\(y'=2021\cdot cos\left(x\sqrt{x}\right)^{2020}\cdot\left(cos\left(x\sqrt{x}\right)\right)'\)

\(=2021\cdot\left(-x\sqrt{x}\right)'\cdot sin\left(x\sqrt{x}\right)\cdot cos\left(x\sqrt{x}\right)^{2020}\)

\(=-2021\cdot\dfrac{\left(x^3\right)'}{2\sqrt{x^3}}\cdot sin\left(x\sqrt{x}\right)\cdot cos^{2020}x\sqrt{x}\)

\(=-2021\cdot\dfrac{3x^2}{2x\sqrt{x}}\cdot sin\left(x\sqrt{x}\right)\cdot cos^{2020}x\sqrt{x}\)

\(=-\dfrac{6063}{2}\sqrt{x}\cdot sin\left(x\sqrt{x}\right)\cdot cos^{2020}x\sqrt{x}\)

\(y=f\left(x\right)=\dfrac{2-3x}{x+2}\left(đk:x\ne-2\right)\)

\(y'=\dfrac{-8}{\left(x+2\right)^2}< 0\forall x\ne-2\)

=> Hàm số f(x) không có cực trị

\(y=f\left(x\right)=\dfrac{cosx}{2}\)

\(f\left(-x\right)=\dfrac{cos\left(-x\right)}{2}=\dfrac{cosx}{2}=f\left(x\right)\)

\(\Rightarrow\) Là hàm số chẵn.

1a.

\(y'=3x^2.f'\left(x^3\right)-2x.g'\left(x^2\right)\)

b.

\(y'=\dfrac{3f^2\left(x\right).f'\left(x\right)+3g^2\left(x\right).g'\left(x\right)}{2\sqrt{f^3\left(x\right)+g^3\left(x\right)}}\)

2.

\(f'\left(x\right)=\left(m-1\right)x^3+\left(m-2\right)x^2-2mx+3=0\)

Để ý rằng tổng hệ số của vế trái bằng 1 nên pt luôn có nghiệm \(x=1\), sử dụng lược đồ Hooc-ne ta phân tích được:

\(\Leftrightarrow\left(x-1\right)\left[\left(m-1\right)x^2+\left(2m-3\right)x-3\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(m-1\right)x^2+\left(2m-3\right)x-3=0\left(1\right)\end{matrix}\right.\)

Xét (1), với \(m=1\Rightarrow x=-3\)

- Với \(m\ne1\Rightarrow\Delta=\left(2m-3\right)^2+12\left(m-1\right)=4m^2-3\)

Nếu \(\left|m\right|< \dfrac{\sqrt{3}}{2}\Rightarrow\) (1) vô nghiệm \(\Rightarrow f'\left(x\right)=0\) có đúng 1 nghiệm

Nếu \(\left|m\right|>\dfrac{\sqrt{3}}{2}\Rightarrow\left(1\right)\) có 2 nghiệm \(\Rightarrow f'\left(x\right)=0\) có 3 nghiệm

1/ L'Hospital:

\(=\lim\limits_{x\rightarrow6}f'\left(x\right)=f'\left(6\right)=2\)

3/ \(=\lim\limits_{x\rightarrow2}\dfrac{\dfrac{3}{2\sqrt{3x+3}}}{1}=\dfrac{1}{2}\Rightarrow2a-b=0\)

4/ \(=\lim\limits_{x\rightarrow1}\dfrac{2f\left(x\right).f'\left(x\right)-f'\left(x\right)}{\dfrac{1}{2\sqrt{x}}}=\dfrac{2.6.5-5}{\dfrac{1}{2}}=110\)

2/ \(x_0=-3\Rightarrow y_0=\dfrac{-3-1}{-3+2}=\dfrac{-4}{-1}=4\)

\(y'=\dfrac{\left(x-1\right)'\left(x+2\right)-\left(x-1\right)\left(x+2\right)'}{\left(x+2\right)^2}=\dfrac{x+2-x+1}{\left(x+2\right)^2}=\dfrac{3}{\left(x+2\right)^2}\)

\(\Rightarrow y'\left(-3\right)=3\)

\(\Rightarrow pttt:y=3\left(x+3\right)+4=3x+13\)

\(x=0\Rightarrow y=13;y=0\Rightarrow x=-\dfrac{13}{3}\)

\(\Rightarrow S=\dfrac{1}{2}.\left|x\right|\left|y\right|=\dfrac{1}{2}.\dfrac{13}{3}.13=\dfrac{169}{6}\left(dvdt\right)\)

P/s: Câu 5,6 bỏ qua nhé, toi ngu hình học :b

1. \(y'=3x^2\sqrt{x}+\dfrac{x^3-5}{2\sqrt{x}}=\dfrac{7x^3-5}{2\sqrt{x}}\)

2. \(y'=3x^5+\dfrac{3}{x^2}+\dfrac{1}{\sqrt{x}}\)

3. \(y'=2-\dfrac{2}{\left(x-2\right)^2}\)

\(y'=\dfrac{-2-m}{\left(x-1\right)^2}\Rightarrow y'\left(2\right)=\dfrac{-2-m}{\left(2-1\right)^2}=-2-m\)

\(\Rightarrow-2-m=-3\Rightarrow m=1\)