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\(A=\left(\frac{1}{\sqrt{x}-1}+\frac{1}{x-\sqrt{x}}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left[\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\frac{\sqrt{x}-1}{\sqrt{x}}=1-\frac{1}{\sqrt{x}}< 1\)
\(M=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}\right)^2-2\sqrt{a}+1}\)
\(=\frac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(=\frac{\sqrt{a}-1}{\sqrt{a}}\)
Mà : \(\sqrt{a}-1< \sqrt{a}\Rightarrow\frac{\sqrt{a}-1}{\sqrt{a}}< 1\)
Vậy \(M< 1\)
a,Với \(a>0;a\ne1\)
\(M=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(=\left(\frac{\sqrt{a}-1+a-\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)^2}\right).\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\frac{a-1}{a+\sqrt{a}}\)
b, Ta có : \(1=\frac{a+\sqrt{a}}{a+\sqrt{a}}\)mà \(a-1=\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\)
\(a+\sqrt{a}=\sqrt{a}\left(\sqrt{a}+1\right)\)vì \(\sqrt{a}-1< \sqrt{a}\)
Vậy \(\frac{a-1}{a+\sqrt{a}}< 1\)hay \(M< 1\)
\(P=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\cdot\left(\sqrt{a}-1\right)}:\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+4}\)
\(=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
Ý tưởng : tử và mẫu có thể đặt nhân tử chung dc, ta rút gọn tử và mẫu cho nha, sau đó làm tiếp...
\(B=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{1}{\sqrt{a}}\)
\(=\left(\frac{\sqrt{a}^3-1}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{a^3+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right):\frac{1}{\sqrt{a}}\)
\(=\left(\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\)\(:\frac{1}{\sqrt{a}}\)
\(=\left(\frac{\sqrt{a}-1}{\sqrt{a}}-\frac{\sqrt{a}+1}{\sqrt{a}}\right):\frac{1}{\sqrt{a}}\)
\(=\frac{\sqrt{a}-1-\sqrt{a}-1}{\sqrt{a}}:\frac{1}{\sqrt{a}}=\frac{-2\sqrt{a}}{\sqrt{a}}=-2\)
\(\left(\frac{1}{1-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(=\left(\frac{1}{1-\sqrt{a}}-\frac{1}{1-\sqrt{a}}\right).\frac{a-2\sqrt{a}+1}{\sqrt{a}+1}\)
\(=\left(\frac{0}{1-\sqrt{a}}\right).\frac{a-2\sqrt{a}+1}{\sqrt{a}+1}\)
\(=0.\frac{a-2\sqrt{a}+1}{\sqrt{a}+1}\)
\(=0\)
\(A=\left(\frac{1}{1-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\) đkxđ:\(a>0;a\ne1\)
\(A=\left(\frac{1}{1-\sqrt{a}}-\frac{1}{1-\sqrt{a}}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}}\)\
\(A=0\)
a) ĐKXĐ : \(a>0;a\ne1\)
\(Q=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}+2}-\frac{\sqrt{a}-2}{\sqrt{a}-1}\right)\)
\(Q=\left(\frac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\sqrt{a}}\right):\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\right)\)
\(Q=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}:\frac{\left(a-1\right)-\left(a-4\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}.\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{3}\)
\(Q=\frac{\sqrt{a}+2}{3\sqrt{a}}\)
b) \(Q=\frac{\sqrt{a}+2}{3\sqrt{a}}>2\Rightarrow\sqrt{a}-6\sqrt{a}+2>0\Rightarrow-5\sqrt{a}>-2\Rightarrow0< \sqrt{a}< \frac{2}{5}\)
\(\Rightarrow0< a< \frac{4}{25}\)
\(A=\left(\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{a-1}+4\sqrt{a}\right)\left(\frac{a+1}{\sqrt{a}}\right)\)
\(A=\left(\frac{4\sqrt{a}}{a-1}+\frac{4\sqrt{a}\left(a-1\right)}{a-1}\right)\left(\frac{a+1}{\sqrt{a}}\right)\)
\(A=\frac{4a\sqrt{a}}{a-1}.\frac{a+1}{\sqrt{a}}=\frac{4a\left(a+1\right)}{a-1}\)
....... Tới đây được chưa bạn?
\(M=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a-1}}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}\right)^2-2\sqrt{a}+1}\)
\(=\frac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(=\frac{\sqrt{a}-1}{\sqrt{a}}\)
Mà \(\sqrt{a}-1< \sqrt{a}\) => \(\frac{\sqrt{a}-1}{\sqrt{a}}< 1\)
Vậy M < 1.
\(M=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
\(=\frac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(=\frac{\sqrt{a}-1}{\sqrt{a}}=\frac{\sqrt{a}}{\sqrt{a}}-\frac{1}{\sqrt{a}}=1-\frac{1}{\sqrt{a}}\)
còn so sánh với 1 nữa, Bạn làm tiếp đi