Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)\left(\frac{1}{2\sqrt{x}}-\frac{\sqrt{x}}{2}\right)^2\)
\(\Leftrightarrow A=\left[\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\left[\left(\frac{1}{2\sqrt{x}}\right)^2-2.\frac{1}{2\sqrt{x}}.\frac{\sqrt{x}}{2}+\left(\frac{\sqrt{x}}{2}\right)^2\right]\)
\(\Leftrightarrow A=\left[\frac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{x-1}\right]\left(\frac{1}{4x}-\frac{1}{2}+\frac{x}{4}\right)\)
\(\Leftrightarrow A=\left(\frac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\right)\left(\frac{1}{4x}-\frac{2x}{4x}+\frac{x^2}{4x}\right)\)
\(\Leftrightarrow A=\frac{-4\sqrt{x}}{x-1}.\frac{\left(1-x\right)^2}{4x}\)
\(\Leftrightarrow A=\frac{4\sqrt{x}}{1-x}.\frac{\left(1-x\right)^2}{4x}\)
\(\Leftrightarrow A=\frac{1-x}{\sqrt{x}}\)
b) \(\frac{A}{\sqrt{x}}>1\)
\(\Leftrightarrow\frac{1-x}{\frac{\sqrt{x}}{\sqrt{x}}}>1\)
\(\Leftrightarrow1-x>1\Leftrightarrow x< 0\)
\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)
_Minh ngụy_
\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)^2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(x\sqrt{x}+x\sqrt{y}-2x\sqrt{y}-2y\sqrt{x}+y\sqrt{x}+y\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )
\(\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)\)
_Minh ngụy_
b) \(A=\sqrt{x}+1\)
Để A \(\in\)Z \(\Leftrightarrow\sqrt{x}+1\in Z\)\(\Leftrightarrow\sqrt{x}\in Z\)
\(\Leftrightarrow\sqrt{x}=a\left(a\in Z;a\ge0\right)\)\(\Leftrightarrow x=a^2\)
vậy x là bình phương 1 số tự nhiên thì A thuộc Z