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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.
\(y'=4\left(x-5\right)^3\)
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/ \(y'=4\left(1+sin^2x\right)^3\left(1+sin^2x\right)'=8.sinx.cosx\left(1+sin^2x\right)^3=4sin2x.\left(1+sin^2x\right)^3\)
1d.
Đề ko rõ
1e.
\(\Leftrightarrow\left(4cos^3x-3cosx\right)^2.cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left(4cos^2x-3\right)^2.cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left(2cos2x-1\right)^2cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left[\left(2cos2x-1\right)^2.cos2x-1\right]=0\)
\(\Leftrightarrow cos^2x\left(4cos^32x-4cos^22x+cos2x-1\right)=0\)
\(\Leftrightarrow cos^2x\left(cos2x-1\right)\left(4cos^22x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos2x=1\end{matrix}\right.\) \(\Leftrightarrow...\)
2b.
Đề thiếu
2c.
Nhận thấy \(cos2x=0\) ko phải nghiệm, chia 2 vế cho \(cos^32x\)
\(\frac{8sin^22x}{cos^22x}=\frac{\sqrt{3}sin2x}{cos2x}.\frac{1}{cos^22x}+\frac{1}{cos^22x}\)
\(\Leftrightarrow8tan^22x=\sqrt{3}tan2x\left(1+tan^22x\right)+1+tan^22x\)
\(\Leftrightarrow\sqrt{3}tan^32x-7tan^22x+\sqrt{3}tan2x+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=\frac{1}{\sqrt{3}}\\tanx=\sqrt{3}-2\\tanx=\sqrt{3}+2\end{matrix}\right.\)
\(\Leftrightarrow...\)
a.
\(cos\left(3x-\frac{\pi}{6}\right)=sin\left(2x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(3x-\frac{\pi}{6}\right)=cos\left(\frac{\pi}{6}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{6}=\frac{\pi}{6}-2x+k2\pi\\3x-\frac{\pi}{6}=2x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\cos3x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}cosx\ne0\\cos2x\ne\frac{1}{2}\end{matrix}\right.\)
\(tan3x-tanx=0\)
\(\Leftrightarrow\frac{sin3x}{cos3x}-\frac{sinx}{cosx}=0\)
\(\Leftrightarrow sin3x.cosx-cos3x.sinx=0\)
\(\Leftrightarrow sin2x=0\)
\(\Leftrightarrow2sinx.cosx=0\)
\(\Leftrightarrow sinx=0\Leftrightarrow x=k\pi\)
c.
\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{2\pi}{5}\right)=\frac{1}{2}-\frac{1}{2}cos\left(4x+\frac{8\pi}{5}\right)\)
\(\Leftrightarrow cos\left(2x-\frac{2\pi}{5}\right)=-cos\left(4x+\frac{3\pi}{5}+\pi\right)\)
\(\Leftrightarrow cos\left(2x-\frac{2\pi}{5}\right)=cos\left(4x+\frac{3\pi}{5}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{3\pi}{5}=2x-\frac{2\pi}{5}+k2\pi\\4x+\frac{3\pi}{5}=\frac{2\pi}{5}-2x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
d.
\(\Leftrightarrow cos^2\left(2x-1\right)=0\)
\(\Leftrightarrow cos\left(2x-1\right)=0\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{1}{2}+\frac{k\pi}{2}\)
10. ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)
\(2cos2x+tanx=\frac{4}{5}\)
\(\Leftrightarrow4cos^2x-2+tanx=\frac{4}{5}\)
\(\Leftrightarrow\frac{4}{1+tan^2x}+tanx-\frac{14}{5}=0\)
Đặt \(tanx=t\)
\(\Rightarrow\frac{20}{1+t^2}+5t-14=0\)
\(\Leftrightarrow5t^3-14t^2+5t+6=0\)
\(\Leftrightarrow\left(t-2\right)\left(5t^2-4t-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=2\\t=\frac{2+\sqrt{19}}{5}\\t=\frac{2-\sqrt{19}}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}tanx=2=tana\\tanx=\frac{2+\sqrt{19}}{5}=tanb\\tanx=\frac{2-\sqrt{19}}{5}=tanc\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=a+k\pi\\x=b+k\pi\\x=c+k\pi\end{matrix}\right.\)
9.
