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16.
\(y'=\frac{\left(cos2x\right)'}{2\sqrt{cos2x}}=\frac{-2sin2x}{2\sqrt{cos2x}}=-\frac{sin2x}{\sqrt{cos2x}}\)
17.
\(y'=4x^3-\frac{1}{x^2}-\frac{1}{2\sqrt{x}}\)
18.
\(y'=3x^2-2x\)
\(y'\left(-2\right)=16;y\left(-2\right)=-12\)
Pttt: \(y=16\left(x+2\right)-12\Leftrightarrow y=16x+20\)
19.
\(y'=-\frac{1}{x^2}=-x^{-2}\)
\(y''=2x^{-3}=\frac{2}{x^3}\)
20.
\(\left(cotx\right)'=-\frac{1}{sin^2x}\)
21.
\(y'=1+\frac{4}{x^2}=\frac{x^2+4}{x^2}\)
22.
\(lim\left(3^n\right)=+\infty\)
11.
\(\lim\limits_{x\rightarrow1^+}\frac{-2x+1}{x-1}=\frac{-1}{0}=-\infty\)
12.
\(y=cotx\Rightarrow y'=-\frac{1}{sin^2x}\)
13.
\(y'=2020\left(x^3-2x^2\right)^{2019}.\left(x^3-2x^2\right)'=2020\left(x^3-2x^2\right)^{2019}\left(3x^2-4x\right)\)
14.
\(y'=\frac{\left(4x^2+3x+1\right)'}{2\sqrt{4x^2+3x+1}}=\frac{8x+3}{2\sqrt{4x^2+3x+1}}\)
15.
\(y'=4\left(x-5\right)^3\)
1/ \(y=x^{-1}+\frac{2}{3}x^{-2}-\frac{2}{3}\Rightarrow y'=-\frac{1}{x^2}-\frac{4}{3x^3}\)
\(3x^3y'+3x+4=3x^3\left(-\frac{1}{x^2}-\frac{4}{3x^3}\right)+3x+4\)
\(=-3x-4+3x+4=0\) (đpcm)
2/ \(y'\le0\)
\(\Leftrightarrow3x^2-10x+7\le0\)
\(\Leftrightarrow1\le x\le\frac{7}{3}\)
a/ \(y'=4\left(2x-3\right)^3.\left(2x-3\right)'=8\left(2x-3\right)^3\)
b/ \(y'=5cos^43x.\left(cos3x\right)'=-15cos^43x.sin3x\)
c/ \(y'=\frac{\left[cos\left(1-2x^2\right)\right]'}{2\sqrt{cos\left(1-2x^2\right)}}=\frac{-sin\left(1-2x^2\right).\left(1-2x^2\right)'}{2\sqrt{cos\left(1-2x^2\right)}}=\frac{2x.sin\left(1-2x^2\right)}{\sqrt{cos\left(1-2x^2\right)}}\)
d/ \(y'=\frac{\left(\frac{x+1}{x-1}\right)'}{2\sqrt{\frac{x+1}{x-1}}}=\frac{\frac{-2}{\left(x-1\right)^2}}{2\sqrt{\frac{x+1}{x-1}}}=-\frac{1}{\left(x-1\right)^2\sqrt{\frac{x+1}{x-1}}}\)
e/ \(y'=4\left(1+sin^2x\right)^3\left(1+sin^2x\right)'=8.sinx.cosx\left(1+sin^2x\right)^3=4sin2x.\left(1+sin^2x\right)^3\)
1: \(\Leftrightarrow4\cdot\dfrac{1+\cos2x}{2}-6\cdot\dfrac{1-\cos2x}{2}+5\sin2x-4=0\)
\(\Leftrightarrow2+2\cos2x-3+3\cos2x+5\sin2x-4=0\)
\(\Leftrightarrow5\sin2x+5\cos2x=5\)
\(\Leftrightarrow\cos2x+\sin2x=1\)
\(\Leftrightarrow\sqrt{2}\cdot\sin\left(2x+\dfrac{\Pi}{4}\right)=1\)
\(\Leftrightarrow\sin\left(2x+\dfrac{\Pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{\Pi}{4}=\dfrac{\Pi}{4}+k2\Pi\\2x+\dfrac{\Pi}{4}=\dfrac{3\Pi}{4}+k2\Pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=k\Pi\\x=\dfrac{\Pi}{4}+k\Pi\end{matrix}\right.\)
2: \(\Leftrightarrow\sqrt{3}\cdot\dfrac{1+\cos2x}{2}+\sin2x-\sqrt{3}\cdot\dfrac{1-\cos2x}{2}-1=0\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2}\cos2x+\sin2x+\sqrt{3}\cdot\dfrac{\cos2x-1}{2}-1=0\)
\(\Leftrightarrow\sin2x+\dfrac{\sqrt{3}}{2}\cos2x+\dfrac{\sqrt{3}}{2}\cos2x-\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}-2}{2}=0\)
\(\Leftrightarrow\sin2x+\sqrt{3}\cos2x=\dfrac{\sqrt{3}-\sqrt{3}+2}{2}=1\)
\(\Leftrightarrow\sin\left(2x+\dfrac{\Pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{\Pi}{3}=\dfrac{\Pi}{6}+k2\Pi\\2x+\dfrac{\Pi}{3}=\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{12}\Pi+k\Pi\\x=\dfrac{\Pi}{4}+k\Pi\end{matrix}\right.\)
a.
