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![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(y=\dfrac{1}{2}\sin4x-1\)
\(-1< =\sin4x< =1\)
\(\Leftrightarrow-\dfrac{1}{2}< =\dfrac{1}{2}\cdot\sin4x< =\dfrac{1}{2}\)
\(\Leftrightarrow-\dfrac{3}{2}< =\dfrac{1}{2}\cdot\sin4x-1< =-\dfrac{1}{2}\)
Do đó: \(y_{max}=\dfrac{-1}{2}\) khi \(4x=\dfrac{\Pi}{2}+k\Pi\)
hay \(x=\dfrac{\Pi}{8}+\dfrac{k\Pi}{4}\)
\(y_{min}=\dfrac{-3}{2}\) khi \(4x=-\dfrac{\Pi}{2}+k\Pi\)
hay \(x=-\dfrac{\Pi}{8}+\dfrac{k\Pi}{4}\)
g: \(0>=-2\left|\cos x\right|>=-2\)
\(\Leftrightarrow5>=-2\left|\cos x\right|+5>=3\)
Do đó: \(y_{max}=5\) khi \(\)\(\cos x=0\)
hay \(x=\dfrac{\Pi}{2}+k\Pi\)
\(y_{min}=3\) khi \(\cos x=-1\)
hay \(x=-\Pi+k2\Pi\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
Kiểm tra lại đề bài, câu này phải là \(\dfrac{sinx+2cosx+3}{2sinx+cosx+3}\) mới đúng
2.a
ĐKXĐ: \(cosx\ne0\)
\(\Leftrightarrow\dfrac{1}{cos^2x}=4tanx+6\)
\(\Leftrightarrow1+tan^2x=4tanx+6\)
\(\Leftrightarrow tan^2x-4tanx-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=arctan\left(5\right)+k\pi\end{matrix}\right.\)
2b.
Đặt \(x-\dfrac{\pi}{4}=t\Rightarrow x=t+\dfrac{\pi}{4}\)
\(sin^3t=\sqrt{2}sin\left(t+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow sin^3t=sint+cost\)
\(\Leftrightarrow sint\left(1-cos^2t\right)=sint+cost\)
\(\Leftrightarrow sint.cos^2t+cost=0\)
\(\Leftrightarrow cost\left(sint.cost+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cost=0\\sin2t=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=0\\sin\left(2x-\dfrac{\pi}{2}\right)=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=0\\cos2x=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow...\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\frac{\sqrt[3]{5x+3}-2+2-\sqrt{2x+2}}{x-1}=\lim\limits_{x\rightarrow1^-}\frac{\frac{5\left(x-1\right)}{\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4}-\frac{2\left(x-1\right)}{2+\sqrt{2x+2}}}{x-1}\)
\(=\lim\limits_{x\rightarrow1^-}\left(\frac{5}{\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4}-\frac{2}{2+\sqrt{2x+2}}\right)=-\frac{1}{12}\)
\(\lim\limits_{x\rightarrow1^+}=\lim\limits_{x\rightarrow1^+}m.sin\left(\frac{\pi x}{2}+2019\right)=\)
Đến đây lại thêm vấn đề nữa, \(sin\left(\frac{\pi x}{2}+2019\right)\) hay \(sin\left(\frac{\pi x}{2}+2019\pi\right)\) bạn?
Bạn ghi đề sai thì phải, nhìn hàm khi \(x< 1\) thì \(\lim\limits_{x\rightarrow1^-}f\left(x\right)\) không tồn tại (ko phải dạng vô định \(\frac{0}{0}\), khi thay x=1 vào tử số ra khác 0)
![](https://rs.olm.vn/images/avt/0.png?1311)
Coi như tất cả các biểu thức cần tính đạo hàm đều xác định.
1.
\(y'=2sin\sqrt{4x+3}.\left(sin\sqrt{4x+3}\right)'=2sin\sqrt{4x+3}.cos\sqrt{4x+3}.\left(\sqrt{4x+3}\right)'\)
\(=sin\left(2\sqrt{4x+3}\right).\dfrac{4}{2\sqrt{4x+3}}=\dfrac{2sin\left(2\sqrt{4x+3}\right)}{\sqrt{4x+3}}\)
2.