\(\Leftrightarrow cos2x-3cosx=2\left(cosx+1\right)\)
\(\Leftrightarrow2cos^2x-1-3cosx=2cosx+2\)
\(\Leftrightarrow2cos^2x-5cosx-3=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=3\left(l\right)\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow x=\pm\frac{2\pi}{3}+k2\pi\)
Câu 2:
\(f'\left(x\right)=\frac{-3}{\left(2x-1\right)^2}\)
a/ \(x_0=-1\Rightarrow\left\{{}\begin{matrix}f'\left(x_0\right)=-\frac{1}{3}\\f\left(x_0\right)=0\end{matrix}\right.\)
Pttt: \(y=-\frac{1}{3}\left(x+1\right)=-\frac{1}{3}x-\frac{1}{3}\)
b/ \(y_0=1\Rightarrow\frac{x_0+1}{2x_0-1}=1\Leftrightarrow x_0+1=2x_0-1\Rightarrow x_0=2\)
\(\Rightarrow f'\left(x_0\right)=-\frac{1}{3}\)
Pttt: \(y=-\frac{1}{3}\left(x-2\right)+1\)
c/ \(x_0=0\Rightarrow\left\{{}\begin{matrix}f'\left(x_0\right)=-3\\y_0=-1\end{matrix}\right.\)
Pttt: \(y=-3x-1\)
d/ \(6x+2y-1=0\Leftrightarrow y=-3x+\frac{1}{2}\)
Tiếp tuyến song song d \(\Rightarrow\) có hệ số góc bằng -3
\(\Rightarrow\frac{-3}{\left(2x_0-1\right)^2}=-3\Rightarrow\left(2x_0-1\right)^2=1\Rightarrow\left[{}\begin{matrix}x_0=0\Rightarrow y_0=-1\\x_0=1\Rightarrow y_0=2\end{matrix}\right.\)
Có 2 tiếp tuyến thỏa mãn: \(\left[{}\begin{matrix}y=-3x-1\\y=-3\left(x-1\right)+2\end{matrix}\right.\)
Làm câu 1,3 trước, câu 2 hơi dài tối rảnh làm sau:
1/ \(\lim\limits\frac{n^2+2n+1}{2n^2-1}=lim\frac{1+\frac{2}{n}+\frac{1}{n^2}}{2-\frac{1}{n^2}}=\frac{1}{2}\)
\(\lim\limits_{x\rightarrow0}\frac{2\sqrt{x+1}-x^2+2x+2}{x}=\frac{2-0+0+2}{0}=\frac{4}{0}=+\infty\)
Chắc bạn ghi nhầm đề, câu này biểu thức tử số là \(...-x^2+2x-2\) thì hợp lý hơn
3/ \(y'=2sin2x.\left(sin2x\right)'=4sin2x.cos2x=2sin4x\)
b/ \(y'=4x^3-4x\)
c/ \(y'=\frac{3\left(x+2\right)-1\left(3x-1\right)}{\left(x+2\right)^2}=\frac{7}{\left(x+2\right)^2}\)
d/ \(y'=10\left(x^2+x+1\right)^9\left(x^2+x+1\right)'=10\left(x^2+x+1\right)^9.\left(2x+1\right)\)
e/ \(y'=\frac{\left(2x^2-x+3\right)'}{2\sqrt{2x^2-x+3}}=\frac{4x-1}{2\sqrt{2x^2-x+3}}\)
d/
\(\Leftrightarrow\sqrt{2}\left(\frac{1}{2}cos\left(\frac{x}{5}-\frac{\pi}{12}\right)-\frac{\sqrt{3}}{2}sin\left(\frac{x}{5}-\frac{\pi}{12}\right)\right)=sin\left(\frac{x}{5}+\frac{2\pi}{3}\right)-sin\left(\frac{3x}{5}+\frac{\pi}{6}\right)\)
\(\Leftrightarrow\sqrt{2}cos\left(\frac{x}{5}-\frac{\pi}{12}+\frac{\pi}{3}\right)=2cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)sin\left(\frac{\pi}{4}-\frac{x}{5}\right)\)
