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/ Đề không rõ ràng bạn

Từ câu b trở đi, dễ dàng nhận ra tất cả các hàm số đều liên tục trên R

b/ Xét \(f\left(x\right)=x^3+3x^2-1\)

Ta có: \(f\left(-3\right)=-1\) ; \(f\left(-2\right)=3\)

\(\Rightarrow f\left(-3\right).f\left(-2\right)< 0\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm trên \(\left(-3;-2\right)\)

\(f\left(0\right)=-1\Rightarrow f\left(-2\right).f\left(1\right)< 0\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm trên \(\left(-2;0\right)\)

\(f\left(1\right)=3\Rightarrow f\left(0\right).f\left(1\right)< 0\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm trên \(\left(0;1\right)\)

\(\Rightarrow f\left(x\right)\) luôn có 3 nghiệm phân biệt

c/\(f\left(x\right)=m\left(x-1\right)^3\left(m^2-4\right)+x^4-3\)

\(f\left(-2\right)=13\) ; \(f\left(1\right)=-2\)

\(\Rightarrow f\left(-2\right).f\left(1\right)< 0\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm trên \(\left(-2;1\right)\)

\(f\left(2\right)=13\Rightarrow f\left(1\right).f\left(2\right)< 0\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm trên \(\left(1;2\right)\)

\(\Rightarrow f\left(x\right)\) luôn có ít nhất 2 nghiệm

d/ \(f\left(x\right)=5sin3x+x-10\)

\(f\left(0\right)=-10\)

\(f\left(4\pi\right)=4\pi-10\)

\(\Rightarrow f\left(0\right).f\left(4\pi\right)=-10\left(4\pi-10\right)< 0\)

\(\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm thuộc \(\left(0;4\pi\right)\) hay \(f\left(x\right)\) luôn có nghiệm

1.

\(-1\le sinx\le1\Rightarrow-6\le y\le4\)

b.

\(y=1-\frac{1}{2}sin^22x\)

Do \(0\le sin^22x\le1\Rightarrow-\frac{1}{2}\le y\le1\)

2.

a.

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{2}+k2\pi\\x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

b. Đề chắc chắn đúng chứ bạn?

\(\Leftrightarrow\sqrt{3}\left(1+tan^2\left(x+1\right)\right)+\left(1-\sqrt{3}\right)sinx-1-\sqrt{3}=0\)

\(\Leftrightarrow\frac{\sqrt{3}}{1-sin^2\left(x+1\right)}+\left(1-\sqrt{3}\right)sin\left(x+1\right)-1-\sqrt{3}=0\)

\(\Leftrightarrow\left(\sqrt{3}-1\right)sin^3\left(x+1\right)+\left(1+\sqrt{3}\right)sin^2\left(x+1\right)+\left(1-\sqrt{3}\right)sin\left(x+1\right)-1=0\)

Pt bậc 3 này ko giải được :)

Nên chắc bạn ghi sai đề

đề giáo viên cho tui vậy mà

Máy 570VN PLUS

Shift -> phim bên phải phím CALC -> nhập y vào -> x=-1 -> = -> 9

A

a: \(-1< =cosx< =1\)

\(\Leftrightarrow-2< =2cosx< =2\)

\(\Leftrightarrow-5< =2cosx-3< =-1\)

\(f\left(x\right)_{min}=-5\) khi cos x=-1

hay \(x=\Pi+k2\Pi\)

\(f\left(x\right)_{max}=-1\) khi cos x=1

hay \(x=k2\Pi\)

b: \(-1< =sinx< =1\)

\(\Leftrightarrow-2< =2sinx< =2\)

\(\Leftrightarrow5< =2sinx+7< =9\)

\(\Leftrightarrow\sqrt{5}< =\sqrt{2sinx+7}< =3\)

\(\Leftrightarrow3\sqrt{5}< =3\sqrt{2sinx+7}< =9\)

\(f\left(x\right)_{min}=3\sqrt{5}\) khi sin x=-1

hay \(x=-\dfrac{\Pi}{2}+k2\Pi\)

\(f\left(x\right)_{max}=9\) khi sin x=1

hay \(x=\dfrac{\Pi}{2}+k2\Pi\)

Lời giải:
a)

