វិញ្ញាសារទី២ និងចម្លើយ

១.a)ស្រាយបញ្ជាក់ថា : 2cosx-1=\dfrac{2cos2x+1}{2cosx+1}

b)ទាញរកការគណនា:

S=(2cosa-1)(2cos2a-1)(cos2a-1)(2cos4a-1)...(2cos2^na-1)

២.គណមាលីមីតខាងក្រោម:

a) \displaystyle\lim_{x\to\ 0^{+}}(arcsinx)^{tgx}

b) \displaystyle\lim_{x\to\ 0}(1-cosx)^{tgx}

៣.គណនាដេរីវេទី n នៃអនុគមន៍ y=xln(1-3x) ,( n ជាចំនួនគត់វិជ្ជមាន)​ ។

៤.គេឲ្យ x, y, z ជាបណ្តាចំនួនវិជ្ជមាននិង \sqrt{xy}+\sqrt{xz}+\sqrt{yz}=1

​​ រកតំលៃតូចបំផុតនៃកន្សោម A=\dfrac{x^2}{x+y}+\dfrac{y^2}{y+z}+\dfrac{z^2}{z+x}

ចំលើយ

១a).ស្រាយថា​ 2cosx-1=\dfrac{2cos2x+1}{2cosx+1} តាងដោយ (B)

យើងមាន: (B) \Leftrightarrow (2cosx-1)(2cosx+1)=2cos2x+1\Leftrightarrow 4cos^2x-1=2(2cos^2x-1)+1

\Leftrightarrow 4cos^2x-1=4cos^2x-1 ពិត

b).អនុវត្តន៍ (B), យើងបាន:

2cosa-1=\dfrac{2cos2a+1}{2cosa+1}, 2cos2a-1=\dfrac{2cos4a+1}{2cos2a+1}

2cos4a-1=\dfrac{2cos2^3a+1}{2cos4a+1},..., 2cos2^na-1=\dfrac{2cos2^{n+1}a+1}{2cos2^na+1}

ដូចនោះ S=\dfrac{2cos2a+1}{2cosa+1}.\dfrac{2cos4a+1}{2cos2a+1}.\dfrac{2cos2^3a+1}{2cos4a+1}...\dfrac{2cos2^{n+1}a+1}{2cosa+1}

S=\dfrac{2cos2^{n+1}a+1}{2cosa+1}

២គណនាលីមីតខាងក្រោម:

a) A=\displaystyle\lim_{x\to\ 0^{+}}(arcsinx)^{tgx}=e^{\displaystyle\lim_{x\to\ 0^{+}}tgx.ln(arcsinx)}=e^{\displaystyle\lim_{x\to\ 0^{+}}\dfrac{ln(arcsinx)}{cotgx} (\dfrac{\infty}{\infty})}

= e^{\displaystyle\lim_{x\to\ 0^{+}}\dfrac{1}{arcsinx}.\dfrac{1}{\sqrt{1-x^2}}.(-sin^2x)}=e^0=1

ព្រោះពេល x\to 0 នោះ \left\{\begin{array}{l} -sin^2x\thicksim -x^2\\arcsinx.\sqrt{1-x^2}\thicksim x\end{array}\right.\quad

ដូចនេះ លីមីត A=1

b). \displaystyle\lim_{x\to\ 0}(1-cosx)^{tgx}=e^{\displaystyle\lim_{x\to\ 0}\dfrac{ln(1-cosx)}{cotgx}(\dfrac{\infty}{\infty})}=e^{\displaystyle\lim_{x\to\ 0}\dfrac{1}{1-cosx}.sinx.(-sin^2x) (\dfrac{\infty}{\infty})}

=e^{\displaystyle\lim_{x\to\ 0}\dfrac{-3sin^2x.cosx}{sinx}}=e^0=1

ដូចនេះ លីមីត​ B=1

៣.គណនាដេរីវេទី n នៃអនុគមន៍: y=xln(1-3x)

តាង u=x, v=ln(1-3x)\Rightarrow y=u.v តាមរូបមន្ត Leibniz​ យើងបាន

y^{(n)}=\displaystyle\sum_{k=0}^{n}C_{n}^{k}.u^{(k)}.v^{(n-k)}=C_{n}^{0}u^{(n)}v+C_{n}^{1}u^{(n-1)}v^{'}+...+C_{n}^{(k)}u^{(n-k)}v^{(k)}+...+C_{n}^{n}uv^{(n)}

យើងមាន u^{'}=1, u^{''}=0\Rightarrow u^{(n)}=0\forall k\geq 2

v^{'}=\dfrac{1}{1-3x}(-3)=(1-3x)^{-1}(-3), v^{''}=(-1)(1-3x)^{-2x}(-3)^2,..., v^{(n)}=(-1)(-2)...(-n+1)(1-3x)^{-n}(-3)^n=\dfrac{-3^n(n-1)!}{(1-3x)^n}

\Rightarrow y^{(n)}=x\dfrac{-3^n(n-1)}{(1-3x)^n}+n\dfrac{-3^{n-1}(n-2)!}{(1-3x)^{n-1}}

ដូចនេះ y^{(n)}=\dfrac{3^{n-1}(n-2)!}{(1-3x)^n}.(3x-n)

៤រកតំលៃធំបំផុតនៃកន្សោម A=\dfrac{x^2}{x+y}+\dfrac{y^2}{y+z}+\dfrac{z^2}{z+x}

យើងមាន: \dfrac{x^2}{x+y}=x-\dfrac{xy}{x+y}\geq x-\dfrac{xy}{2\sqrt{xy}}=x-\dfrac{\sqrt{xy}}{2} (1)

ដូចគ្នាដែរយើងបាន​: \dfrac{y^2}{y+z}\geq y-\dfrac{\sqrt{yz}}{2} (2), \dfrac{z^2}{x+z}\geq z-\dfrac{\sqrt{xz}}{2} (3)

ពី (1), (2), (3) ទាញបាន: A\geq x+y+z-\dfrac{1}{2}

ម្យ៉ាងវិញទៀតយើងឃើញថា: x+y+z\geq \sqrt{xy}+\sqrt{yz}+\sqrt{zx}\Rightarrow x+y+z\geq 1

មានន័យថា: A\geq 1-\dfrac{1}{2}=\dfrac{1}{2} ។ សមភាពកើតមានពេល x=y=z=\dfrac{1}{3}

ដូចនេះ Amin=\dfrac{1}{2}