\(y'=3x^3+\dfrac{17}{x\sqrt{x}}\)
3.
\(y'=\dfrac{1}{2\sqrt{\dfrac{sin4x}{cos\left(x^2+2\right)}}}.\left(\dfrac{sin4x}{cos\left(x^2+2\right)}\right)'\)
\(=\dfrac{1}{2\sqrt{\dfrac{sin4x}{cos\left(x^2+2\right)}}}.\dfrac{4cos4x.cos\left(x^2+2\right)+2x.sin4x.sin\left(x^2+2\right)}{cos^2\left(x^2+2\right)}\)
4.
\(y'=-\dfrac{\left(\sqrt{sin^2\left(6-x\right)+4x}\right)'}{sin^2\left(6-x\right)+4x}=-\dfrac{\left[sin^2\left(6-x\right)+4x\right]'}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}\)
\(=-\dfrac{2sin\left(6-x\right).\left[sin\left(6-x\right)\right]'+4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}=-\dfrac{-2sin\left(6-x\right).cos\left(6-x\right)+4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}\)
\(=\dfrac{sin\left(12-2x\right)-4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}\)
5.
\(y'=sin^2\left(\dfrac{2x-1}{4-x}\right)+2x.sin\left(\dfrac{2x-1}{4-x}\right).\left[sin\left(\dfrac{2x-1}{4-x}\right)\right]'\)
\(=sin^2\left(\dfrac{2x-1}{4-x}\right)+2x.sin\left(\dfrac{2x-1}{4-x}\right).cos\left(\dfrac{2x-1}{4-x}\right).\left(\dfrac{2x-1}{4-x}\right)'\)
\(=sin^2\left(\dfrac{2x-1}{4-x}\right)+x.sin\left(\dfrac{4x-2}{4-x}\right).\dfrac{7}{\left(4-x\right)^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1, \(y=2-sin\left(\dfrac{3x}{2}+x\right).cos\left(x+\dfrac{\pi}{2}\right)\)
\(y=2-\left(-cosx\right).\left(-sinx\right)\)
y = 2 - sinx.cosx
y = \(2-\dfrac{1}{2}sin2x\)
Max = 2 + \(\dfrac{1}{2}\) = 2,5
Min = \(2-\dfrac{1}{2}\) = 1,5
2, y = \(\sqrt{5-\dfrac{1}{2}sin^22x}\)
Min = \(\sqrt{5-\dfrac{1}{2}}=\dfrac{3\sqrt{2}}{2}\)
Max = \(\sqrt{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
Chắc đề là \(sin\left[\pi sin2x\right]=1?\)
\(\Leftrightarrow\pi.sin2x=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow sin2x=\dfrac{1}{2}+2k\) (1)
Do \(-1\le sin2x\le1\Rightarrow-1\le\dfrac{1}{2}+2k\le1\)
\(\Rightarrow-\dfrac{3}{4}\le k\le\dfrac{1}{4}\Rightarrow k=0\)
Thế vào (1)
\(\Rightarrow sin2x=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{6}+n2\pi\\2x=\dfrac{5\pi}{6}+m2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+n\pi\\x=\dfrac{5\pi}{12}+m\pi\end{matrix}\right.\)
2.
