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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 \(cos^32x\)
\(\frac{8sin^22x}{cos^22x}=\frac{\sqrt{3}sin2x}{cos2x}.\frac{1}{cos^22x}+\frac{1}{cos^22x}\)
\(\Leftrightarrow8tan^22x=\sqrt{3}tan2x\left(1+tan^22x\right)+1+tan^22x\)
\(\Leftrightarrow\sqrt{3}tan^32x-7tan^22x+\sqrt{3}tan2x+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=\frac{1}{\sqrt{3}}\\tanx=\sqrt{3}-2\\tanx=\sqrt{3}+2\end{matrix}\right.\)
\(\Leftrightarrow...\)
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.
\(\Leftrightarrow cos2x-3cosx=2\left(cosx+1\right)\)
\(\Leftrightarrow2cos^2x-1-3cosx=2cosx+2\)
\(\Leftrightarrow2cos^2x-5cosx-3=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=3\left(l\right)\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow x=\pm\frac{2\pi}{3}+k2\pi\)
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.
\(\Leftrightarrow cos^2\left(2x-1\right)=0\)
\(\Leftrightarrow cos\left(2x-1\right)=0\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{1}{2}+\frac{k\pi}{2}\)
Cho e hỏi là vì sao khúc cuối có dấu bằng mà trên đề k có dấu bằng ạ?
Vì mình lấy giá trị nguyên bạn
Chính xác là \(-\frac{1}{4}< k< \frac{2020-\frac{\pi}{2}}{2\pi}\)
\(\Rightarrow-0,25< k< 321,243\) (1)
Nhưng k nguyên nên chỉ cần lấy khoảng ở số nguyên gần nhất, tức là \(0\le k\le321\)
a/ Thiếu đề, sau dấu "-" hình như còn gì đó
b/ \(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}=sin\left(\frac{\pi}{4}\right)\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
c/ \(\Rightarrow sin2x=-sinx\Leftrightarrow sin2x=sin\left(-x\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x=-x+k2\pi\\2x=\pi+x+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{k2\pi}{3}\\x=\pi+k2\pi\end{matrix}\right.\)
d/ \(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2\left(sinx.cosx\right)^2=1\)
\(\Leftrightarrow sinx.cosx=0\Leftrightarrow sin2x=0\)
\(\Rightarrow2x=k\pi\Rightarrow x=\frac{k\pi}{2}\)
e/ f/ Thiếu đề
g/ \(\Leftrightarrow\left[{}\begin{matrix}cos3x=cos2x\\cos3x=-cos2x\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}cos3x=cos2x\\cos3x=cos\left(\pi-2x\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=2x+k2\pi\\3x=-2x+k2\pi\\3x=\pi-2x+k2\pi\\3x=2x-\pi+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k2\pi\\x=\frac{k2\pi}{5}\\x=\frac{\pi}{5}+\frac{k2\pi}{5}\\x=-\pi+k2\pi\end{matrix}\right.\)
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.
\(SA=SC\Rightarrow SO\perp AC\)
\(SB=SD\Rightarrow SO\perp BD\)
\(\Rightarrow SO\perp\left(ABCD\right)\)
20.
\(\left(SAB\right)\perp\left(ABCD\right);\left(SAD\right)\perp\left(ABCD\right)\Rightarrow SA\perp\left(ABCD\right)\)
\(\Rightarrow SA\perp BD\)
\(\Rightarrow\) Mệnh đề D sai vì nếu D đúng \(\left\{{}\begin{matrix}BD\perp SA\\BD\perp AK\end{matrix}\right.\) \(\Rightarrow BD\perp\left(SAD\right)\)
\(\Rightarrow BD\perp AD\) (vô lý)
21.
