Ta có : \(M_{kk}=\dfrac{20.32+80.28}{20+80}=28,8\) ( đvc )

\(\Rightarrow d_{\dfrac{KK}{H2}}=\dfrac{M_{kk}}{M_{h2}}=\dfrac{28,8}{2}=14,4\)

Vậy ...

16, Ta có : \(2020^{\left(x-2\right)\left(2x+3\right)}=1=2020^0\)

\(\Leftrightarrow\left(x-2\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)

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c, NaCl

Ta có : \(\left(\sin\dfrac{x}{2}+\cos\dfrac{x}{2}\right)^2+\sqrt{3}\cos x=2\)

\(\Leftrightarrow\sin^2\dfrac{x}{2}+2\sin\dfrac{x}{2}.\cos\dfrac{x}{2}+\cos^2\dfrac{x}{2}+\sqrt{3}\cos x-2=0\)

\(\Leftrightarrow1+\sin x+\sqrt{3}\cos x-2=0\)

\(\Leftrightarrow\sin x+\sqrt{3}\cos x=1\)

\(\Leftrightarrow\sin x.\cos\dfrac{\pi}{3}+\cos x.\sin\dfrac{\pi}{3}=\sin\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\x+\dfrac{\pi}{3}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\) \(\left(K\in Z\right)\)

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

Vậy ...

Ta có : \(\cos\left(2x+\dfrac{\pi}{6}\right)=m+1,x\in\left(\dfrac{7\pi}{24};\dfrac{3\pi}{4}\right)\)

Thấy \(x\in\left(\dfrac{7\pi}{24};\dfrac{3\pi}{4}\right)\)

\(\Rightarrow2x+\dfrac{\pi}{6}\in\left(\dfrac{3\pi}{4};\dfrac{5\pi}{3}\right)\)

\(\Rightarrow\cos\left(2x+\dfrac{\pi}{6}\right)\in\left(-1;\dfrac{1}{2}\right)\)

\(\Rightarrow-1< m+1< \dfrac{1}{2}\)

\(\Rightarrow-2< m< -\dfrac{1}{2}\)

Vậy ...

Câu 101 :

Ta có : \(\sin^2x-4\sin x.\cos x+4\cos^2x=5\)

\(\Leftrightarrow\left(\sin x-2\cos x\right)^2=5\)

Đặt \(\cos a=\dfrac{1}{\sqrt{5}}\)

\(\Rightarrow\left(\sqrt{5}\sin\left(x-a\right)\right)^2=5\)

\(\Leftrightarrow\sin^2\left(x-a\right)=1\)

\(\Leftrightarrow\left[{}\begin{matrix}\sin\left(x-a\right)=1\\\sin\left(x-a\right)=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=a+\dfrac{\pi}{2}+k2\pi\\x=a-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\) \(\left(K\in Z\right)\)

Vậy có 2 điểm biểu diễn nghiệm phương trình trên đường tròn .