Helllppppppppppppppp

Đáp án:22. Không có đáp án

23.B

24.D

Giải thích các bước giải:

$\begin{align} & 22.y=\frac{2\text{x}+\sin 2\text{x}}{1-\cos 2x} \\ & \Rightarrow y'=\frac{(2\text{x}+\sin 2\text{x) }\!\!'\!\!\text{ (}1-\cos 2x)-(1-\cos 2x)'(2\text{x}+\sin 2\text{x})}{{{(1-\cos 2x)}^{2}}} \\ & =\frac{(2+2\cos 2x)(1-\cos 2x)-2\sin 2x(2x+\sin 2x)}{{{(1-\cos 2x)}^{2}}} \\ & =\frac{2-2\cos 2x+2\cos 2x-2{{\cos }^{2}}2x-4x\sin 2x-2{{\sin }^{2}}2x}{{{(1-\cos 2x)}^{2}}} \\ & =\frac{2-2({{\cos }^{2}}2x+{{\sin }^{2}}2x)-4x\sin 2x}{{{\sin }^{2}}2x} \\ & =\frac{2-2-4x\sin 2x}{{{\sin }^{2}}2x} \\ & =\frac{-4x\sin 2x}{{{\sin }^{2}}2x} \\ & 23.y=2{{x}^{3}}-6x\sqrt{x}+11 \\ & \Rightarrow y'=2.3.{{x}^{2}}-6.\sqrt{x}-6x\frac{1}{2\sqrt{x}} \\ & =6{{x}^{2}}-6\sqrt{x}-3\frac{x}{\sqrt{x}} \\ & 24.y=\sqrt{{{x}^{2}}-2x+3} \\ & \Rightarrow y'=\frac{({{x}^{2}}-2x+3)'}{2\sqrt{{{x}^{2}}-2x+3}} \\ & =\frac{2x-2}{2\sqrt{{{x}^{2}}-2x+3}} \\=\frac{x-1}{\sqrt{{{x}^{2}}-2x+3}} \end{align}$