Môn Toán Lớp 12 Ai giải giúp mình câu 6 vs 7 ạ
Question
Eloise
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Trả lời ( )
$6)\cos x-1=-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+o(x^4)\\ \ln\cos x=\ln(\cos x-1+1)\\ =\cos x-1-\dfrac{(\cos x-1)^2}{2}+\dfrac{(\cos x-1)^3}{3}-\dfrac{(\cos x-1)^4}{4}+o(x^4)\\ =-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{\left(-\dfrac{x^2}{2!}\right)^2}{2}+o(x^4)\\ =-\dfrac{x^2}{2!}-\dfrac{x^4}{12}+o(x^4)\\ 7)x\cos x=x\left(1-\dfrac{x^2}{2!}+o(x^2)\right)=x-\dfrac{x^3}{2!}+o(x^3)\\ e^{x\cos x}=1+x\cos x+\dfrac{(x\cos x)^2}{2!}+\dfrac{(x\cos x)^3}{3!}+o(x^3)\\ =1+x-\dfrac{x^3}{2!}+o(x^3)$