Pt \(\Leftrightarrow6\left(sinx-cosx\right)-sinx.cosx-6=0\)

Đặt \(t=sinx-cosx;t\in\left[-\sqrt{2};\sqrt{2}\right]\)

\(\Rightarrow\dfrac{t^2-1}{2}=-sinx.cosx\)

Pttt:\(6t-\dfrac{t^2-1}{2}-6=0\)

\(\Leftrightarrow-t^2+12t-11=0\Leftrightarrow\left[{}\begin{matrix}t=11\left(ktm\right)\\t=1\left(tm\right)\end{matrix}\right.\)

\(\Rightarrow sinx-cosx=1\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=\pi+k\pi2\end{matrix}\right.\)\(\left(k\in Z\right)\)

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a)Pt\(\Leftrightarrow sin^25x=1\)

\(\Leftrightarrow\left[{}\begin{matrix}sin5x=1\\sin5x=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{10}+\dfrac{k2\pi}{5}\\x=-\dfrac{\pi}{10}+\dfrac{k2\pi}{5}\end{matrix}\right.\)\(\left(k\in Z\right)\)

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b)Pt\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\cos2x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2sin2x.cos2x=0\\cos2x=0\end{matrix}\right.\)\(\Rightarrow2.sin2x.cos2x=0\)\(\Leftrightarrow sin4x=0\Leftrightarrow x=\dfrac{k\pi}{4}\)\(\left(k\in Z\right)\)

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\(f\left(x\right)=2x^2+3x+1=0\)

\(\Leftrightarrow2x^2+2x+x+1=0\)

\(\Leftrightarrow2x\left(x+1\right)+\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-1\end{matrix}\right.\)

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Gọi \(\Delta\) là đường thẳng đi qua điểm A(-2;3) và nhận \(\overrightarrow{AB}\left(6;-4\right)\) là vecto chỉ phương

\(\Rightarrow\Delta:\left\{{}\begin{matrix}x=-2+6t\\y=3-4t\end{matrix}\right.\)

Gọi \(d\) là đường thẳng đi qua điểm A(-2;3) và nhận \(\overrightarrow{n}\left(4;6\right)\) là vecto pháp tuyến

\(\Rightarrow\)\(\left(d\right):4\left(x+2\right)+6\left(y-3\right)=0\) \(\Leftrightarrow4x+6y-10=0\Leftrightarrow2x+3y-5=0\)

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Để bpt luôn đúng với mọi \(x\in R\Leftrightarrow\left\{{}\begin{matrix}a=1>0\left(lđ\right)\\\Delta\le0\end{matrix}\right.\)

\(\Leftrightarrow9-4\left(m-2\right)\le0\)\(\Leftrightarrow m\ge\dfrac{17}{4}\)

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a)\(a-5\sqrt{a}=\sqrt{a}\left(\sqrt{a}-5\right)\)

b)\(a-7=\left(\sqrt{a}-\sqrt{7}\right)\left(\sqrt{a}+\sqrt{7}\right)\)

c)\(a+4\sqrt{a}+4=\left(\sqrt{a}+2\right)^2\)

d)\(\sqrt{xy}-4\sqrt{x}+3\sqrt{y}-12=\sqrt{x}\left(\sqrt{y}-4\right)+3\left(\sqrt{y}-4\right)=\left(\sqrt{x}+3\right)\left(\sqrt{y}-4\right)\)

Đk:\(x\ne\dfrac{k\pi}{2};k\in Z\)

\(sinx+cosx+\dfrac{1}{cosx}+\dfrac{1}{sinx}=\dfrac{10}{3}\)

\(\Leftrightarrow sinx+cosx+\dfrac{sinx+cosx}{sinx.cosx}=\dfrac{10}{3}\)

Đặt \(t=sinx+cosx;t\in\left[-\sqrt{2};\sqrt{2}\right]\)

\(\Rightarrow\dfrac{t^2-1}{2}=sinx.cosx\)

Pttt:\(t+\dfrac{t}{\dfrac{t^2-1}{2}}=\dfrac{10}{3}\)

\(\Leftrightarrow t+\dfrac{2t}{t^2-1}=\dfrac{10}{3}\)\(\Leftrightarrow t^3-\dfrac{10}{3}t^2+t+\dfrac{10}{3}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=2\left(ktm\right)\\t=\dfrac{2+\sqrt{19}}{3}\left(ktm\right)\\t=\dfrac{2-\sqrt{19}}{3}\left(tm\right)\end{matrix}\right.\)\(\Rightarrow sinx+cosx=\dfrac{2-\sqrt{19}}{3}\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{-\sqrt{38}+2\sqrt{2}}{6}\)

\(\Leftrightarrow x=\dfrac{\pi}{4}\pm arc.cos\left(\dfrac{-\sqrt{38}+2\sqrt{2}}{6}\right)+k2\pi\) \(\left(k\in Z\right)\)(tm)

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Đk:\(\left\{{}\begin{matrix}5-x>0\\2x^2-3x+1\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x< 5\\x\in R\backslash\left(\dfrac{1}{2};1\right)\end{matrix}\right.\)

\(\Rightarrow x\in\left(-\infty;\dfrac{1}{2}\right)\cup\)\([1;5)\)

\(cos^2\alpha=1-sin^2\alpha=1-\left(0,8\right)^2=0,36\)

\(\Rightarrow cos\alpha=0,6\)

\(1+cot^2\alpha=\dfrac{1}{sin^2\alpha}\Rightarrow cot^2\alpha=\dfrac{1}{sin^2\alpha}-1=\dfrac{9}{16}\)

\(\Rightarrow cot\alpha=0,75\)

\(tan\alpha=\dfrac{1}{cot\alpha}=\dfrac{1}{0,75}=\dfrac{4}{3}\)

a)\(\sqrt{400.0,81}=\sqrt{4.81}=\sqrt{2^2.9^2}=2.9=18\)

b)\(\sqrt{\dfrac{5}{27}.\dfrac{3}{20}}=\sqrt{\dfrac{5}{3^3}.\dfrac{3}{2^2.5}}=\sqrt{\dfrac{1}{3^2.2^2}}=\dfrac{1}{3.2}=\dfrac{1}{6}\)

c)\(\sqrt{\left(-5\right)^2.3^2}=\sqrt{5^2.3^2}=5.3=15\)

d)\(\sqrt{\left(2-\sqrt{5}\right)^2\left(2+\sqrt{5}\right)^2}=\sqrt{\left[2^2-\left(\sqrt{5}\right)^2\right]^2}=\sqrt{\left(-1\right)^2}=1\)