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10.
\(\dfrac{sin3x-cos3x}{sinx+cosx}=\dfrac{3sinx-4sin^3x-\left(4cos^3x-3cosx\right)}{sinx+cosx}\)
\(=\dfrac{3\left(sinx+cosx\right)-4\left(sin^3x+cos^3x\right)}{sinx+cosx}\)
\(=\dfrac{3\left(sinx+cosx\right)-4\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)}{sinx+cosx}\)
\(=\dfrac{3\left(sinx+cosx\right)-4\left(sinx+cosx\right)\left(1-sinx.cosx\right)}{sinx+cosx}\)
\(=\dfrac{\left(sinx+cosx\right)\left(3-4+4sinx.cosx\right)}{sinx+cosx}\)
\(=-1+4sinx.cosx\)
\(=2sin2x-1\)
11.
\(tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)\dfrac{1+cos\left(\dfrac{\pi}{2}+x\right)}{sin\left(\dfrac{\pi}{2}+x\right)}=tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).\dfrac{1+sin\left(-x\right)}{cos\left(-x\right)}\)
\(=tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).\dfrac{1-sinx}{cosx}=tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)\dfrac{sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}-2sin\dfrac{x}{2}cos\dfrac{x}{2}}{cos^2\dfrac{x}{2}-sin^2\dfrac{x}{2}}\)
\(=tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).\dfrac{\left(cos\dfrac{x}{2}-sin\dfrac{x}{2}\right)^2}{\left(cos\dfrac{x}{2}-sin\dfrac{x}{2}\right)\left(cos\dfrac{x}{2}+sin\dfrac{x}{2}\right)}\)
\(=tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).\dfrac{cos\dfrac{x}{2}-sin\dfrac{x}{2}}{cos\dfrac{x}{2}+sin\dfrac{x}{2}}\)
\(=tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).\dfrac{cos\left(\dfrac{x}{2}+\dfrac{\pi}{4}\right)}{sin\left(\dfrac{x}{2}+\dfrac{\pi}{4}\right)}\)
\(=tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).cot\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)\)
\(=1\)
Lời giải:
$\overrightarrow{MA}=(1-x, 3-y), \overrightarrow{MB}=(4-x, 2-y)$
Để $MAB$ là tam giác vuông cân tại $M$ thì:
\(\left\{\begin{matrix}
\overrightarrow{MA}.\overrightarrow{MB}=0\\
MA^2=MB^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
(1-x)(4-x)+(3-y)(2-y)=0\\
(1-x)^2+(3-y)^2=(4-x)^2+(2-y)^2\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x^2-5x+y^2-5y+10=0\\ 6x-2y-10=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x^2-5x+y^2-5y+10=0\\ y=3x-5\end{matrix}\right.\)
\(\Leftrightarrow (x,y)=(2,1), (3,4)\)
Tham khảo tại link sau:
https://hoc24.vn/cau-hoi/ai-giup-em-cau-2-voi-a.3401576227354
Bài 6:
ĐK: \(9a< \dfrac{4}{a}\Leftrightarrow a^2< \dfrac{4}{9}\Leftrightarrow-\dfrac{2}{3}< a< \dfrac{2}{3}\)
Bài 7:
ĐK: \(a=\dfrac{4}{a}\Leftrightarrow a^2=4\Leftrightarrow\left[{}\begin{matrix}a=2\\a=-2\end{matrix}\right.\)
\(A=\left\{1;2;5;10;17\right\}\)
\(\left(x^2-4\right)\left(2x^2-x-1\right)=0\Rightarrow\left[{}\begin{matrix}x^2-4=0\\2x^2-x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm2\\x=1\\x=-\dfrac{1}{2}\notin Z\end{matrix}\right.\) \(\Rightarrow B=\left\{-2;1;2\right\}\)
\(\Rightarrow A\backslash B=\left\{5;10;17\right\}\)
\(P\cap Q=\varnothing\Leftrightarrow\left[{}\begin{matrix}a+1< -5\\a-1>4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a< -6\\a>5\end{matrix}\right.\)
\(\Rightarrow P\cap Q\ne\varnothing\Leftrightarrow-6\le a\le5\)
Vậy: \(a\in\left[-6;5\right]\).
Bài 18 :
\(\left(-\infty;9a\right)\cap\left(\dfrac{4}{a};+\infty\right)\ne\varnothing\) \(\left(a< 0\right)\)
\(\Rightarrow9a>\dfrac{4}{a}\)
\(\Leftrightarrow9a-\dfrac{4}{a}>0\)
\(\Leftrightarrow\dfrac{9a^2-4}{a}>0\)
\(\Leftrightarrow9a^2-4< 0\left(a< 0\right)\)
\(\Leftrightarrow9a^2< 4\)
\(\Leftrightarrow a^2< \dfrac{4}{9}=\left(\dfrac{2}{3}\right)^2\)
\(\Leftrightarrow-\dfrac{2}{3}< a< \dfrac{2}{3}\)
mà \(a< 0\)
\(\Leftrightarrow-\dfrac{2}{3}< a< 0\)