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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: \(\Leftrightarrow\sqrt{3}\cdot\dfrac{1+\cos2x}{2}+\sin2x-\sqrt{3}\cdot\dfrac{1-\cos2x}{2}-1=0\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2}\cos2x+\sin2x+\sqrt{3}\cdot\dfrac{\cos2x-1}{2}-1=0\)
\(\Leftrightarrow\sin2x+\dfrac{\sqrt{3}}{2}\cos2x+\dfrac{\sqrt{3}}{2}\cos2x-\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}-2}{2}=0\)
\(\Leftrightarrow\sin2x+\sqrt{3}\cos2x=\dfrac{\sqrt{3}-\sqrt{3}+2}{2}=1\)
\(\Leftrightarrow\sin\left(2x+\dfrac{\Pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{\Pi}{3}=\dfrac{\Pi}{6}+k2\Pi\\2x+\dfrac{\Pi}{3}=\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{12}\Pi+k\Pi\\x=\dfrac{\Pi}{4}+k\Pi\end{matrix}\right.\)
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. \(4\cos^2\dfrac{x}{2}+\dfrac{1}{2}\sin x+3\sin^2\dfrac{x}{2}=3\)
\(\Leftrightarrow4\cos^2\dfrac{x}{2}+\dfrac{1}{2}\sin x+3\sin^2\dfrac{x}{2}-3\left(\cos^2\dfrac{x}{2}+\sin^2\dfrac{x}{2}\right)=0\)
\(\Leftrightarrow\cos^2\dfrac{x}{2}+\dfrac{1}{2}\sin x=0\)
\(\Leftrightarrow\dfrac{1+\cos x}{2}+\dfrac{1}{2}\sin x=0\)
\(\Leftrightarrow\cos x+\sin x=-1\)
\(A=x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
\(\Rightarrow A_{min}=\frac{3}{4}\) khi \(x=\frac{1}{2}\), \(A_{max}\) ko tồn tại
\(B=x^2+2x+5=\left(x+1\right)^2+4\ge4\)
\(\Rightarrow B_{min}=4\) khi \(x=-1\) , \(B_{max}\) ko tồn tại
\(C=\left(5-x\right)^2+2017\ge2017\)
\(\Rightarrow C_{min}=2017\) khi \(x=5\) , \(C_{max}\) ko tồn tại
\(D=20-\left(2x+1\right)^2\le20\)
\(\Rightarrow D_{max}=20\) khi \(x=-\frac{1}{2}\), \(D_{min}\) ko tồn tại
\(E=x^2+2\left|y+1\right|+7\ge7\)
\(\Rightarrow E_{min}=7\) khi \(\left\{{}\begin{matrix}x=0\\y=-1\end{matrix}\right.\) , \(E_{max}\) ko tồn tại
:v bn ns v là bn bik hết là dạng gì rr mà lm ko đc á :))
Bài 1. Ta có:
\(\begin{array}{l} S = \sum\limits_{k = 1}^n {{x^{2k}}} + \sum\limits_{k = 1}^n {\dfrac{1}{{{x^{2k}}}} + 2n} \\ = {x^2}\dfrac{{1 - {x^{2n}}}}{{1 - {x^2}}} + \dfrac{1}{{{x^2}}}.\dfrac{{1 - \dfrac{1}{{{x^{2n}}}}}}{{1 - \dfrac{1}{{{x^2}}}}} + 2n\\ = \dfrac{{\left( {1 - {x^{2n}}} \right)\left( {{x^{2n + 2}} - 1} \right)}}{{\left( {1 - {x^2}} \right){x^{2n}}}} + 2n \end{array}\)
Bài 2.