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{2}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{\pi}{4}+k2\pi\\\dfrac{\pi}{2}cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{\pi}{4}+k_12\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}+4k\\cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2}+4k_1\end{matrix}\right.\) (2)
Do \(-1\le cos\left(x-\dfrac{\pi}{4}\right)\le1\Rightarrow\left\{{}\begin{matrix}-1\le\dfrac{1}{2}+4k\le1\\-1\le-\dfrac{1}{2}+4k_1\le1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}k=0\\k_1=0\end{matrix}\right.\)
Thế vào (2):
\(\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\\cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow...\) chắc bạn tự giải tiếp được
![](https://rs.olm.vn/images/avt/0.png?1311)
\(y'=-3x^2+6x+2m-1=-3x^2+6x-3+2m+2\)
\(y'=-3\left(x-1\right)^2+2m+2\le2m+2\)
\(\Rightarrow\) Hệ số góc lớn nhất của tiếp tuyến \(\left(C_m\right)\) là \(k=2m+2\)
Để tiếp tuyến song song với \(x-2y-4=0\Rightarrow y=\frac{1}{2}x-2\)
\(\Rightarrow k=\frac{1}{2}\Rightarrow2m+2=\frac{1}{2}\Rightarrow m=-\frac{3}{4}\)
Câu 2:
\(y'=\frac{\left(2x+\frac{1}{2\sqrt{x+1}}\right)x-\left(x^2-1+\sqrt{x+1}\right)}{x^2}\)
\(=\frac{4x^2\sqrt{x+1}+x-2x^2\sqrt{x+1}+2\sqrt{x+1}-2\left(x+1\right)}{2x^2\sqrt{x+1}}\)
\(=\frac{2x^2\sqrt{x+1}+2\sqrt{x+1}-x-2}{2x^2\sqrt{x+1}}\)
Hoặc làm thế này cũng được:
\(y=x-\frac{1}{x}+\frac{\sqrt{x+1}}{x}\)
\(\Rightarrow y'=1+\frac{1}{x^2}+\frac{\frac{x}{2\sqrt{x+1}}-\sqrt{x+1}}{x^2}\)
\(=1+\frac{1}{x^2}-\frac{x+2}{2x^2\sqrt{x+1}}\)
Sau đó quy đồng sẽ có kết quả giống bên trên
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>2sin(x+pi/3)=-1
=>sin(x+pi/3)=-1/2
=>x+pi/3=-pi/6+k2pi hoặc x+pi/3=7/6pi+k2pi
=>x=-1/2pi+k2pi hoặc x=2/3pi+k2pi
b: =>2sin(x-30 độ)=-1
=>sin(x-30 độ)=-1/2
=>x-30 độ=-30 độ+k*360 độ hoặc x-30 độ=180 độ+30 độ+k*360 độ
=>x=k*360 độ hoặc x=240 độ+k*360 độ
c: =>2sin(x-pi/6)=-căn 3
=>sin(x-pi/6)=-căn 3/2
=>x-pi/6=-pi/3+k2pi hoặc x-pi/6=4/3pi+k2pi
=>x=-1/6pi+k2pi hoặc x=3/2pi+k2pi
d: =>2sin(x+10 độ)=-căn 3
=>sin(x+10 độ)=-căn 3/2
=>x+10 độ=-60 độ+k*360 độ hoặc x+10 độ=240 độ+k*360 độ
=>x=-70 độ+k*360 độ hoặc x=230 độ+k*360 độ
e: \(\Leftrightarrow2\cdot sin\left(x-15^0\right)=-\sqrt{2}\)
=>\(sin\left(x-15^0\right)=-\dfrac{\sqrt{2}}{2}\)
=>x-15 độ=-45 độ+k*360 độ hoặc x-15 độ=225 độ+k*360 độ
=>x=-30 độ+k*360 độ hoặc x=240 độ+k*360 độ
f: \(\Leftrightarrow sin\left(x-\dfrac{pi}{3}\right)=-\dfrac{1}{\sqrt{2}}\)
=>x-pi/3=-pi/4+k2pi hoặc x-pi/3=5/4pi+k2pi
=>x=pi/12+k2pi hoặc x=19/12pi+k2pi
g) \(3+\sqrt[]{5}sin\left(x+\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=-\dfrac{3}{\sqrt[]{5}}\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=sin\left[arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)\right]\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)+k2\pi\\x+\dfrac{\pi}{3}=\pi-arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{2\pi}{3}-arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)+k2\pi\end{matrix}\right.\)
h) \(1+sin\left(x-30^o\right)=0\)
\(\Leftrightarrow sin\left(x-30^o\right)=-1\)
\(\Leftrightarrow sin\left(x-30^o\right)=sin\left(-90^o\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-30^o=-90^0+k360^o\\x-30^o=180^o+90^0+k360^o\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-60^0+k360^o\\x=300^0+k360^o\end{matrix}\right.\)
\(\Leftrightarrow x=-60^0+k360^o\)
Áp dụng BĐT \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\):
\(y^2=\left(\sqrt{sinx}+\sqrt{1-sinx}\right)^2\le sinx+1-sinx=1\)
\(\Rightarrow-1\le y\le1\)
\(\Rightarrow M^4-m^4=0\)