\(f=qvB.sin\alpha=1,6.10^{-19}.5.10^6.10^{-12}.sin30^o=4.10^{-25}\left(N\right)\)

Ta có : \(f'\left(x\right)=9mx^8+6x^5\left(m^2-3m+2\right)+4x^3\left(2m^3-m^2-m\right)\)

\(=x^3\left[9mx^5+6x^2\left(m^2-3m+2\right)+4\left(2m^3-m^2-m\right)\right]\)

f'(x) = 0 \(\Leftrightarrow\left[{}\begin{matrix}x=0\\9mx^5+6x^2\left(m^2-3m+2\right)+4\left(2m^3-m^2-m\right)=0\left(1\right)\end{matrix}\right.\)

Vì f(x) có : \(f'\left(x\right)\ge0\forall x\in R\Rightarrow\) f(x) đồng biến trên R

Khi đó : (1) nhận x = 0 là no . Suy ra : \(2m^3-m^2-m=0\)

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

m = 0 \(\Rightarrow f\left(x\right)=2x^6\Rightarrow f'\left(x\right)=12x^5\)  ko \(\ge0\forall x\in R\)  (L)

Làm tương tự : với m = 1 (t/m) và m = -1/2 (L)

Vậy ... 

Câu 1 : \(a.f'\left(x\right)=6x-1;g'\left(x\right)=x'sinx+x\left(sinx\right)'=sinx+xcosx\)

b. \(y=x^4-2x^2\Rightarrow y'=4x^3-4x\) 

Với x = -2 \(\Rightarrow y'=-24;y=8\)

PTTT của đths tại điểm có h/đ = -2 : \(y=-24\left(x+2\right)+8=-24x-40\)

\(\left(...\right)=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{1+2x}-x-1\right)+x+1-\sqrt[3]{1+3x}}{x^2}\)  

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1+2x-x^2-2x-1}{\sqrt{1+2x}+x+1}+\dfrac{x^3+3x^2+3x+1-1-3x}{...}}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{-x^2}{\sqrt{1+2x}+x+1}+\dfrac{x^2\left(x+3\right)}{...}}{x^2}\)    

\(=\lim\limits_{x\rightarrow0}\dfrac{-1}{\sqrt{1+2x}+x+1}+\dfrac{x+3}{\left(x+1\right)^2+\left(\sqrt[3]{1+3x}\right)^2+\left(x+1\right)\sqrt[3]{1+3x}}\)

\(=\dfrac{-1}{2}+\dfrac{3}{1+1+1}=1-\dfrac{1}{2}=\dfrac{1}{2}\)

a . ĐKXĐ : \(x>0;x\ne1\)

Ta có : \(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\)

\(=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\)  . \(\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}\)

\(=\left[\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\right].\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)

\(=2.\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) . Vậy ,,,

b. \(A< 0\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow0< x< 1\)

c. \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\)  A \(\in Z\Leftrightarrow\dfrac{2}{\sqrt{x}-1}\in Z\)

Mà \(\sqrt{x}-1>-1\) . Do đó : \(\sqrt{x}-1\in\left\{1;2\right\}\Leftrightarrow x\in\left\{4;9\right\}\)

Vậy ... 

ĐKXĐ : \(1\ne x\ge0\)

\(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}< 0\)  \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}>0\\\sqrt{x}-1< 0\end{matrix}\right.\)  \(\Leftrightarrow0< x< 1\)

Vậy ...