\(a,ĐK:x>0;x\ne4\\ E=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-2}{2}=\dfrac{\sqrt{x}-2}{2\sqrt{x}}\\ b,x=19-8\sqrt{3}=\left(4-\sqrt{3}\right)^2\\ \Leftrightarrow E=\dfrac{4-\sqrt{3}-2}{2\left(4-\sqrt{3}\right)}=\dfrac{\left(2-\sqrt{3}\right)\left(4+\sqrt{3}\right)}{26}=\dfrac{5-2\sqrt{3}}{26}\\ c,E=-1\Leftrightarrow\sqrt{x}-2=-2\sqrt{x}\\ \Leftrightarrow3\sqrt{x}=2\Leftrightarrow\sqrt{x}=\dfrac{2}{3}\Leftrightarrow x=\dfrac{4}{9}\left(tm\right)\\ d,E=\dfrac{1}{\sqrt{x}}\Leftrightarrow\dfrac{\sqrt{x}-2}{2}=1\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\left(tm\right)\)

\(e,E>0\Leftrightarrow\sqrt{x}-2>0\left(2\sqrt{x}>0\right)\Leftrightarrow x>4\\ f,E=\dfrac{\sqrt{x}-2}{2\sqrt{x}}=\dfrac{1}{2}-\dfrac{1}{\sqrt{x}}< \dfrac{1}{2}\left(-\dfrac{1}{\sqrt{x}}< 0\right)\\ g,\dfrac{1}{E}=\dfrac{2\sqrt{x}}{\sqrt{x}-2}=\dfrac{2\left(\sqrt{x}-2\right)+4}{\sqrt{x}-2}\in Z\\ \Leftrightarrow\sqrt{x}-2\inƯ\left(4\right)=\left\{-1;0;1;2;4\right\}\left(\sqrt{x}-2>-2\right)\\ \Leftrightarrow\sqrt{x}\in\left\{1;2;3;4;6\right\}\\ \Leftrightarrow x\in\left\{1;9;16;36\right\}\left(x\ne4\right)\\ h,x>4\Leftrightarrow\sqrt{x}-2>0\\ \Leftrightarrow E=\dfrac{\sqrt{x}-2}{2\sqrt{x}}>0\Leftrightarrow E\ge\sqrt{E}\)

\(\int e^2dx=e^2.x+C\)

C. f(x) = \(e^2\)x+C

a) Với x bất kì và \(h = x - {x_0}\), ta có:

\(\begin{array}{l}f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {{x_0} + h} \right) - f\left( {{x_0}} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{{x_0} + h}} - {e^{{x_0}}}}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{{x_o}}}\left( {{e^h} - 1} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} {e^{{x_0}}}.\mathop {\lim }\limits_{h \to 0} \frac{{{e^h} - 1}}{h} = {e^{{x_0}}}\end{array}\)

Vậy hàm số \(y = {e^x}\)  có đạo hàm là hàm số \(y' = {e^x}\)

b) Ta có \({a^x} = {e^{x\ln a}}\,\)nên \(\left( {{a^x}} \right)' = \left( {{e^{x\ln a}}} \right)' = \left( {x\ln a} \right)'.{e^{x\ln a}} = {e^{x\ln a}}\ln a = {a^x}\ln a\)

a/

\(x\in\left\{0;1;2;3;4;5\right\}\)

b/

\(x\in\left\{6;7;8;9\right\}\)

c/

\(x\in\left\{7;8;9;10;11;12\right\}\)

d/

\(x\in\left\{8;9;10;11;12\right\}\)

e/

\(x\in\left\{7;8;9;10;11;12\right\}\)

a)A={0; 1; 2; 3; 4; 5}

b)B={6; 7; 8; 9}

c)C={7; 8; 9; 10; 11; 12}

d)D={8; 9; 10; 11}

e)E={7; 8; 9; 10; 11; 12}

A

A

A

tính đạo hàm của y ạ.ko phải tính giới hạn đâu a

ta có

\(y=\frac{\left(e^x+e^{-x}\right)\left(e^x+e^{-x}\right)-\left(e^x-e^{-x}\right)\left(e^x-e^{-x}\right)}{\left(e^x+e^{-x}\right)^2}=\frac{\left(e^x+e^{-x}\right)^2-\left(e^x-e^{-x}\right)^2}{\left(e^x+e^x\right)^2}=\frac{\left(e^x+e^{-x}+e^x-e^{-x}\right)\left(e^x+e^{-x}-e^x+e^{-x}\right)}{\left(e^x+e^{-x}\right)^2}=2\frac{e^x-e^{-x}}{\left(e^x+e^{-x}\right)^2}=\frac{2}{e^x+e^{-x}}\)

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