Ta có:
\(\begin{array}{l} T = \dfrac{1}{2} + \dfrac{3}{{{2^2}}} + \dfrac{5}{{{2^3}}} + ... + \dfrac{{2n - 1}}{{{2^n}}}\left( 1 \right)\\ \dfrac{1}{2}T = \dfrac{1}{{{2^2}}} + \dfrac{3}{{{2^3}}} + \dfrac{5}{{{2^4}}} + ... + \dfrac{{2n - 3}}{{{2^n}}} + \dfrac{{2n - 1}}{{{2^{n + 1}}}}\left( 2 \right) \end{array}\)
\((1)-(2)\)\(\Rightarrow \dfrac{1}{2}T = \dfrac{1}{2} + \dfrac{2}{{{2^2}}} + \dfrac{2}{{{2^3}}} + ... + \dfrac{2}{{{2^n}}} - \dfrac{{2n - 1}}{{{2^{n + 1}}}}\)
\(\begin{array}{l} \Rightarrow T = 2\left[ {\dfrac{1}{2} + \dfrac{1}{2}\dfrac{{1 - {{\left( {\dfrac{1}{2}} \right)}^{n - 1}}}}{{1 - \dfrac{1}{2}}} - \dfrac{{2n - 1}}{{{2^{n + 1}}}}} \right]\\ = 1 + \dfrac{{{2^{n - 1}} - 1}}{{{2^{n - 2}}}} - \dfrac{{2n - 1}}{{{2^n}}} \end{array}\)
\(S=x^2+\frac{1}{x^2}+2+x^4+\frac{1}{x^4}+2+...+x^{2n}+\frac{1}{x^{2n}}+2\)
\(=\left(x^2+x^4+...+x^{2n}\right)+\left(\frac{1}{x^2}+\frac{1}{x^4}+...+\frac{1}{x^{2n}}\right)+2n\)
\(=x^2.\frac{\left(x^2\right)^{n-1}-1}{x^2-1}+\frac{1}{x^2}.\frac{\left(\frac{1}{x^2}\right)^{n-1}-1}{\frac{1}{x^2}-1}+2n\)
\(=\frac{x^{2n}-x^2}{x^2-1}+\frac{x^{2-2n}-1}{1-x^2}+2n\)
\(T=\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+...+\frac{2n-3}{2^{n-1}}+\frac{2n-1}{2^n}\)
\(\Rightarrow2T=1+\frac{3}{2}+\frac{5}{2^2}+...+\frac{2n-1}{2^{n-1}}\)
\(\Rightarrow T=1+\frac{2}{2}+\frac{2}{2^2}+\frac{2}{2^3}+...+\frac{2}{2^{n-1}}-\frac{2n-1}{2^n}\)
\(T=1+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{n-2}}-\frac{2n-1}{2^n}\)
\(T=1+1.\frac{\left(\frac{1}{2}\right)^{n-2}-1}{\frac{1}{2}-1}-\frac{2n-1}{2^n}=3-\frac{1}{2^{n-1}}-\frac{2n-1}{2^n}=3-\frac{1}{2^n}-\frac{n}{2^{n-1}}\)
e.
\(3\left(1-sin^2x\right)-5sinx-1=0\)
\(\Leftrightarrow-3sin^2x-5sinx+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{3}\\sinx=-2\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(\frac{1}{3}\right)+k2\pi\\x=\pi-arcsin\left(\frac{1}{3}\right)+k2\pi\end{matrix}\right.\)
f.
\(2\left(2cos^2x-1\right)-cosx+7=0\)
\(\Leftrightarrow4cos^2x-cosx+5=0\)
Phương trình vô nghiệm
g.
\(\Leftrightarrow\sqrt{2}sin\left(4x+\frac{\pi}{4}\right)=2\)
\(\Leftrightarrow sin\left(4x+\frac{\pi}{4}\right)=\sqrt{2}>1\)
Phương trình vô nghiệm
h.
\(\Leftrightarrow\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx=\frac{1}{2}\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{6}=\frac{\pi}{6}+k2\pi\\x-\frac{\pi}{6}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
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/ \(y'=4\left(1+sin^2x\right)^3\left(1+sin^2x\right)'=8.sinx.cosx\left(1+sin^2x\right)^3=4sin2x.\left(1+sin^2x\right)^3\)
Chọn D.
Đầu tiên sử dụng quy tắc nhân.
y’ = [(x2 – x + 1)]’(x2 + x + 1)2 + [(x2 x + 1)2]/(x2 – x + 1)3.
Sau đó sử dụng công thức u a '
y' = 3(x2 – x + 1)2(x2 – x + 1)’(x2 + x + 1) + 2(x2 + x + 1)(x2 + x + 1)’(x2 – x + 1)3
y’ = 3(x2 – x + 1)2(2x – 1) (x2 + x + 1)2 + 2(x2 + x + 1)(2x + 1)(x2 – x + 1)3
y’ = (x2 – x + 1)2(x2 + x + 1)[3(2x – 1)(x2 + x + 1) + 2(2x + 1)(x2 – x + 1)].