Khẳng định sau đây sai là
A. $\displaystyle \sqrt{ab}=\sqrt{a}\sqrt{b}$$\displaystyle \forall a,b$
B. $\displaystyle \sqrt[2n]{{{a}^{2n}}}\ge 0$$\displaystyle \forall a$,$\displaystyle n$ nguyên dương$\displaystyle \left( n\ge 2 \right)$
C. $\displaystyle \sqrt[2n]{{{a}^{2n}}}=\left| a \right|$$\displaystyle \forall a$,$\displaystyle n$ nguyên dương$\displaystyle \left( n\ge 2 \right)$
D. $\displaystyle \sqrt[4]{{{a}^{2}}}=\sqrt{a}$$\displaystyle \forall a\ge 0$
A. $\displaystyle \sqrt{ab}=\sqrt{a}\sqrt{b}$$\displaystyle \forall a,b$
B. $\displaystyle \sqrt[2n]{{{a}^{2n}}}\ge 0$$\displaystyle \forall a$,$\displaystyle n$ nguyên dương$\displaystyle \left( n\ge 2 \right)$
C. $\displaystyle \sqrt[2n]{{{a}^{2n}}}=\left| a \right|$$\displaystyle \forall a$,$\displaystyle n$ nguyên dương$\displaystyle \left( n\ge 2 \right)$
D. $\displaystyle \sqrt[4]{{{a}^{2}}}=\sqrt{a}$$\displaystyle \forall a\ge 0$
Loga
Hóa Học lớp 12
