1,
$\oplus$ Ta có:
$P=\dfrac{x}{\sqrt{1-x}} + \dfrac{y}{\sqrt{1-y}} = \dfrac{x^2}{x\sqrt{1-x}} + \dfrac{y^2}{y\sqrt{1-y}} \overset{C-S}{\ge} \dfrac{(x+y)^2}{x\sqrt{1-x}+y\sqrt{1-y}} \overset{Bunyakovsky}{\ge} \dfrac{1^2}{\sqrt{(x^2+y^2)[2-(x+y)]}} = \dfrac{1}{\sqrt{x^2+y^2}}$
$\oplus$ Ta có bđt: $\sqrt{a} + \sqrt{b} \ge \sqrt{a+b}$
$\Longrightarrow$ $\sqrt{a+b} \leq \sqrt{a} + \sqrt{b}$
$\oplus$ Áp dụng bđt phụ , ta có:
$P \ge \dfrac{1}{\sqrt{x^2+y^2}} \ge \dfrac{1}{\sqrt{x^2} + \sqrt{y^2}} = \dfrac{1}{x+y} = 1$
_______________________
Câu $2,$ tương tự nhé
Bài viết đã được chỉnh sửa nội dung bởi Tienanh tx: 06-06-2013 - 13:01
$\cdot$ $( - 1) = {( - 1)^5} = {( - 1)^{2.\frac{5}{2}}} = {\left[ {{{( - 1)}^2}} \right]^{\frac{5}{2}}} = {1^{\frac{5}{2}}} =\sqrt{1}= 1$
$\cdot$ $\dfrac{0}{0}=\dfrac{100-100}{100-100}=\dfrac{10.10-10.10}{10.10-10.10}=\dfrac{10^2-10^2}{10(10-10)}=\dfrac{(10-10)(10+10)}{10(10-10)}=\dfrac{20}{10}=2$
$\cdot$ $\pi\approx 2^{5^{0,4}}-0,6-\left(\frac{0,3^{9}}{7}\right)^{0,8^{0,1}}$
$\cdot$ $ - 2 = \sqrt[3]{{ - 8}} = {( - 8)^{\frac{1}{3}}} = {( - 8)^{\frac{2}{6}}} = {\left[ {{{( - 8)}^2}} \right]^{\frac{1}{6}}} = {64^{\frac{1}{6}}} = \sqrt[6]{{64}} = 2$