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ADCT: \(\sqrt{u}'=\dfrac{u'}{2\sqrt{u}}\); \(\left(\dfrac{u}{v}\right)'=\dfrac{u'.v-u.v'}{v^2}\)
y'=\(\dfrac{\left(\dfrac{x^3}{x-1}\right)'}{2\sqrt{\dfrac{x^3}{x-1}}}\)
\(\left(\dfrac{x^3}{x-1}\right)'=\dfrac{\left(x^3\right)'.\left(x-1\right)-\left(x-1\right)'.x^3}{\left(x-1\right)^2}\)
=\(\dfrac{3x^2.\left(x-1\right)-x^3}{\left(x-1\right)^2}\)=\(\dfrac{2x^3-3x^2}{\left(x-1\right)^2}\)
=>y'\(\dfrac{2x^3-3x^2}{\left(x-1\right)^2.\sqrt{\dfrac{x^3}{x-1}}}\)=\(\dfrac{2x^3-3x^2}{\sqrt{\left(\dfrac{x}{x-1}\right)^3}}\)
\(sin^4\left(x+\dfrac{\pi}{2}\right)-sin^4x=sin4x\)
\(\Rightarrow cos^4x-sin^4x=sin4x\)
\(\Rightarrow\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right)=sin4x\)
\(\Rightarrow cos^2x-sin^2x=4sinx.cosx.cos2x\)
......
1.
\(\Leftrightarrow sin^2x\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cos^2x\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cosx\right)\left(1+cosx\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cosx\right)\left(sinx+cosx+sinx.cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\Leftrightarrow...\\sinx+cosx+sinx.cosx-1=0\left(1\right)\end{matrix}\right.\)
Xét (1):
Đặt \(sinx+cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
\(\Leftrightarrow t+\frac{t^2-1}{2}-1=0\)
\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
2.
\(\Leftrightarrow\sqrt{3}sinx.cosx+\sqrt{2}cos^2x+\sqrt{6}cosx=0\)
\(\Leftrightarrow cosx\left(\sqrt{3}sinx+\sqrt{2}cosx+\sqrt{6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Leftrightarrow...\\\sqrt{3}sinx+\sqrt{2}cosx=-\sqrt{6}\left(1\right)\end{matrix}\right.\)
Xét (1):
Do \(\sqrt{3}^2+\sqrt{2}^2< \left(-\sqrt{6}\right)^2\) nên (1) vô nghiệm