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\(a=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}{\left(x-1\right)\left(x^2+x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x+1\right)\left(x^2+1\right)}{x^2+x-1}=\frac{4}{1}=4\)
\(b=\lim\limits_{x\rightarrow-1}\frac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\lim\limits_{x\rightarrow-1}\frac{x^4-x^3+x^2-x+1}{x^2-x+1}=\frac{5}{3}\)
\(c=\lim\limits_{x\rightarrow3}\frac{\left(x+1\right)\left(x-3\right)^2}{\left(x^2+1\right)\left(x^2-9\right)}=\lim\limits_{x\rightarrow3}\frac{\left(x+1\right)\left(x-3\right)}{\left(x^2+1\right)\left(x+3\right)}=\frac{0}{60}=0\)
\(d=\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+x}{x^2-2x+1}=\lim\limits_{x\rightarrow1}\frac{24x^5-25x^4+1}{2x-2}=\lim\limits_{x\rightarrow1}\frac{120x^4-100x^3}{2}=10\)
\(e=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}\)
\(f=\lim\limits_{x\rightarrow-2}\frac{\left(x+2\right)\left(x-2\right)\left(x^2+4\right)}{\left(x+2\right)x^2}=\lim\limits_{x\rightarrow-2}\frac{\left(x-2\right)\left(x^2+4\right)}{x^2}=-8\)
Hai câu d, e khai triển thì dài quá nên làm biếng sử dụng L'Hopital
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.
\(y'=4\left(x-5\right)^3\)
3.
\(x-2y+1=0\Leftrightarrow y=\frac{1}{2}x+\frac{1}{2}\)
\(y'=\frac{2}{\left(x+1\right)^2}\Rightarrow\frac{2}{\left(x+1\right)^2}=\frac{1}{2}\)
\(\Rightarrow\left(x+1\right)^2=4\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=1\\x=-3\Rightarrow y=3\end{matrix}\right.\)
Có 2 tiếp tuyến: \(\left[{}\begin{matrix}y=\frac{1}{2}\left(x-1\right)+1\\y=\frac{1}{2}\left(x+3\right)+3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y=\frac{1}{2}x+\frac{1}{2}\left(l\right)\\y=\frac{1}{2}x+\frac{9}{2}\end{matrix}\right.\)
4.
\(\lim\limits\frac{\sqrt{2n^2+1}-3n}{n+2}=\lim\limits\frac{\sqrt{2+\frac{1}{n^2}}-3}{1+\frac{2}{n}}=\sqrt{2}-3\)
\(\Rightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\)
5.
\(\lim\limits_{x\rightarrow a}\frac{2\left(x^2-a^2\right)+a\left(a+1\right)-\left(a+1\right)x}{\left(x-a\right)\left(x+a\right)}=\lim\limits_{x\rightarrow a}\frac{\left(x-a\right)\left(2x+2a\right)-\left(a+1\right)\left(x-a\right)}{\left(x-a\right)\left(x+a\right)}\)
\(=\lim\limits_{x\rightarrow a}\frac{\left(x-a\right)\left(2x+a-1\right)}{\left(x-a\right)\left(x+a\right)}=\lim\limits_{x\rightarrow a}\frac{2x+a-1}{x+a}=\frac{3a-1}{2a}\)
1.
\(f'\left(x\right)=-3x^2+6mx-12=3\left(-x^2+2mx-4\right)=3g\left(x\right)\)
Để \(f'\left(x\right)\le0\) \(\forall x\in R\) \(\Leftrightarrow g\left(x\right)\le0;\forall x\in R\)
\(\Leftrightarrow\Delta'=m^2-4\le0\Rightarrow-2\le m\le2\)
\(\Rightarrow m=\left\{-1;0;1;2\right\}\)
2.
\(f'\left(x\right)=\frac{m^2-20}{\left(2x+m\right)^2}\)
Để \(f'\left(x\right)< 0;\forall x\in\left(0;2\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}m^2-20< 0\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{20}< m< \sqrt{20}\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow m=\left\{1;2;3;4\right\}\)
a/ \(y'=3x^2-4x\Rightarrow y''=6x-4\)
b/ \(y'=x^3-2x+1\Rightarrow y''=3x^2-2\)
c/ \(y'=-x^4+12x^3-6x^2+8x-1\)
\(\Rightarrow y''=-4x^3+36x^2-12x+8\)
a) Dễ thấy cosx = 0 không thỏa mãn phương trình đã cho nên chiaw phương trình cho cos2x ta được phương trình tương đương 2tan2x + tanx - 3 = 0.
