt{11\mathrm{x}}$

$\mathrm{x}.\sqrt{\frac{-29}{\mathrm{x}}}=\sqrt{-29\mathrm{x}}\left(\mathrm{x}<0\right)$

$\mathrm{x}.\sqrt{\frac{-29}{\mathrm{x}}}=-\sqrt{29\mathrm{x}}\left(\mathrm{x}<0\right)$

$\mathrm{x}.\sqrt{\frac{-29}{\mathrm{x}}}=\sqrt{29\mathrm{x}}\left(\mathrm{x}<0\right)$

$\mathrm{x}.\sqrt{\frac{-29}{\mathrm{x}}}=-\sqrt{-29\mathrm{x}}\left(\mathrm{x}<0\right)$

$\sqrt{\frac{3}{\mathrm{x}}}=\frac{3\sqrt{\mathrm{x}}}{\mathrm{x}}\left(\mathrm{x}>0\right)$

$\sqrt{\frac{7}{2}}=\frac{\sqrt{14}}{2}$

Cả 2 đáp án trên đều đúng

Cả 2 đáp án trên đều sai

$\frac{13\left(4+\sqrt{10}\right)}{3}$

$\frac{13\left(4-\sqrt{10}\right)}{3}$

$-\frac{13\left(4+\sqrt{10}\right)}{3}$

$-\frac{13\left(4-\sqrt{10}\right)}{3}$

$\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}=\frac{{\left(\sqrt{7}+\sqrt{5}\right)}^2}{2}$

$\frac{5}{\sqrt{3\mathrm{x}}}=-\frac{5.\sqrt{3\mathrm{x}}}{3\mathrm{x}}$

$\frac{9}{\sqrt{5}}=\frac{9\sqrt{5}}{5}$

$\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}=\frac{{\left(\sqrt{7}-\sqrt{5}\right)}^2}{2}$

$\frac{\mathrm{b}}{3+\sqrt{\mathrm{b}}}=-\frac{\mathrm{b}\left(3-\sqrt{\mathrm{b}}\right)}{9-\mathrm{b}}$

$\frac{\mathrm{b}}{3+\sqrt{\mathrm{b}}}=\frac{\mathrm{b}\left(3+\sqrt{\mathrm{b}}\right)}{9-\mathrm{b}}$

$\frac{\mathrm{b}}{3+\sqrt{\mathrm{b}}}=\frac{\mathrm{b}\left(3-\sqrt{\mathrm{b}}\right)}{9-\mathrm{b}}$

$\frac{\mathrm{b}}{3+\sqrt{\mathrm{b}}}=\frac{\mathrm{b}\left(3-\sqrt{\mathrm{b}}\right)}{9+\mathrm{b}}$

$\frac{\sqrt{2\mathrm{x}}-\sqrt{\mathrm{y}}}{2\mathrm{x}+\mathrm{y}}$ (với x > 0; y > 0; $2\mathrm{x}\ne \mathrm{y}$)

$\frac{\sqrt{2\mathrm{x}}-\sqrt{\mathrm{y}}}{2\mathrm{x}-\mathrm{y}}$ (với x > 0; y > 0; $2\mathrm{x}\ne \mathrm{y}$)

$\frac{\sqrt{2\mathrm{x}}+\sqrt{\mathrm{y}}}{2\mathrm{x}+\mathrm{y}}$ (với x > 0; y > 0; $2\mathrm{x}\ne \mathrm{y}$)

$\frac{\sqrt{2\mathrm{x}}+\sqrt{\mathrm{y}}}{2\mathrm{x}-\mathrm{y}}$ (với x > 0; y > 0; $2\mathrm{x}\ne \mathrm{y}$)