\(\lim\limits_{x\to-1}\frac{\sqrt[3]{x}+1}{2x^2+5x+3}=\lim\limits_{x\to-1}\frac{x+1}{\left(\sqrt[3]{x^2}-\sqrt[3]{x}+1\right)\left(x+1\right)\left(2x+3\right)}\)

\(\lim\limits_{x\to-1}\frac{1}{\left(\sqrt[3]{x^2}-\sqrt[3]{x}+1\right)\left(2x+3\right)}=\frac{1}{\left(\sqrt[3]{\left(-1\right)^2}-\sqrt[3]{-1}+1\right)\left(2.-1+3\right)}=\frac{1}{3}\)

b)

\(\lim\limits_{x\to1}\frac{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}{\left(x-1\right)^2}=\lim\limits_{x\to1}\frac{\left(\sqrt[3]{x}-1\right)^2}{\left(x-1\right)^2}=\lim\limits_{x\to1}\frac{\left(x-1\right)^2}{\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)^2\left(x-1\right)^2}\)

\(=\lim\limits_{x\to1}\frac{1}{\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)^2}=\frac{1}{\left(1+1+1\right)^2}=\frac{1}{9}\)

c)

\(\lim_{x\to 1}\frac{\sqrt[4]{x}-1}{x^3+x^2-2}=\lim_{x\to 1}\frac{\sqrt[4]{x}-1}{(x-1)(x^2+2x+2)}=\lim_{x\to 1}\frac{x-1}{(\sqrt{x}+1)(\sqrt[4]{x}+1)(x-1)(x^2+2x+2)}\)

\(=\lim_{x\to 1}\frac{1}{(\sqrt{x}+1)(\sqrt[4]{x}+1)(x^2+2x+2)}=\frac{1}{(1+1)(1+1)(1+2.1+2)}=\frac{1}{20}\)

d)

\(\lim_{x\to -2}\frac{\sqrt[3]{2x+12}+x}{x^2+2x}=\lim_{x\to -2}\frac{2x+12+x^3}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}\)

\(=\lim_{x\to -2}\frac{(x+2)(x^2-2x+6)}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}=\lim_{x\to -2}\frac{x^2-2x+6}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x}\)

\(=\frac{-7}{12}\)

Lời giải:
a)

\(\lim\limits_{x\to-1}\frac{\sqrt[3]{x}+1}{2x^2+5x+3}=\lim\limits_{x\to-1}\frac{x+1}{\left(\sqrt[3]{x^2}-\sqrt[3]{x}+1\right)\left(x+1\right)\left(2x+3\right)}\)

\(\lim\limits_{x\to-1}\frac{1}{\left(\sqrt[3]{x^2}-\sqrt[3]{x}+1\right)\left(2x+3\right)}=\frac{1}{\left(\sqrt[3]{\left(-1\right)^2}-\sqrt[3]{-1}+1\right)\left(2.-1+3\right)}=\frac{1}{3}\)

b)

\(\lim\limits_{x\to1}\frac{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}{\left(x-1\right)^2}=\lim\limits_{x\to1}\frac{\left(\sqrt[3]{x}-1\right)^2}{\left(x-1\right)^2}=\lim\limits_{x\to1}\frac{\left(x-1\right)^2}{\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)^2\left(x-1\right)^2}\)

\(=\lim\limits_{x\to1}\frac{1}{\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)^2}=\frac{1}{\left(1+1+1\right)^2}=\frac{1}{9}\)

c)

\(\lim_{x\to 1}\frac{\sqrt[4]{x}-1}{x^3+x^2-2}=\lim_{x\to 1}\frac{\sqrt[4]{x}-1}{(x-1)(x^2+2x+2)}=\lim_{x\to 1}\frac{x-1}{(\sqrt{x}+1)(\sqrt[4]{x}+1)(x-1)(x^2+2x+2)}\)

\(=\lim_{x\to 1}\frac{1}{(\sqrt{x}+1)(\sqrt[4]{x}+1)(x^2+2x+2)}=\frac{1}{(1+1)(1+1)(1+2.1+2)}=\frac{1}{20}\)

d)

\(\lim_{x\to -2}\frac{\sqrt[3]{2x+12}+x}{x^2+2x}=\lim_{x\to -2}\frac{2x+12+x^3}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}\)

\(=\lim_{x\to -2}\frac{(x+2)(x^2-2x+6)}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}=\lim_{x\to -2}\frac{x^2-2x+6}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x}\)

\(=\frac{-7}{12}\)