1 Hình như thiếu đề

2. \(f\left(x\right)=\left[{}\begin{matrix}2x\left(x^3-3x\right),x\in\left[-\sqrt{3};0\right]\cup[\sqrt{3};+vc)\\2x\left(3x-x^3\right);x\in\left(-vc,-\sqrt{3}\right)\cup\left(0;\sqrt{3}\right)\end{matrix}\right.\)

\(f'\left(x\right)=\left[{}\begin{matrix}8x^3-12x\\12x-8x^3\end{matrix}\right.\) 

Xét \(f'\left(x\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pm\sqrt{6}}{2}\\x=0\end{matrix}\right.\)

Mk làm theo kiểu gộp cả hai biểu thức của f(x) vào chung BBT

4 cực trị

(Cách xét dấu: trong khoảng \(\left[-\sqrt{3};0\right]\cup[\sqrt{3};+vc)\) xét \(f'\left(x\right)=8x^3-12x\) với nghiệm \(x=-\dfrac{\sqrt{6}}{2};x=0\)

trong khoảng \(\left(-vc,-\sqrt{3}\right)\cup\left(0;\sqrt{3}\right)\)xét \(f'\left(x\right)=12x-8x^3\) với nghiệm \(x=\dfrac{\sqrt{6}}{2}\)

a. 125.27

=5³.3³

=(5.3)³

b. 243.32

= 3⁵.2⁵

=(3.2)⁵

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