\(1-sin^2x+1-2sin^2x+sinx+2=0\)
\(\Leftrightarrow-3sin^2x+sinx+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\frac{4}{3}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=-\frac{\pi}{2}+k2\pi\)
b. ĐKXĐ; ...
\(5tanx-\frac{2}{tanx}-3=0\)
\(\Leftrightarrow5tan^2x-3tanx-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=-\frac{2}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=arctan\left(-\frac{2}{5}\right)+k\pi\end{matrix}\right.\)
e.
Ko rõ vế phải
f.
\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)
\(\Leftrightarrow1-2sin^22x=0\)
\(\Leftrightarrow cos4x=0\)
\(\Leftrightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}\)
1) đặc : \(f\left(x\right)=y=cot4x\)
điều kiện xác định : \(sin4x\ne0\Leftrightarrow4x\ne k\pi\Leftrightarrow x\ne\dfrac{k\pi}{4}\)
\(\Rightarrow x\in D\) thì \(-x\in D\)
ta có : \(f\left(-x\right)=cot\left(-4x\right)=-cot4x=-f\left(x\right)\)
\(\Rightarrow\) hàm này là hàm lẽ
2) đặc : \(f\left(x\right)=y=\left|cotx\right|\)
điều kiện xác định : \(sinx\ne0\Leftrightarrow x\ne k\pi\)
\(\Rightarrow x\in D\) thì \(-x\in D\)
ta có : \(f\left(-x\right)=\left|cot\left(-x\right)\right|=\left|-cotx\right|=\left|cotx\right|=f\left(x\right)\)
\(\Rightarrow\) hàm này là hàm chẳn
3) đặc : \(f\left(x\right)=y=1-sin^2x=cos^2x\)
điều kiện xác định : \(D=R\)
\(\Rightarrow x\in D\) thì \(-x\in D\)
ta có : \(f\left(-x\right)=cos^2\left(-x\right)=cos^2x=f\left(x\right)\)
\(\Rightarrow\) hàm này là hàm chẳn
4) đặc : \(f\left(x\right)=y=sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{sinx+cosx}{\sqrt{2}}\)
điều kiện xác định : \(D=R\)
\(\Rightarrow x\in D\) thì \(-x\in D\)
ta có : \(f\left(-x\right)=\dfrac{sin\left(-x\right)+cos\left(-x\right)}{\sqrt{2}}=\dfrac{-sinx+cosx}{\sqrt{2}}\ne f\left(x\right);-f\left(x\right)\)
\(\Rightarrow\) hàm này là hàm không chẳn không lẽ
mấy bài còn lại bn làm tương tự cho quen nha
1. \(4\cos^2x-6\sin^2x+5\sin2x-4=0\)
\(\Leftrightarrow4\cos^2x-6\sin^2x+10\sin x\cos x-4\left(\cos^2x+\sin^2x\right)=0\)
\(\Leftrightarrow10\sin x\cos x-10\sin^2x=0\)
\(\Leftrightarrow10\sin x\left(\cos x-\sin x\right)=0\)
2. \(\sqrt{3}\cos^2x+2\sin x\cos x-\sqrt{3}\sin^2x-1=0\)
\(\Leftrightarrow\left(\sqrt{3}\cos^2x+\sin x\cos x\right)+\left(\sin x\cos x-\sqrt{3}\sin^2x\right)-1=0\)
\(\Leftrightarrow2\cos x\left(\dfrac{\sqrt{3}}{2}\cos x+\dfrac{1}{2}\sin x\right)+2\sin x\left(\dfrac{1}{2}\cos x-\dfrac{\sqrt{3}}{2}\sin x\right)-1=0\)
\(\Leftrightarrow2\cos x.\cos\left(\dfrac{\Pi}{6}-x\right)+2\sin x.\sin\left(\dfrac{\Pi}{6}-x\right)-1=0\)
\(\Leftrightarrow\cos\dfrac{\Pi}{6}+\cos\left(2x-\dfrac{\Pi}{6}\right)+\cos\left(2x-\dfrac{\Pi}{6}\right)-\cos\dfrac{\Pi}{6}-1=0\)
\(\Leftrightarrow\cos\left(2x-\dfrac{\Pi}{6}\right)=\dfrac{1}{2}\)
3. \(2\sin^22x-3\sin2x\cos2x+\cos^22x=2\)
\(\Leftrightarrow2\sin^22x-3\sin2x\cos2x+\cos^22x-2\left(\sin^22x+\cos^22x\right)=0\)
\(\Leftrightarrow3\sin2x\cos2x+\cos^22x=0\)
\(\Leftrightarrow\cos2x\left(3\sin2x+\cos2x\right)=0\)
-TH1: ...
- TH2: \(\cos2x=-3\sin2x\) mà \(\cos^22x+\sin^22x=1\) suy ra ...
4. \(4\cos^2\dfrac{x}{2}+\dfrac{1}{2}\sin x+3\sin^2\dfrac{x}{2}=3\)
\(\Leftrightarrow4\cos^2\dfrac{x}{2}+\dfrac{1}{2}\sin x+3\sin^2\dfrac{x}{2}-3\left(\cos^2\dfrac{x}{2}+\sin^2\dfrac{x}{2}\right)=0\)
\(\Leftrightarrow\cos^2\dfrac{x}{2}+\dfrac{1}{2}\sin x=0\)
\(\Leftrightarrow\dfrac{1+\cos x}{2}+\dfrac{1}{2}\sin x=0\)
\(\Leftrightarrow\cos x+\sin x=-1\)
Chọn D.
Ta có:
⇔ y’ = x3 – x2 + x – 1.