\(A=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right):\left(1-\frac{\sqrt{xy}-\sqrt{x}}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\right)\)
a, Rút gọn
b, Cho \(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=6\) tìm Max A
Những câu hỏi liên quan
A=\(\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right):\left(1-\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\right)\)
a, Rút gọn biểu thức
b, Cho \(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=6\)Tìm Max A
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 ........................................................
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cho biểu thức: Pleft(frac{sqrt{x}+1}{sqrt{xy}+1}+frac{sqrt{xy}+sqrt{x}}{1-sqrt{xy}}+1right):left(1-frac{sqrt{xy}+sqrt{x}}{sqrt{xy}-1}-frac{sqrt{x}+1}{sqrt{xy}+1}right) Pleft(frac{sqrt{x}+1}{sqrt{xy}+1}+frac{sqrt{xy}+sqrt{x}}{1-sqrt{xy}}+1right):left(1-frac{sqrt{xy}+1}{sqrt{xy}-1}-frac{sqrt{x}+1}{sqrt{xy}+1}right).backslash với x,yge0;x,yne1 a) Rút gọn Pb) Tính P khi xsqrt[3]{4-2sqrt{6}}+sqrt[3]{4+2sqrt{6}}và yx^2+6
Đọc tiếp
cho biểu thức: \(P=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right):\left(1-\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\right)\) \(P=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right):\left(1-\frac{\sqrt{xy}+1}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\right).\backslash\ \)với \(x,y\ge0;x,y\ne1\)
a) Rút gọn P
b) Tính P khi \(x=\sqrt[3]{4-2\sqrt{6}}+\sqrt[3]{4+2\sqrt{6}}\)và \(y=x^2+6\)
a/ \(P=\frac{1}{\sqrt{xy}}\)
b/ \(x^3=8-6x\)
\(\Rightarrow P=\frac{1}{\sqrt{x\left(x^2+6\right)}}=\frac{1}{\sqrt{x^3+6x}}=\frac{1}{\sqrt{8-6x+6x}}=\frac{1}{2\sqrt{2}}\)
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Cho \(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=6\) Tìm GTLN của \(A=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right):\left(1-\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\frac{\sqrt{x}+\sqrt{1}}{1+\sqrt{xy}}\right)\)
Rút gọn\(A=\left(\frac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\frac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}\right):\left(1+\frac{x+y+2xy}{1-xy}\right)\)
Rút gọn biểu thức A:
\(A=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right):\left(1-\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\right)\)với \(x>0,\)\(y>0,\)\(xy\ne1\)
Rút gọn:
\(A=\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\left(\frac{1}{x}+\frac{1}{y}\right).\frac{1}{x+y+2\sqrt{xy}}+\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}.\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\right]\)
\(x=\sqrt{2-\sqrt{3}};y=\sqrt{2+\sqrt{3}}\)
Cho \(A=\left(2-\frac{2\sqrt{xy}+1}{\sqrt{xy}+1}+\frac{1}{1-\sqrt{xy}}+\frac{2\sqrt{x}}{1-xy}\right):\left(\frac{\sqrt{xy}-\sqrt{x}}{\sqrt{xy}+1}-\frac{\sqrt{xy+\sqrt{x}}}{\sqrt{xy}-1}\right)\)
a, Cho \(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=12\) Chứng minh \(A\le36\) b, Cho \(x^2+9y^2=18\) . Tính GTNN của A
Cho biểu thức : \(A=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right):\left(1-\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\right)\).
a) Rút gọn A .
b) Cho \(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=6\). Tính giá trị lớn nhất của A .
\(D=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-1\right)\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}-\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}+1\right)\)
Rút gọn D
\(D=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}-\sqrt{xy}+1}{\sqrt{xy}-1}\right)\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}-\frac{\sqrt{xy}+\sqrt{x}-\sqrt{xy}+1}{\sqrt{xy}-1}\right)\)
\(D=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{x}+1}{\sqrt{xy}-1}\right)\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}-\frac{\sqrt{x}+1}{\sqrt{xy}-1}\right)\)
\(D=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{xy}+1\right)^2}-\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{xy}-1\right)^2}\)
\(D=\left(\sqrt{x}+1\right)^2\left(\frac{1}{\left(\sqrt{xy}+1\right)^2}-\frac{1}{\left(\sqrt{xy}-1\right)^2}\right)\)
\(D=\left(\sqrt{x}+1\right)^2\cdot\frac{xy+1-2\sqrt{xy}-xy-1-2\sqrt{xy}}{\left(xy-1\right)^2}\)
\(D=\frac{\left(\sqrt{x}+1\right)^2\cdot\left(-4\sqrt{xy}\right)}{\left(xy-1\right)^2}\)
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