a) $A=\sqrt[3]{a}+\sqrt{b^{4}}$

Q: a) $A=\sqrt[3]{a}+\sqrt{b^{4}}$

A. True B. False Ta có: $A=\left(a^{3} \cdot a^{\frac{1}{2}}\right)^{\log _{a} b}+\left(b^{\frac{2}{3}}\right)^{\log _{b} a}=\left(a^{\frac{7}{2}}\right)^{\log _{a} b}+\left(b^{\frac{2}{3}}\right)^{\log _{b} a}$$=\left(a^{\log _{a} b}\right)^{\frac{7}{2}}+\left(b^{\log _{b} a}\right)^{\frac{2}{3}}=b^{\frac{7}{2}}+a^{\frac{2}{3}}=\sqrt[3]{a^{2}}+\sqrt{b^{7}}$.Ta có: $B=\log \left(\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{d}\right)-\log \frac{a}{d}=\log \left(\frac{a}{d}: \frac{a}{d}\right)=\log 1=0$.

Question

Accepted Answer

A. True
B. False

Ta có: $A=\left(a^{3} \cdot a^{\frac{1}{2}}\right)^{\log _{a} b}+\left(b^{\frac{2}{3}}\right)^{\log _{b} a}=\left(a^{\frac{7}{2}}\right)^{\log _{a} b}+\left(b^{\frac{2}{3}}\right)^{\log _{b} a}$$=\left(a^{\log _{a} b}\right)^{\frac{7}{2}}+\left(b^{\log _{b} a}\right)^{\frac{2}{3}}=b^{\frac{7}{2}}+a^{\frac{2}{3}}=\sqrt[3]{a^{2}}+\sqrt{b^{7}}$.

Ta có: $B=\log \left(\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{d}\right)-\log \frac{a}{d}=\log \left(\frac{a}{d}: \frac{a}{d}\right)=\log 1=0$.

Đề thi giữa kì 2 (Cấu trúc mới) - Đề số 20 - MĐ 10904