\(cos\left(SI;BC\right)=\frac{\left|\overrightarrow{BC}.\overrightarrow{SI}\right|}{BC.SI}=\frac{\left|\overrightarrow{BC}\left(\overrightarrow{SB}+\overrightarrow{SA}\right)\right|}{2BC.SI}=\frac{\left|\overrightarrow{BC}.\overrightarrow{SB}\right|}{2BC.SI}=\frac{a\sqrt{2}.a.cos45^0}{\frac{2.a\sqrt{2}.a\sqrt{2}}{2}}=\frac{1}{2}\)
\(\Rightarrow\left(SI;BC\right)=60^0\)
22.
\(MC=\frac{a\sqrt{3}}{2}\Rightarrow tan\alpha=\frac{CC'}{MC}=\frac{2\sqrt{3}}{3}\)
16.
\(y=x^2\Rightarrow dy=2x.dx\)
17.
Mệnh đề C sai
\(BC\perp\left(SAB\right)\Rightarrow BC\perp AB\)
Điều này chỉ đúng khi tam giác ABC vuông tại B chứ ko đúng với tam giác bất kì
18.
\(SA=SC\Rightarrow\Delta SAC\) cân tại S \(\Rightarrow SO\perp AC\) (1)
\(SB=SD\Rightarrow\Delta SBD\) cân tại S \(\Rightarrow SO\perp BD\) (2)
(1);(2) \(\Rightarrow SO\perp\left(ABCD\right)\)
19.
Khẳng định A sai
\(CD\perp\left(SBC\right)\Rightarrow CD\perp BC\Rightarrow ABCD\) là hình chữ nhật
\(\Rightarrow BC=AD\) (vô lý do BC=a; CD=2a)
15.
Phương trình đường thẳng d có hệ số góc bằng 1 qua giao của 2 tiệm cận:
\(y=1\left(x-2\right)+2\Leftrightarrow y=x\)
Hoành độ giao điểm của d và (C): \(x=\frac{2x-3}{x-2}\Leftrightarrow x^2-4x+3=0\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}M\left(1;1\right)\\N\left(3;3\right)\end{matrix}\right.\)
\(y=-\frac{1}{\left(x-2\right)^2}\Rightarrow\left[{}\begin{matrix}y'\left(1\right)=-1\\y'\left(3\right)=-1\end{matrix}\right.\)
Có 2 tiếp tuyến thỏa mãn: \(\left[{}\begin{matrix}y=-\left(x-1\right)+1=-x+2\\y=-1\left(x-3\right)+3=-x+6\end{matrix}\right.\)
Giao của 2 tiếp tuyến với 2 đường thẳng đã cho là: \(\left[{}\begin{matrix}A\left(2;0\right);B\left(0;2\right)\\C\left(2;4\right);D\left(4;2\right)\end{matrix}\right.\)
\(\Rightarrow\) Tổng các hoành độ tiếp điểm bằng 4 hoặc 6 (đáp án thiếu 1 trường hợp)
Câu 12:
\(y'=-6x^2+12x\Rightarrow\left\{{}\begin{matrix}y'\left(3\right)=-18\\y\left(3\right)=-5\end{matrix}\right.\)
Tiếp tuyến: \(y=-18\left(x-3\right)-5=-18x+49\)
13.
\(y'=3x^2\Rightarrow y'\left(1\right)=3\)
14.
\(y'=-x^2-4x-3=-\left(x+2\right)^2+1\le1\)
\(\Rightarrow k_{max}=1\)
Phần còn lại tối rảnh làm tiếp
9.
\(y'=3x^2+6x\)
\(y'\left(1\right)=9;y\left(1\right)=2\)
Tiếp tuyến: \(y=9\left(x-1\right)+2=9x-7\)
10.
\(y'=\frac{-1}{\left(x-1\right)^2}=2019\)
\(\Leftrightarrow\left(x-1\right)^2=-\frac{1}{2019}< 0\)
Phương trình vô nghiệm \(\Rightarrow\) có 0 điểm thỏa mãn
11.
\(y'=\frac{3}{\left(x+2\right)^2}\Rightarrow\left\{{}\begin{matrix}y'\left(-3\right)=3\\y\left(-3\right)=4\end{matrix}\right.\)
Tiếp tuyến: \(y=3\left(x+3\right)+4=3x+13\)
6.
