វិញ្ញាសារទី១ និងចម្លើយ

១.ចូរប្រៀបធៀបចំនួន A និង B: A=\dfrac{0,246246246}{(0,123123123)^2+1} ;B=\dfrac{0,246246248}{(0,123123124)^2+1}

២.គណនាលីមីត \displaystyle\lim_{n\to\ 0}\dfrac{\sqrt[3]{1+3x}.\sqrt[4]{1+4x}-1}{\sqrt{1-x}-1}

៣.រកតំលៃធំបំផុតនិងតូចបំផុតនៃអនុគមន៍ f(x)=2\pi+3x^2-6arccotgx^2 ចំពោះ -\sqrt[4]{3}\leq x\leq 1

៤.គណនាអាំងតេក្រាលខាងក្រោមនេះ :

a) A=\int\dfrac{dx}{x^2-a^2}

b) B=\int\dfrac{dx}{a^2+x^2}

c) C=\int\dfrac{dx}{\sqrt{a^2-x^2}}

ចំលើយ

១ប្រៀបធៀបចំនួន A និង B :

តាង​ a=0,123123123; b=0,123123124 យើងបាន :

A=\dfrac{2.0,123123123}{(0,123123123)^2+1}=\dfrac{2a}{a^2+1}

B=\dfrac{2.0,123123124}{(0,123123124)^2+1}=\dfrac{2b}{b^2+1}, (0<a<b<1)

A-B=\dfrac{2a}{a^2+1}-\dfrac{2b}{b^2+1}=2\dfrac{a(b^2+1)-b(a^2+1)}{(a^2+1)(b^2+1)}

=2\dfrac{ab^2+a-a^2b-b}{(a^2+1)(b^2+1)}=2\dfrac{(a-b)-ab(a-b)}{(a^2+1)(b^2+1)}

=2\dfrac{(a-b)(1-ab)}{(a^2+1)(b^2+1)}<0 ព្រោះ 0<a<b<1\Rightarrow a-b0

ដោយ A-B<0\Rightarrow A<B

ដូចនេះ A<B

២គណនាលីមីត:

A=\displaystyle\lim_{x\to\ 0}\dfrac{\sqrt[3]{1+3x}.\sqrt[4]{1+4x}-1}{\sqrt{1-x}-1}=\displaystyle\lim_{x\to\ 0}\dfrac{\sqrt[3]{1+3x}(\sqrt[4]{1+4x}-1)}{\sqrt{1-x}-1}+\displaystyle\lim_{x\to\ 0}\dfrac{\sqrt[3]{1+3x}-1}{\sqrt{1-x}-1}

ពេល x\to 0, \sqrt[n]{1+ax}-1\thicksim \dfrac{ax}{n}\Rightarrow

\Rightarrow A=\displaystyle\lim_{x\to\ 0}\dfrac{\sqrt[3]{1+3x}.x}{\dfrac{-x}{2}}+\displaystyle\lim_{x\to\ 0}\dfrac{x}{\dfrac{-x}{2}}=-2-2=-4

ដូចនេះ លីមីត A=-4

៣.រកតំលៃធំបំផុតនិងតូចបំផុតនៃអនុគមន៍: f(x)=2\pi+3x^2-6arccotgx^2 \forall x\in[-\sqrt[4]{3},1]

យើងមាន f^{'}(x)=6x-6.\dfrac{2x}{1+x^4}=6x.\dfrac{x^4-1}{x^4+1}=0\Leftrightarrow x=0, x=1, x=-1

ពីនោះយើងបាន f^{'}(0)=2\pi, f^{'}(1)=f^{'}(-1)=3+\dfrac{\pi}{2}, f^{'}(-\sqrt[4]{3})=3\sqrt{3}

ប្រៀបធៀបបណ្តាលទ្ធផលខាងលើដោយសង់តារាងអថេរភាពសិក្សារកចំនុចអតិនិងអប្បរបស់អនុគមន៍ខាងលើ(សូមអ្នកអានគូសដោយខ្លួនឯង)

យើងបាន maxf(x)=2\pi, minf(x)=3+\dfrac{\pi}{2}

៤គណនាអាំងតេក្រាលខាងក្រោម:

a) A=\int\dfrac{dx}{x^2-a^2}=\dfrac{1}{2a}.\int (\dfrac{x+a}{x^2-a^2}-\dfrac{x-a}{x^2-a^2})dx

=\dfrac{1}{2a}\int (\dfrac{1}{x-a}-\dfrac{1}{x+a})dx=\dfrac{1}{2a}\int (\dfrac{d(x-a)}{x-a}-\dfrac{d(x+a)}{x+a})

= \dfrac{1}{2a}(ln|x-a|-ln|x+a|)+C=\dfrac{1}{2a}ln|\dfrac{x-a}{x+a}|+C

b) B=\int\dfrac{dx}{a^2+x^2}

តាង x=atgt\Rightarrow t=arctg\dfrac{x}{a}\Rightarrow dx=\dfrac{a}{cos^2t}.dt

\Rightarrow \dfrac{1}{x^2+a^2}=\dfrac{1}{a^2tg^2t+a^2}=\dfrac{1}{a^2(tg^2t+1)}=\dfrac{cos^2t}{a^2}

\Rightarrow B=\int\dfrac{cos^2t}{a^2}.\dfrac{a}{cos^2t}dt=\int\dfrac{1}{a}dt=\dfrac{1}{a}t+C=\dfrac{1}{a}arctg\dfrac{x}{a}+C

c) C=\int\dfrac{dx}{\sqrt{a^2-x^2}}