Hey bro, ko phải nam nữ đâu, hs khá,giỏi đấy

C25)

Pt\(\Leftrightarrow cosx=cos\left(\dfrac{\pi}{2}-2x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}-2x+k2\pi\\x=2x-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\left(1\right)\\x=\dfrac{\pi}{2}-k2\pi\left(2\right)\end{matrix}\right.\)\(\left(k\in Z\right)\)

Do \(x\in\left[-2\pi;\dfrac{\pi}{2}\right]\)\(\Rightarrow\left[{}\begin{matrix}-2\pi\le\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\le\dfrac{\pi}{2}\\-2\pi\le\dfrac{\pi}{2}-k2\pi\le\dfrac{\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow\left[{}\begin{matrix}-\dfrac{13}{4}\le k\le\dfrac{1}{2}\\\dfrac{5}{4}\ge k\le0\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow\left[{}\begin{matrix}k=\left\{-3;-2;-1;0\right\}\\k=\left\{1;0\right\}\end{matrix}\right.\)

Tại k=-3 thay vào (1)\(\Rightarrow x=-\dfrac{11\pi}{6}\)

Tại k=-2 thay vào (1)\(\Rightarrow x=-\dfrac{7\pi}{6}\)

Tại k=-1 thay vào (1)\(\Rightarrow x=-\dfrac{\pi}{2}\)

Tại k=0 thay vào (1)\(\Rightarrow x=\dfrac{\pi}{6}\)

Tại k=1 thay vào (2)\(\Rightarrow x=-\dfrac{3\pi}{2}\)

Tại k= 0 thay vào (2)\(\Rightarrow x=\dfrac{\pi}{2}\)

Vậy...

Đk:\(x>0;x\ne1\)

\(B=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(x-1\right)}\)

\(=\dfrac{\sqrt{x}+1+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{1}{\sqrt{x}-1}\)

\(B=\dfrac{1}{2}\Leftrightarrow\dfrac{1}{\sqrt{x}-1}=\dfrac{1}{2}\Leftrightarrow\sqrt{x}-1=2\)\(\Leftrightarrow x=9\) (tm)

Vậy..