Cho a, b là hai số dương khác 1. Trong các khẳng định sau, khẳng định nào đúng ?
\(ba^{\frac{2}{\log_ba}+1}-2a^{\log_ab+1}b^{\log_ba+1}+ab^{\frac{2}{\log_ab}+1}=ab\left(a-b\right)^2\) \(ba^{\frac{2}{\log_ba}+1}-2a^{\log_ab+1}b^{\log_ba+1}+ab^{\frac{2}{\log_ab}+1}=ab\left(a+b\right)^2\) \(ba^{\frac{2}{\log_ba}+1}-2a^{\log_ab+1}b^{\log_ba+1}+ab^{\frac{2}{\log_ab}+1}=ab\left(\frac{1}{a}-\frac{1}{b}\right)^2\) \(ba^{\frac{2}{\log_ba}+1}-2a^{\log_ab+1}b^{\log_ba+1}+ab^{\frac{2}{\log_ab}+1}=\frac{\left(a-b\right)^2}{ab}\) Hướng dẫn giải:Có:
\(ba^{\frac{2}{\log_ba}+1}=b.a^{2\log_ab+1}=b.\left(a^{2\log_ab}\right)a=b.a^{\log_ab^2}.a=b.b^2.a=b^3a\)
\(a^{\log_ab+1}b^{\log_ba+1}=b.a.a.b=a^2b^2\)
\(ab^{\frac{2}{\log_ab}+1}=ab^{2\log_ba}.b=a.a^2.b=a^3b\)
Vậy:
\(ba^{\frac{2}{\log_ba}+1}-2a^{\log_ab+1}b^{\log_ba+1}+ab^{\frac{2}{\log_ab}+1}=b^3a+2a^2b^2+a^3b\)
\(=ab\left(b^2+2ab+a^2\right)=ab\left(a+b\right)^2\)
