12 câu Trắc nghiệm Vi phân có đáp án (Phần 2)

$dy=(1+{\tan }^{2}2x)dx$

$dy=(1-{\tan }^{2}2x)dx$

$dy=2(1-{\tan }^{2}2x)dx$

$dy=2(1+{\tan }^{2}2x)dx$

$\text{d}f\left(x\right)=\frac{-\sin 4x}{2\sqrt{1+{\cos }^{2}2x}}\text{d}x$

$\text{d}f\left(x\right)=\frac{-\sin 4x}{\sqrt{1+{\cos }^{2}2x}}\text{d}x$

$\text{d}f\left(x\right)=\frac{\cos 2x}{\sqrt{1+{\cos }^{2}2x}}\text{d}x$

$\text{d}f\left(x\right)=\frac{-\sin 2x}{2\sqrt{1+{\cos }^{2}2x}}\text{d}x$

$\text{d}y=\frac{\text{d}x}{{\left(x-1\right)}^2}$

$\text{d}y=\frac{3\text{d}x}{{\left(x-1\right)}^2}$

$\text{d}y=\frac{-3\text{d}x}{{\left(x-1\right)}^2}$

$\text{d}y=-\frac{\text{d}x}{{\left(x-1\right)}^2}$

$\text{d}y=\left(x\cos x–\sin x\right)\text{d}x$

$\text{d}y=\left(x\cos x\right)\text{d}x$

$\text{d}y=\left(\cos x–\sin x\right)\text{d}x$

$\text{d}y=\left(x\sin x\right)\text{d}x$

- 4

11

1,1

-0,4

4,0313

4,0312

4,0311

4,0314

$\frac{\sqrt{3}}{3}-\frac{\pi }{1200}$

$\frac{\sqrt{3}}{2}-\frac{\pi }{1440}$

$\frac{\sqrt{3}}{2}+\frac{\pi }{140}$

$\frac{\sqrt{3}}{3}+\frac{\pi }{100}$

$dy=(3x^2-4x)dx$

$dy=(3x^2+x)dx$

$dy=(3x^2+2x)dx$

$dy=(3x^2+4x)dx$

$dy=\frac{3}{2\sqrt{3x+2}}dx$

$dy=\frac{1}{2\sqrt{3x+2}}dx$

$dy=\frac{1}{\sqrt{3x+2}}dx$

$dy=\frac{3}{\sqrt{3x+2}}dx$

$dy=\left(\cos 2x+3{\sin }^{2}x\cos x\right)dx$

$dy=\left(2\cos 2x+3{\sin }^{2}x\cos x\right)dx$

$dy=\left(2\cos 2x+{\sin }^{2}x\cos x\right)dx$

$dy=\left(\cos 2x+{\sin }^{2}x\cos x\right)dx$

$\text{d}y=-\frac{x^2-2x-2}{{(x-1)}^2}\text{d}x$

$\text{d}y=\frac{2x+1}{{(x-1)}^2}\text{d}x$

$\text{d}y=-\frac{2x+1}{{(x-1)}^2}\text{d}x$

$\text{d}y=\frac{x^2-2x-2}{{(x-1)}^2}\text{d}x$