ដោះស្រាយសមីការខាងក្រោមលើសំនុំចំនួនពិតៈ
\[\sqrt{1+\sqrt{1-x^2}}[\sqrt{(1+x)^3}-\sqrt{(1-x)^3}]=2+\sqrt{1-x^2}\]
ដំណោះស្រាយ
លក្ខខណ្ឌៈ $-1\le x\le 1$
តាង $x=\cos t, 0\le t\le \pi$
យើងបាន សមីការៈ $\sqrt{1+\sin t}[\sqrt{(1+\cos t)^3}-\sqrt{(1-cos t)^3}]=2+\sin t$
$\Leftrightarrow \sqrt{(cos\frac{t}{2}+\sin\frac{t}{2})^2}[\cos^3\frac{t}{2}-\sin^3\frac{t}{2}]2\sqrt{2}=2+\sin t$
$\Leftrightarrow (\cos^2\frac{t}{2}-\sin^2\frac{t}{2})(1+\cos\frac{t}{2}\sin\frac{t}{2})2\sqrt{2}=2+\sin t$
$\Leftrightarrow \cos t(2+\sin t)\sqrt{2}=2+\sin t \Leftrightarrow (2+\sin t)(\sqrt{2}\cos t-1)=0$
$\Leftrightarrow \sqrt{2}\cos t-1=0\Leftrightarrow \cos t=\frac{1}{\sqrt{2}}\Leftrightarrow x=\frac{1}{\sqrt{2}}$
\[\sqrt{1+\sqrt{1-x^2}}[\sqrt{(1+x)^3}-\sqrt{(1-x)^3}]=2+\sqrt{1-x^2}\]
ដំណោះស្រាយ
លក្ខខណ្ឌៈ $-1\le x\le 1$
តាង $x=\cos t, 0\le t\le \pi$
យើងបាន សមីការៈ $\sqrt{1+\sin t}[\sqrt{(1+\cos t)^3}-\sqrt{(1-cos t)^3}]=2+\sin t$
$\Leftrightarrow \sqrt{(cos\frac{t}{2}+\sin\frac{t}{2})^2}[\cos^3\frac{t}{2}-\sin^3\frac{t}{2}]2\sqrt{2}=2+\sin t$
$\Leftrightarrow (\cos^2\frac{t}{2}-\sin^2\frac{t}{2})(1+\cos\frac{t}{2}\sin\frac{t}{2})2\sqrt{2}=2+\sin t$
$\Leftrightarrow \cos t(2+\sin t)\sqrt{2}=2+\sin t \Leftrightarrow (2+\sin t)(\sqrt{2}\cos t-1)=0$
$\Leftrightarrow \sqrt{2}\cos t-1=0\Leftrightarrow \cos t=\frac{1}{\sqrt{2}}\Leftrightarrow x=\frac{1}{\sqrt{2}}$
No comments :
Post a Comment