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\(=\frac{x-1}{2\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)^2}{x-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1-\sqrt{x}-1\right)\left(\sqrt{x}-1+\sqrt{x}+1\right)}{2\sqrt{x}}\)
\(=\frac{-2.2\sqrt{x}}{2}\)
\(=-2\sqrt{x}\)
Thank bạn bài vừa rồi đã k cho mk^^
\(VT=\frac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}:\left(\frac{a}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}+\frac{b}{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}-\frac{a+b}{\sqrt{ab}}\right)\)
\(=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\frac{a\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-b\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)-\left(a+b\right)\left(a-b\right)}{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\frac{a^2-a\sqrt{ab}-b^2-b\sqrt{ab}-a^2+b^2}{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{a+b}{\sqrt{a}+\sqrt{b}}.\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{-\left(a+b\right)\sqrt{ab}}=\sqrt{b}-\sqrt{a}=VP\)
Vậy đẳng thức được chứng minh
a) Với \(\left\{{}\begin{matrix}a>0\\a\notin\left\{1;4\right\}\end{matrix}\right.\) thì \(\left\{{}\begin{matrix}\sqrt{a}-1\ne0\\\sqrt{a}>0\\\sqrt{a}-2\ne0\\\sqrt{a}-1\ne0\end{matrix}\right.\)và \(\frac{\sqrt{a}+1}{\sqrt{a}-2}\ne\frac{\sqrt{a}+2}{\sqrt{a}-1}\)
nên Q được xác định(đpcm)
b) Ta có: \(Q=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\left(\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\right)\)
\(=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-\left(a-4\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+4}\)
\(=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
Để Q dương thì \(Q=\frac{\sqrt{a}-2}{3\sqrt{a}}>0\)
\(\Leftrightarrow\sqrt{a}-2\) và \(3\sqrt{a}\) cùng dấu
mà \(3\sqrt{a}>0\forall a\) thỏa mãn ĐKXĐ
nên \(\sqrt{a}-2>0\)
\(\Leftrightarrow\sqrt{a}>2\)
\(\Leftrightarrow\left|a\right|>4\)
\(\Leftrightarrow\left[{}\begin{matrix}a< -4\\a>4\end{matrix}\right.\)
Kết hợp ĐKXĐ, ta được: a>4
Vậy: Khi a>4 thì Q dương
a/ ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(J=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right):\left(1-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(=\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\frac{\left(\sqrt{x}+1-\sqrt{x}+1\right)\left(\sqrt{x}+1+\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{2}{\sqrt{x}+1}\)
\(=\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}+1}{2}\)
\(=\frac{2\sqrt{x}}{\sqrt{x}-1}\)
Vậy...
b/ ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(K=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{1+\sqrt{x}}+\frac{2}{x-1}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x-1}}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\frac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\frac{x-1}{\sqrt{x}}\)
Vậy...
c/ Tương tự
\(A=\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{\sqrt{a}+1}{a}\)
\(A=\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right).\frac{a}{\sqrt{a}+1}\)
\(A=\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right).\frac{a}{\sqrt{a}+1}\)
\(A=\frac{a-1}{\sqrt{a}}.\frac{a}{\sqrt{a}+1}\)
\(A=\left(\sqrt{a}-1\right).\sqrt{a}\)
\(A=a-\sqrt{a}\)
A=\(\left(\frac{\sqrt{a}\left(\sqrt{a}\right)-1}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right):\frac{\sqrt{a}+1}{a}\)= \(\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\):\(\frac{\sqrt{a}+1}{a}\)=
=\(\left(\frac{a-1}{\sqrt{a}}\right)\). \(\frac{a}{\sqrt{a}+1}\)= \(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}\)\(\frac{a}{\sqrt{a}+1}\)= \(\frac{\sqrt{a}-1}{\sqrt{a}}\)