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1: \(y=x+\dfrac{4}{\left(x-2\right)^2}\)
\(\Leftrightarrow y'=1+\left(\dfrac{4}{\left(x-2\right)^2}\right)'\)
=>\(y'=1+\dfrac{4'\left(x-2\right)^2-4\left[\left(x-2\right)^2\right]'}{\left(x-2\right)^4}\)
=>\(y'=1+\dfrac{-4\cdot2\cdot\left(x-2\right)'\left(x-2\right)}{\left(x-2\right)^4}\)
=>\(y'=1-\dfrac{8}{\left(x-2\right)^3}\)
Đặt y'=0
=>\(\dfrac{8}{\left(x-2\right)^3}=1\)
=>\(\left(x-2\right)^3=8\)
=>x-2=2
=>x=4
Đặt \(f\left(x\right)=x+\dfrac{4}{\left(x-2\right)^2}\)
\(f\left(4\right)=4+\dfrac{4}{\left(4-2\right)^2}=4+1=5\)
\(f\left(0\right)=0+\dfrac{4}{\left(0-2\right)^2}=0+\dfrac{4}{4}=1\)
\(f\left(5\right)=5+\dfrac{4}{\left(5-2\right)^2}=5+\dfrac{4}{9}=\dfrac{49}{9}\)
Vì f(0)<f(4)<f(5)
nên \(f\left(x\right)_{max\left[0;5\right]\backslash\left\{2\right\}}=f\left(5\right)=\dfrac{49}{9}\) và \(f\left(x\right)_{min\left[0;5\right]\backslash\left\{2\right\}}=1\)
2: \(y=cos^22x-sinx\cdot cosx+4\)
\(=1-sin^22x-\dfrac{1}{2}\cdot sin2x+4\)
\(=-sin^22x-\dfrac{1}{2}\cdot sin2x+5\)
\(=-\left(sin^22x+\dfrac{1}{2}\cdot sin2x-5\right)\)
\(=-\left(sin^22x+2\cdot sin2x\cdot\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{81}{16}\right)\)
\(=-\left(sin2x+\dfrac{1}{4}\right)^2+\dfrac{81}{16}\)
\(-1< =sin2x< =1\)
=>\(-\dfrac{3}{4}< =sin2x+\dfrac{1}{4}< =\dfrac{5}{4}\)
=>\(0< =\left(sin2x+\dfrac{1}{4}\right)^2< =\dfrac{25}{16}\)
=>\(0>=-\left(sin2x+\dfrac{1}{4}\right)^2>=-\dfrac{25}{16}\)
=>\(\dfrac{81}{16}>=-sin\left(2x+\dfrac{1}{4}\right)^2+\dfrac{81}{16}>=-\dfrac{25}{16}+\dfrac{81}{16}=\dfrac{7}{2}\)
=>\(\dfrac{81}{16}>=y>=\dfrac{7}{2}\)
\(y_{min}=\dfrac{7}{2}\) khi \(sin2x+\dfrac{1}{4}=\dfrac{5}{4}\)
=>\(sin2x=1\)
=>\(2x=\dfrac{\Omega}{2}+k2\Omega\)
=>\(x=\dfrac{\Omega}{4}+k\Omega\)
\(y_{max}=\dfrac{81}{16}\) khi sin 2x=-1
=>\(2x=-\dfrac{\Omega}{2}+k2\Omega\)
=>\(x=-\dfrac{\Omega}{4}+k\Omega\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-\dfrac{\pi}{2}+k2\pi\\x\ne\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\\end{matrix}\right.\)
\(\dfrac{cosx-2sinx.cosx}{2cos^2x-1-sinx}=\sqrt{3}\)
\(\Leftrightarrow\dfrac{cosx-sin2x}{cos2x-sinx}=\sqrt{3}\)
\(\Rightarrow cosx-sin2x=\sqrt{3}cos2x-\sqrt{3}sinx\)
\(\Leftrightarrow cosx+\sqrt{3}sinx=\sqrt{3}cos2x+sin2x\)
\(\Leftrightarrow\dfrac{1}{2}cosx+\dfrac{\sqrt{3}}{2}sinx=\dfrac{\sqrt{3}}{2}cos2x+\dfrac{1}{2}sin2x\)
\(\Leftrightarrow cos\left(x-\dfrac{\pi}{3}\right)=cos\left(2x-\dfrac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{6}=x-\dfrac{\pi}{3}+k2\pi\\2x-\dfrac{\pi}{6}=\dfrac{\pi}{3}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\left(loại\right)\end{matrix}\right.\)
Vậy \(x=-\dfrac{\pi}{6}+k2\pi\)
Biến đổi :
\(4\sin^2x+1=5\sin^2x+\cos^2x=\left(a\sin x+b\cos x\right)\left(\sqrt{3}\sin x+\cos x\right)+c\left(\sin^2x+\cos^2x\right)\)
\(=\left(a\sqrt{3}+c\right)\sin^2x+\left(a+b\sqrt{3}\right)\sin x.\cos x+\left(b+c\right)\cos^2x\)
Đồng nhấtheej số hai tử số
\(\begin{cases}a\sqrt{3}+c=5\\a+b\sqrt{3}=0\\b+c=1\end{cases}\)
\(\Leftrightarrow\) \(\begin{cases}a=\sqrt{3}\\b=-1\\c=2\end{cases}\)
\(=\int\left(6x^2-\dfrac{4}{x}+sin3x-cos4x+e^{2x+1}+9^{x-1}+\dfrac{1}{cos^2x}-\dfrac{1}{sin^2x}\right)dx\)
\(=2x^3-4ln\left|x\right|-\dfrac{1}{3}cos3x-\dfrac{1}{4}sin4x+\dfrac{1}{2}e^{2x+1}+\dfrac{9^{x-1}}{ln9}+tanx+cotx+C\)
1) Đặt \(t=1+\sqrt{x-1}\Leftrightarrow x=\left(t-1\right)^2+1\forall t\ge1\Rightarrow dx=d\left(t-1\right)^2=2dt\)
\(\Rightarrow I_1=\int\frac{\left(t-1\right)^2+1}{t}\cdot2dt=2\int\frac{t^2-2t+2}{t}dt=2\int\left(t-2+\frac{2}{t}\right)dt\\ =t^2-4t+4lnt+C\)
Thay x vào ta có...
2) \(I_2=\int\frac{2sinx\cdot cosx}{cos^3x-\left(1-cos^2x\right)-1}dx=\int\frac{-2cosx\cdot d\left(cosx\right)}{cos^3x+cos^2x-2}=\int\frac{-2t\cdot dt}{t^3+t-2}\)
\(I_2=\int\frac{-2t}{\left(t-1\right)\left(t^2+2t+2\right)}dt=-\frac{2}{5}\int\frac{dt}{t-1}+\frac{1}{5}\int\frac{2t+2}{t^2+2t+2}dt-\frac{6}{5}\int\frac{dt}{\left(t+1\right)^2+1}\)
Ta có:
\(\int\frac{2t+2}{t^2+2t+2}dt=\int\frac{d\left(t^2+2t+2\right)}{t^2+2t+2}=ln\left(t^2+2t+2\right)+C\)
\(\int\frac{dt}{\left(t+1\right)^2+1}=\int\frac{\frac{1}{cos^2m}}{tan^2m+1}dm=\int dm=m+C=arctan\left(t+1\right)+C\)
Thay x vào, ta có....
