\(#TyHM\)

\(21,\\ e,PT\Leftrightarrow\left|2x-5\right|=5-2x\Leftrightarrow\left[{}\begin{matrix}2x-5=5-2x\left(x\ge\dfrac{5}{2}\right)\\5-2x=5-2x\left(x< \dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(tm\right)\\0x=0\left(tm\right)\end{matrix}\right.\\ \Leftrightarrow x\in R\\ f,\Leftrightarrow\left|x-\dfrac{1}{4}\right|=\dfrac{1}{4}-x\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{4}=\dfrac{1}{4}-x\left(x\ge\dfrac{1}{4}\right)\\\dfrac{1}{4}-x=\dfrac{1}{4}-x\left(x< \dfrac{1}{4}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\left(tm\right)\\0x=0\left(tm\right)\end{matrix}\right.\\ \Leftrightarrow x\in R\)

21:

\(y'=\dfrac{\left(x^2-3x+5\right)'\left(x+2\right)-\left(x+2\right)'\left(x^2-3x+5\right)}{\left(x+2\right)^2}\)

\(=\dfrac{\left(2x-3\right)\left(x+2\right)-\left(x^2-3x+5\right)}{\left(x+2\right)^2}\)

\(=\dfrac{2x^2+4x-3x-6-x^2+3x-5}{\left(x+2\right)^2}=\dfrac{x^2+4x-11}{\left(x+2\right)^2}\)

17:

Khi x<>0 thì \(\lim\limits_{x\rightarrow0}f\left(x\right)=\lim\limits_{x\rightarrow0}\dfrac{1+4x-1}{\sqrt{1+4x}+1}\cdot\dfrac{1}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{4}{\sqrt{1+4x}+1}=\dfrac{4}{1+1}=\dfrac{4}{2}=2\)

=>Chọn B

Câu 84:

$\sin 3x+2\cos ^2x=1$

$\sin 3x=1-2\cos ^2x=-\cos 2x=\sin (2x-\frac{\pi}{2})$

\(\Rightarrow \left[\begin{matrix} 3x=2x-\frac{\pi}{2}+2k\pi\\ 3x=\frac{3}{2}\pi-2x+2k\pi\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=(2k+\frac{3}{2})\pi\\ x=\frac{2k+\frac{3}{2}}{5}\pi\end{matrix}\right.\) với $k$ nguyên 

Nghiệm âm lớn nhất của pt:

$x=\frac{2(-1)+\frac{3}{2}}{5}\pi =\frac{-\pi}{10}$

84.

\(sin3x+2cos^2x=1\)

\(\Leftrightarrow sin3x+cos2x=0\)

\(\Leftrightarrow cos\left(\dfrac{\pi}{2}-3x\right)+cos2x=0\)

\(\Leftrightarrow2cos\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right).cos\left(\dfrac{\pi}{4}-\dfrac{5x}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)=0\\cos\left(\dfrac{\pi}{4}-\dfrac{5x}{2}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{4}-\dfrac{x}{2}=\dfrac{\pi}{2}+k\pi\\\dfrac{\pi}{4}-\dfrac{5x}{2}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{2}-k2\pi\\x=-\dfrac{\pi}{10}-\dfrac{k2\pi}{5}\end{matrix}\right.\)

\(x=-\dfrac{\pi}{2}-k2\pi< 0\Leftrightarrow k>-\dfrac{1}{4}\Rightarrow k=0\Rightarrow x=-\dfrac{\pi}{2}\)

\(x=-\dfrac{\pi}{10}-k2\pi< 0\Leftrightarrow k>-\dfrac{1}{20}\Rightarrow k=0\Rightarrow x=-\dfrac{\pi}{10}\)

Vậy \(x=-\dfrac{\pi}{10}\) là nghiệm âm lớn nhất

Câu 85:

ĐKXĐ: $\cos 2x; \cos 3x\neq 0$

$\tan 2x\tan 3x=1$
$\Leftrightarrow \sin 2x\sin 3x=\cos 2x\cos 3x$
$\Leftrightarrow 2\sin 2x\sin 3x=2\cos 2x\cos 3x$

$\Leftrightarrow \cos 5x+\cos x=\cos x-\cos 5x$

$\Leftrightarrow 2\cos 5x=0$

$\Leftrihgtarrow \cos 5x=0$

$\Leftrightarrow x=\frac{1}{5}(\frac{\pi}{2}+k\pi$

$=\frac{2k+1}{10}\pi$

Nghiệm âm lớn nhất: $\frac{-2+1}{10}\pi =\frac{-\pi}{10}$

ĐKXĐ: \(1-3m\ge0\Rightarrow m\le\dfrac{1}{3}\) (1)

Theo điều kiện có nghiệm của pt lượng giác bậc nhất:

\(m^2+\left(1-3m\right)\ge\left(m-2\right)^2\)

\(\Leftrightarrow1-3m\ge-4m+4\Rightarrow m\ge3\) (2)

Kết hợp (1); (2) \(\Rightarrow\) không tồn tại m thỏa mãn