Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt \(SA=x;SB=y\)
\(S_{\Delta SAB}=\dfrac{1}{2}SA.SB=\dfrac{xy}{2}\)
\(V=\dfrac{SA.SB.SC}{6}.\sqrt{1+2.cos90^0.cos60^0.cos120^0-cos^290^0-cos^260^0-cos^2120^0}=\dfrac{axy}{6}\)
\(\Rightarrow d\left(C;\left(SAB\right)\right)=\dfrac{3V}{S}=\dfrac{axy}{2.\dfrac{xy}{2}}=a\)
\(V=\dfrac{a.a\sqrt{3}.a\sqrt{2}}{6}.\sqrt{1+2cos90^0.cos60^0.cos120^0-cos^290-cos^260-cos^2120}=\dfrac{a^3\sqrt{3}}{6}\)
1: \(Q=\dfrac{ab\left(a-b\right)}{ab}\cdot\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\left(\sqrt{a}+\sqrt{b}\right)^2=a+2\sqrt{ab}+b\)
2: \(=\dfrac{-1+\sqrt{5}-\sqrt{5}+\sqrt{9}-...-\sqrt{2001}+\sqrt{2005}}{4}\)
\(=\dfrac{\sqrt{2005}-1}{4}\)
3.3 d)
\(\sin8x-\cos6x=\sqrt{3}\left(\sin6x+\cos8x\right)\\ \Leftrightarrow\sin8x-\sqrt{3}\cos8x=\sqrt{3}\sin6x+\cos6x\\ \Leftrightarrow\sin\left(8x-\dfrac{\pi}{3}\right)=\sin\left(6x+\dfrac{\pi}{6}\right)\\ \Leftrightarrow\left[{}\begin{matrix}8x-\dfrac{\pi}{3}=6x+\dfrac{\pi}{6}+k2\pi\\8x-\dfrac{\pi}{3}=\pi-\left(6x+\dfrac{\pi}{6}\right)+k2\pi\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{12}+k\dfrac{\pi}{7}\end{matrix}\right.\)
3.4 a)
\(2sin\left(x+\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \Leftrightarrow2cos\left(\dfrac{\pi}{2}-x-\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \Leftrightarrow2cos\left(-x+\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \Leftrightarrow2cos\left(x-\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \)
Chia hai vế cho \(\sqrt{2^2+4^2}=2\sqrt{5}\)
Ta được:
\(\dfrac{1}{\sqrt{5}}cos\left(x-\dfrac{\pi}{4}\right)+\dfrac{2}{\sqrt{5}}sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3}{4}\\ \)
Gọi \(\alpha\) là góc có \(cos\alpha=\dfrac{1}{\sqrt{5}}\)và \(sin\alpha=\dfrac{2}{\sqrt{5}}\)
Phương trình tương đương:
\(cos\left(x-\dfrac{\pi}{4}-\alpha\right)=\dfrac{3}{4}\\ \Leftrightarrow x=\pm arscos\left(\dfrac{3}{4}\right)+\dfrac{\pi}{4}+\alpha+k2\pi\)
\(S_{\Delta ACD}=\dfrac{1}{2}AC.AD.sin\widehat{CAD}=\dfrac{a^2\sqrt{3}}{4}\)
\(V=\dfrac{AB.AC.AD}{6}.\sqrt{1+2cos90^0.cos60^0.cos120^0-cos^290^0-cos^260^0-cos^2120^0}=\dfrac{a^3\sqrt{2}}{12}\)
\(\Rightarrow d\left(B;\left(ACD\right)\right)=\dfrac{3V}{S}=\dfrac{a\sqrt{6}}{3}\)