T
thaison901
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ta có: xét hàm số mũ [tex] y=a^x [/tex]
[tex] (a^x)'= a^x.lna [/tex]
mặt khác [tex] (a^x) = x.a^{x-1} [/tex]
\Rightarrow [tex] a^x.lna = x.a^{x-1} [/tex]
\Leftrightarrow[tex] a.lna=x [/tex]
mọi người thấy vô lí không /
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[tex] (a^x)'= a^x.lna [/tex]
mặt khác [tex] (a^x) = x.a^{x-1} [/tex]
\Rightarrow [tex] a^x.lna = x.a^{x-1} [/tex]
\Leftrightarrow[tex] a.lna=x [/tex]
mọi người thấy vô lí không /
thank cái nha