\(\Leftrightarrow cos\left(\frac{x}{5}-\frac{\pi}{4}\right)=\sqrt{2}cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)cos\left(\frac{x}{5}-\frac{\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(\frac{x}{5}-\frac{\pi}{4}\right)=0\\cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{5}-\frac{\pi}{4}=\frac{\pi}{2}+k\pi\\\frac{2x}{5}+\frac{5\pi}{12}=\frac{\pi}{4}+k2\pi\\\frac{2x}{5}+\frac{5\pi}{12}=-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{15\pi}{4}+k5\pi\\x=-\frac{5\pi}{12}+k5\pi\\x=-\frac{5\pi}{3}+k5\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow\sqrt{3}sin\left(x-\frac{\pi}{3}\right)+cos\left(\frac{\pi}{3}-x\right)=2sin1972x\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin\left(x-\frac{\pi}{3}\right)+\frac{1}{2}cos\left(x-\frac{\pi}{3}\right)=sin1972x\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{3}+\frac{\pi}{6}\right)=sin1972x\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=sin1972x\)
\(\Leftrightarrow\left[{}\begin{matrix}1972x=x-\frac{\pi}{6}+k2\pi\\1972x=\frac{7\pi}{6}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{11826}+\frac{k2\pi}{1971}\\x=\frac{7\pi}{11838}+\frac{k2\pi}{1973}\end{matrix}\right.\)
25.
H là hình chiếu của S lên (ABC)
Do \(SA=SB=SC\Rightarrow HA=HB=HC\)
\(\Rightarrow\) H là tâm đường tròn ngoại tiếp tam giác ABC
26.
\(\left\{{}\begin{matrix}AB\perp BC\\AB\perp CD\end{matrix}\right.\) \(\Rightarrow AB\perp\left(BCD\right)\) \(\Rightarrow AB\perp BD\)
\(\Rightarrow\Delta ABD\) vuông tại B
Pitago tam giác vuông BCD (vuông tại C):
\(BC^2+CD^2=BD^2\Rightarrow BD^2=b^2+c^2\)
Pitago tam giác vuông ABD:
\(AD^2=AB^2+BC^2=a^2+b^2+c^2\)
\(\Rightarrow AD=\sqrt{a^2+b^2+c^2}\)
23.
Gọi H là chân đường cao hạ từ S xuống BC
\(\Rightarrow BH=SB.cos30^0=3a\) ; \(SH=SB.sin30^0=a\sqrt{3}\) ; \(CH=4a-3a=a\)
\(\Rightarrow BC=4HC\Rightarrow d\left(B;\left(SAC\right)\right)=4d\left(H;\left(SAC\right)\right)\)
Từ H kẻ \(HE\perp AC\) ; từ H kẻ \(HF\perp SE\Rightarrow HF\perp\left(SAC\right)\)
\(\Rightarrow HF=d\left(H;\left(SAC\right)\right)\)
\(HE=CH.sinC=\frac{CH.AB}{AC}=\frac{a.3a}{5a}=\frac{3a}{5}\)
\(\frac{1}{HF^2}=\frac{1}{HE^2}+\frac{1}{SH^2}\Rightarrow HF=\frac{HE.SH}{\sqrt{HE^2+SH^2}}=\frac{3a\sqrt{7}}{14}\)
\(\Rightarrow d\left(B;\left(SAC\right)\right)=4HF=\frac{6a\sqrt{7}}{7}\)
24.
\(SA=SC\Rightarrow SO\perp AC\)
\(SB=SD\Rightarrow SO\perp BD\)
\(\Rightarrow SO\perp\left(ABCD\right)\)