Cho \(A=\frac{1}{\sqrt{xy}}\).
Tính Max A khi \(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=6\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
22 tháng 10 2016
a/ Bạn tự tìm ĐKXĐ
\(A=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{x}\left(\sqrt{y}+1\right)}{1-\sqrt{xy}}+1\right):\left(1-\frac{\sqrt{x}\left(\sqrt{y}+1\right)}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\right)\)
Xét
- \(=\frac{\left(\sqrt{x}+1\right)\left(1-\sqrt{xy}\right)+\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)+\left(1+\sqrt{xy}\right)\left(1-\sqrt{xy}\right)}{\left(1+\sqrt{xy}\right)\left(1-\sqrt{xy}\right)}\)
\(=\frac{\sqrt{x}-x\sqrt{y}+1-\sqrt{xy}+xy+\sqrt{xy}+x\sqrt{y}+\sqrt{x}+1-xy}{\left(1+\sqrt{xy}\right)\left(1-\sqrt{xy}\right)}\)
\(=\frac{2\sqrt{x}+2}{\left(1+\sqrt{xy}\right)\left(1-\sqrt{xy}\right)}\)
- \(1-\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\)
\(=\frac{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)-\left(\sqrt{xy}+\sqrt{x}\right)\left(\sqrt{xy}+1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}\)
\(=\frac{xy-1-xy-\sqrt{xy}-x\sqrt{y}-\sqrt{x}-x\sqrt{y}+\sqrt{x}-\sqrt{xy}+1}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}\)
\(=\frac{-2\sqrt{xy}-2x\sqrt{y}}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}=\frac{-2\sqrt{xy}\left(\sqrt{x}+1\right)}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}\)
\(\Rightarrow A=\frac{2\left(\sqrt{x}+1\right)}{\left(1+\sqrt{xy}\right)\left(1-\sqrt{xy}\right)}:\frac{2\sqrt{xy}\left(\sqrt{x}+1\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}=\frac{1}{\sqrt{xy}}\)
b/ Áp dụng BĐT \(\left(a+b\right)^2\ge4ab\) với \(a=\frac{1}{\sqrt{x}},b=\frac{1}{\sqrt{y}}\) được :
\(A=\frac{1}{\sqrt{x}.\sqrt{y}}\le\frac{1}{4}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)^2=\frac{1}{4}.6^2=9\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\sqrt{x}=\sqrt{y}\\\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=6\end{cases}}\Leftrightarrow x=y=\frac{1}{9}\)
Vậy ........................................................
NP
1
16 tháng 10 2016
6 = \(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\ge\frac{2}{\sqrt[4]{xy}}\)
<=> \(\frac{1}{\sqrt{xy}}\le9\)
Vậy max là 9 khi x = y = \(\frac{1}{9}\)
VH
10 tháng 10 2019
1.
a,
\(A\text{ xác định }\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne0\\x-\sqrt{x}\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\\x\ne0\end{matrix}\right.\)
\(\text{Vậy A xác định }\Leftrightarrow x>0\text{ và }x\ne1\)
\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-\sqrt{x}}\right):\frac{1}{\sqrt{x}-1}\)
\(=\left(\frac{\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\sqrt{x}-1\right)\)
\(=\frac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)=\frac{x-2}{\sqrt{x}}\)
VH
10 tháng 10 2019
b, \(x=3-2\sqrt{2}=2-2\sqrt{2}+1=\left(\sqrt{2}-1\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{2}-1\right)^2}=\left|\sqrt{2}-1\right|=\sqrt{2}-1\)
\(A=\frac{x-2}{\sqrt{x}}=\frac{3-2\sqrt{2}-2}{\sqrt{2}-1}\)
\(=\frac{1-2\sqrt{2}}{\sqrt{2}-1}=-\frac{\left(\sqrt{2}-1\right)\left(2+\sqrt{2}+1\right)}{\sqrt{2}-1}=-3-\sqrt{2}\)
10 tháng 1 2019
a/ \(P=\frac{1}{\sqrt{xy}}\)
b/ \(x^3=8-6x\)
\(\Rightarrow P=\frac{1}{\sqrt{x\left(x^2+6\right)}}=\frac{1}{\sqrt{x^3+6x}}=\frac{1}{\sqrt{8-6x+6x}}=\frac{1}{2\sqrt{2}}\)
\(6=\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\ge\frac{2}{\sqrt[4]{xy}}\)\(\Leftrightarrow\)\(\frac{1}{\sqrt{xy}}\le9\)
Dấu "=" xảy ra khi \(x=y=\frac{1}{9}\)