Khi $\log _{2} x=2$ thì $A=1$

Q: Khi $\log _{2} x=2$ thì $A=1$

A. True B. False Ta có:\[\begin{aligned}A & =\log _{2} x^{2}+\log _{\frac{1}{2}} x^{3}+\log _{4} x \\& =2 \log _{2} x+3 \log _{2^{-1}} x+\log _{2^{2}} x=2 \log _{2} x-3 \log _{2} x+\frac{1}{2} \log _{2} x=-\frac{1}{2} \log _{2} x\end{aligned}\]

Question

Accepted Answer

A. True
B. False

Ta có:

$$\begin{aligned}A & =\log _{2} x^{2}+\log _{\frac{1}{2}} x^{3}+\log _{4} x \& =2 \log _{2} x+3 \log _{2^{-1}} x+\log _{2^{2}} x=2 \log _{2} x-3 \log _{2} x+\frac{1}{2} \log _{2} x=-\frac{1}{2} \log _{2} x\end{aligned}$$

Đề thi giữa kì 2 (Cấu trúc mới) - Đề số 31 - MĐ 10930