Đặt t = tanx thì phương trình này trở thành
2t2 + t - 3 = 0 ⇔ t ∈ {1 ; 
}.
Vậy 
b) Thay 2 = 2(sin2x + cos2x), phương trình đã cho trở thành
3sin2x - 4sinxcosx + 5cos2x = 2sin2x + 2cos2x
⇔ sin2x - 4sinxcosx + 3cos2x = 0
⇔ tan2x - 4tanx + 3 = 0
⇔ 
⇔ x = 
+ kπ ; x = arctan3 + kπ, k ∈ Z.
c) Thay sin2x = 2sinxcosx ; 
= (sin2x + cos2x) vào phương trình đã cho và rút gọn ta được phương trình tương đương
sin2x + 2sinxcosx - 
cos2x = 0 ⇔ tan2x + 4tanx - 5 = 0 ⇔ 
⇔ x = + kπ ; x = arctan(-5) + kπ, k ∈ Z.
d) 2cos2x - 3√3sin2x - 4sin2x = -4
⇔ 2cos2x - 3√3sin2x + 4 - 4sin2x = 0
⇔ 6cos2x - 6√3sinxcosx = 0 ⇔ cosx(cosx - √3sinx) = 0
⇔ 
Lời giải:
\(\lim\limits _{x\to +\infty}\sqrt{\frac{3x^4+4x^5+2}{9x^5+5x^4+4}}=\lim\limits _{x\to +\infty}\sqrt{\frac{\frac{3}{x}+4+\frac{2}{x^5}}{9+\frac{5}{x}+\frac{4}{x^5}}}=\sqrt{\frac{4}{9}}=\frac{2}{3}\)
Đáp án B.
a.
\(\lim\limits_{x\to 1+}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}=\lim_{x\to 1+}\frac{2x^4-5x^3+3x^2+1}{(x-1)^3(3x+1)}=\lim\limits _{x\to 1+}\frac{2x^4-5x^3+3x^2+1}{3x+1}.\lim\limits_{x\to 1+}\frac{1}{(x-1)^3}\)
\(=\frac{1}{4}.(+\infty)=+\infty \)
Hoàn toàn tương tự:
\(\lim\limits_{x\to 1-}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}=-\infty \)
Do đó: \(\lim\limits_{x\to 1+}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}\neq \lim\limits_{x\to 1-}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}\) nên không tồn tại \(\lim\limits_{x\to 1}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}\)
b.
\(\lim\limits_{x\to 1+}\frac{x^3-3x^2+2}{x^4-4x+3}=\lim\limits_{x\to 1+}\frac{(x-1)(x^2-2x-2)}{(x-1)^2(x^2+2x+3)}=\lim\limits_{x\to 1+}\frac{x^2-2x-2}{(x-1)(x^2+2x+3)}\)
\(=\lim\limits_{x\to 1+}\frac{x^2-2x-2}{x^2+2x+3}.\lim\limits_{x\to 1+}\frac{1}{x-1}=\frac{-1}{2}.(+\infty)=-\infty \)
Tương tự \(\lim\limits_{x\to 1-}\frac{x^3-3x^2+2}{x^4-4x+3}=+\infty \)
Do đó không tồn tại \(\lim\limits_{x\to 1}\frac{x^3-3x^2+2}{x^4-4x+3}\)
c.
\(\lim\limits_{x\to 1}\frac{x^3-2x-1}{x^5-2x-1}=\frac{1^3-2.1-1}{1^5-2.1-1}=1\)
d.
\(\lim\limits_{x\to -1}\frac{(x+2)^2-1}{x^2-1}=\lim\limits_{x\to -1}\frac{(x+2-1)(x+2+1)}{(x-1)(x+1)}=\lim\limits_{x\to -1}\frac{x+3}{x-1}=-1\)
a) cosx - √3sinx = √2 ⇔ cosx - tan
sinx = √2
⇔ coscosx - sinsinx = √2cos ⇔ cos(x + ) = 
⇔ 
b) 3sin3x - 4cos3x = 5 ⇔ 
sin3x - 
cos3x = 1.
Đặt α = arccos thì phương trình trở thành
cosαsin3x - sinαcos3x = 1 ⇔ sin(3x - α) = 1 ⇔ 3x - α = 
+ k2π
⇔ x = 
, k ∈ Z (trong đó α = arccos).
Chọn C