វិញ្ញាសារទី៣ និងចម្លើយ

១.a).ស្រាយបញ្ជាក់ថា 1+\dfrac{1}{cos2x}=\dfrac{tg2x}{tgx}

b).ចូរគណនាកន្សោម P=1+(\dfrac{1}{cosa}).(1+\dfrac{1}{cos2a})(1+\dfrac{1}{cos2^2a})...(1+\dfrac{1}{cos2^na})

២.គណនាលីមីតខាងក្រោមនេះ :

a) \displaystyle\lim_{x\to\ 0^{+}}(arctgx)^{sinx}

b) \displaystyle\lim_{x\to\dfrac{\pi}{2}}(1-sinx)^{cotgx}

៣.គណនាដេរីវេទី n នៃអនុគមន៍ y=\dfrac{1-2x}{e^{2x}} ( n ជាចំនួនគត់វិជ្ជមាន​ ) ។

៤.គេឲ្យ x, y, z>0 ។ រកតំលៃធំបំផុតនៃកន្សោម A=\dfrac{xy}{x^2+xy+yz}+\dfrac{yz}{y^2+yz+xz}+\dfrac{xz}{z^2+xz+yx}

ចំលើយ

១.a) ស្រាយថា 1+\dfrac{1}{cos2x}=\dfrac{tg2x}{tgx} តាងដោយ (A)

យើងមាន ( A )\Leftrightarrow\dfrac{cos2x+1}{cos2x}=\dfrac{tg2x}{tgx}

\Leftrightarrow\dfrac{2cos^2x}{cos2x}=\dfrac{tg2x}{tgx}\Leftrightarrow 2cos^2xtgx=sin2x
\Leftrightarrow 2cosx.(cosxtgx)=sin2x\Leftrightarrow 2cosxsinx=sin2x ពិត

b)គណនាកន្សោម P:

អនុវត្តន៍សមភាព ( A ) យើងបាន:1+\dfrac{1}{cosa}=\dfrac{tga}{tg\dfrac{a}{2}}; 1+\dfrac{1}{cos2a}=\dfrac{tg2a}{tga}

1+\dfrac{1}{cos2^2a}=\dfrac{tg2^2a}{tg2a}; ...; 1+\dfrac{1}{cos2^na}=\dfrac{tg2^na}{tg2^{n-1}a}

ពីនោះទាញបាន P=\dfrac{tga}{tg\dfrac{a}{2}}.\dfrac{tg2a}{tga}.\dfrac{tg2^2a}{tg2a}....\dfrac{tg2^{n-1}a}{tg2^{n-2}a}.\dfrac{tg2^na}{tg2^{n-1}a}=\dfrac{tg2^na}{tg\dfrac{a}{2}}

ដូចនេះ P=\dfrac{tg2^na}{tg\dfrac{a}{2}}

២. គណនាលីមីតខាងក្រោម:

a) A=\displaystyle\lim_{x\to\ 0^{+}}(arctgx)^{sinx}=e^{\displaystyle\lim_{x\to\ 0^{+}}sinx.ln(arctgx)}=e^{\displaystyle\lim_{x\to\ 0{+}}\dfrac{ln(arctgx)}{\dfrac{1}{sinx}}(\dfrac{\infty}{\infty})}

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

ព្រោះពេល x\to o នោះ \left\{\begin{array}{l} -sin^2x\thicksim -x^2\\arctgx.(1+x^2)cosx\thicksim x\end{array}\right.\quad

ដូចនេះ លីមីត A=1

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

=e^{\displaystyle\lim_{x\to\dfrac{\pi}{2}}\dfrac{-3cos^2x.(-sinx)}{-cosx}}=e^{\displaystyle\lim_{x\to\dfrac{\pi}{2}}-3cosxsinx}=e^{0}=1

ដូចនេះ លីមីត B=1

៣.គណនាដេរីវេទី n នៃអនុគមន៍: y=\dfrac{1-2x}{e^{2x}}

តាង u=1-2x, v=e^{-2x}\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^{'}=2, u^{''}=0\Rightarrow u^{(n)}=0\forall k\geq 2

v^{'}=(-2)e^{-2x}, v^{''}=(-2)^{2}.e^{-2x},..., v^{(n)}=(-2)^n.e^{-2x}

y^{(n)}=(1-2x).(-2)^n.e^{-2x}+n.(-2).(-2)^{n-1}.e^{-2x}

=(-2)^n.e^{-2x}.(n+1-2x)

ដូចនេះ y^{(n)}=(-2)^n.e^{-2x}.(n+1-2x)

៤.រកតំលៃធំំបំផុតនៃកន្សោម A=\dfrac{xy}{x^2+xy+yz}+\dfrac{yz}{y^2+yz+xz}+\dfrac{xz}{z^2+xz+yx}

សរសេរឡើងវិញ A=\dfrac{1}{\dfrac{x}{y}+\dfrac{z}{x}+1}+\dfrac{1}{\dfrac{y}{z}+\dfrac{x}{y}+1}+\dfrac{1}{\dfrac{z}{x}+\dfrac{y}{z}+1}

តាង: \dfrac{x}{y}=a, \dfrac{y}{z}=b, \dfrac{z}{x}=c, យើងបាន abc=1 និង a, b, c > 0

ដូចនោះ A=\dfrac{1}{a+b+1}+\dfrac{1}{c+b+1}+\dfrac{1}{a+c+1}

ដោយ a, b>0 នោះ: a+b=(\sqrt[3]{a}+\sqrt[3]{b})(\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2})\geq \sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b})

\Rightarrow a+b+1\geq \sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b})+\sqrt[3]{abc}

\Rightarrow {a+b+1}\geq \sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c})=\dfrac{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}{\sqrt[3]{c}}

\Rightarrow \dfrac{1}{a+b+1}\leq\dfrac{\sqrt[3]{c}}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}

ដូចគ្នាដែរគេមាន \dfrac{1}{c+b+1}\leq\dfrac{\sqrt[3]{a}}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}; \dfrac{1}{a+c+1}\leq\dfrac{\sqrt[3]{b}}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}

ដូចនេះ A\leq 1, សញ្ញាសមភាពកើតមានកាលណា x=y=z, ពេលនោះយើងបាន

maxA=1