Chắc đoạn \(\left(3m+2\right)\) bạn ghi thiếu x đằng sau
\(y'=\left(m+1\right)x^3-2\left(m+1\right)x+3m+2\)
\(y'\le0;\forall x\in R\) \(\Leftrightarrow\left\{{}\begin{matrix}m+1< 0\\\Delta'=\left(m+1\right)^2-\left(m+1\right)\left(3m+2\right)\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m< -1\\\left(m+1\right)\left(-2m-1\right)\le0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m< -1\\\left[{}\begin{matrix}m\ge-1\\m\le-\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow m\le-\frac{1}{2}\)
7.
\(y'=-3x^2+1\Rightarrow y'\left(-2\right)=-11\)
8.
\(f'\left(x\right)=-3x^2+6mx-12\)
\(\Delta'=9m^2-36\le0\Rightarrow-2\le m\le2\)
\(\Rightarrow\) Có 5 giá trị nguyên của m thỏa mãn
4.
\(y'=\frac{-6}{\left(x-2\right)^2}\)
Phương trình tiếp tuyến d: \(y=-\frac{6}{\left(x_0-2\right)^2}\left(x-x_0\right)+y_0\)
Gọi A là giao d với Ox \(\Rightarrow A\left(\frac{y_0\left(x_0-2\right)^2}{6}+x_0;0\right)\)
B là giao d với Oy \(\Rightarrow B\left(0;\frac{6x_0}{\left(x_0-2\right)^2}+y_0\right)\)
\(\Rightarrow\left|2x_0y_0+\frac{y_0^2\left(x_0-2\right)^2}{6}+6\left(\frac{x_0}{x_0-2}\right)^2\right|=\frac{3}{2}\)
\(\Leftrightarrow\left(x_0+\frac{2x_0}{x_0-2}\right)^2=1\Leftrightarrow\frac{x_0^2}{x_0-2}=-1\)
\(\Rightarrow x_0=-2\Rightarrow y_0=\frac{3}{2}\Rightarrow x_0+2y_0=1\)
5.
Đặt \(g\left(x\right)=f\left(x\right)+f\left(2x\right)\Rightarrow g'\left(x\right)=f'\left(x\right)+2f'\left(2x\right)\)
\(\Rightarrow\left\{{}\begin{matrix}f'\left(1\right)+2f'\left(2\right)=18\\f'\left(2\right)+2f'\left(4\right)=1000\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}f'\left(1\right)+2f'\left(2\right)=18\\2f'\left(2\right)+4f'\left(4\right)=2000\end{matrix}\right.\)
\(\Rightarrow f'\left(1\right)-4f'\left(4\right)=-1982\)
Đặt \(h\left(x\right)=f\left(x\right)-f\left(4x\right)\Rightarrow h'\left(x\right)=f'\left(x\right)-4f'\left(4x\right)\)
\(\Rightarrow h'\left(1\right)=f'\left(1\right)-4f'\left(4\right)=-1982\)
Cả 4 đáp án đều sai
1.
\(y'=\sqrt{x^2-2x}+x.\frac{\left(x^2-2x\right)'}{2\sqrt{x^2-2x}}=\sqrt{x^2-2x}+\frac{x^2-x}{\sqrt{x^2-2x}}\)
\(=\frac{x^2-2x+x^2-x}{\sqrt{x^2-2x}}=\frac{2x^2-3x}{\sqrt{x^2-2x}}\)
2.
\(f\left(x\right)=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=1-\frac{1}{2}sin^22x\)
\(g\left(x\right)=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\)
\(=1-\frac{3}{4}sin^22x\)
\(\Rightarrow3f\left(x\right)-2g\left(x\right)+2=3-\frac{3}{2}sin^22x-2+\frac{3}{2}sin^22x+2=3\)
3.
\(y'=\frac{-3\left(x-2\right)-\left(-3x+4\right)}{\left(x-2\right)^2}=\frac{2}{\left(x-2\right)^2}\)