HOC24
Lớp học
Môn học
Chủ đề / Chương
Bài học
\(-2\Pi\)
M là trung điểm BC.
\(\overrightarrow{CI}=\overrightarrow{CA}+\overrightarrow{AI}=-\overrightarrow{AC}+\dfrac{1}{3}\overrightarrow{AM}=-\overrightarrow{AC}+\dfrac{1}{6}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)=\dfrac{1}{6}\overrightarrow{AB}-\dfrac{5}{6}\overrightarrow{AC}\)
\(\overrightarrow{CK}=\overrightarrow{CA}+\overrightarrow{AK}=-\overrightarrow{AC}+\dfrac{1}{5}\overrightarrow{AB}=\dfrac{6}{5}\overrightarrow{CI}\)
Suy ra C, I, K thẳng hàng.
1/ \(pt\Leftrightarrow\left(3cos^2x-sin^2x\right)\left(cos^2x-sin^2x\right)=0\)
\(\Leftrightarrow\left(\dfrac{3}{2}\left(1+cos2x\right)-\dfrac{1}{2}\left(1-cos2x\right)\right)\left(\dfrac{1}{2}\left(1+cos2x\right)-\dfrac{1}{2}\left(1-cos2x\right)\right)=0\)
\(\Leftrightarrow\left(2cos2x+1\right)cos2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=-\dfrac{1}{2}\end{matrix}\right.\)
2/ \(pt\Leftrightarrow\left(sinx-1\right)\left(sin^2x+sinx+6\right)=0\)
\(\Leftrightarrow sinx=1\)
3/ \(pt\Leftrightarrow\dfrac{1-cos2x}{2}-4sin2x+\dfrac{7}{2}\left(1+cos2x\right)=0\)
\(\Leftrightarrow3cos2x-4sin2x=-4\)
\(\Leftrightarrow5\left(\dfrac{3}{5}cos2x-\dfrac{4}{5}sin2x\right)=-4\)
\(\Leftrightarrow cos\left(2x+arccos\dfrac{3}{5}\right)=-\dfrac{4}{5}\)
4,5 giải tương tự câu 3
a) làm tương tự 2 bài mk đã giải nha.
b) \(y=2\cos^2x-2\sqrt{3}\sin x\cos x+1\)
\(=1-\left(\cos2x+\sqrt{3}\sin2x\right)\)
Lại có \(-2\le\cos2x+\sqrt{3}\sin2x\le2\) \(\Rightarrow-1\le y\le3\)
c) Vì \(\left\{{}\begin{matrix}0\le\sqrt[4]{\sin x}\le1\\0\le\sqrt{\cos x}\le1\end{matrix}\right.\)
Do đó \(-1\le y\le1\)
\(pt\Leftrightarrow2\sin x+2\sin x\cos x-\cos x-\cos^2x-\sin^2x=0\)
\(\Leftrightarrow2\sin x+2\sin x\cos x-\cos x-1=0\)
\(\Leftrightarrow\left(2\sin x-1\right)\left(\cos x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sin x=\dfrac{1}{2}\\\cos x=-1\end{matrix}\right.\)
Đến đây tự làm típ nha....
Cauchy: \(a+b\le\sqrt{2\left(a^2+b^2\right)}\)
AM-GM
\(y=\sin x\sqrt{2-\sin^2x}\le\dfrac{\sin^2x+2-\sin^2x}{2}=1\)
\("="\Leftrightarrow\sin x=\sqrt{2-\sin^2x}\Leftrightarrow\sin x=1\Leftrightarrow x=\dfrac{\Pi}{2}+k2\Pi\)
1. Do \(\cos x+2>0\forall x\in R\) \(\Rightarrow\) Hàm số xác định \(\forall x\in R\)
\(y=\dfrac{\sin x+1}{\cos x+2}\)
\(\Leftrightarrow\)\(y\cos x-\sin x=1-2y\)
pt có nghiệm \(\Leftrightarrow y^2+\left(-1\right)^2\ge\left(1-2y\right)^2\)
\(\Leftrightarrow3y^2-4y\le0\)
\(\Leftrightarrow0\le y\le\dfrac{4}{3}\)
2. \(y=\dfrac{\cos x+2\sin x+3}{2\cos x-\sin x+4}\)
\(\Leftrightarrow\left(2y-1\right)\cos x-\left(y+2\right)\sin x=3-4y\)
pt có nghiệm \(\Leftrightarrow\left(2y-1\right)^2+\left(y+2\right)^2\ge\left(3-4y\right)^2\)
\(\Leftrightarrow11y^2-24y+4\le0\)
\(\Leftrightarrow\dfrac{2}{11}\le y\le2\)
kiểm tra giúp mình xem có sai sót gì không...
Xem câu b ấy bn
Dùng hđt \(\sqrt[3]{a}-\sqrt[3]{b}=\dfrac{a-b}{\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}}\) và \(\sqrt[3]{a}+\sqrt[3]{b}=\dfrac{a+b}{\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2}}\)
Ta có:
\(\sqrt[3]{3x^2-x+2001}-\sqrt[3]{3x^2-7x+2002}=\sqrt[3]{6x+2003}+\sqrt[3]{2002}=0\)
\(\Leftrightarrow\dfrac{6x-1}{\sqrt[3]{\left(3x^2-x+2001\right)^2}+\sqrt[3]{\left(3x^2-x+2001\right)\left(3x^2-7x+2002\right)}+\sqrt[3]{\left(3x^2-7x+2002\right)^2}}=\dfrac{6x-1}{\sqrt[3]{\left(6x+2003\right)^2}-\sqrt[3]{2002.\left(6x+2003\right)}+\sqrt[3]{2002^2}}\)
\(\Leftrightarrow x=\dfrac{1}{6}\)