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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) \(Q=\frac{\sqrt{a}+2}{3\sqrt{a}}>2\Rightarrow\sqrt{a}-6\sqrt{a}+2>0\Rightarrow-5\sqrt{a}>-2\Rightarrow0< \sqrt{a}< \frac{2}{5}\)
\(\Rightarrow0< a< \frac{4}{25}\)
\(\left(\frac{1}{1-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(=\left(\frac{1}{1-\sqrt{a}}-\frac{1}{1-\sqrt{a}}\right).\frac{a-2\sqrt{a}+1}{\sqrt{a}+1}\)
\(=\left(\frac{0}{1-\sqrt{a}}\right).\frac{a-2\sqrt{a}+1}{\sqrt{a}+1}\)
\(=0.\frac{a-2\sqrt{a}+1}{\sqrt{a}+1}\)
\(=0\)
\(A=\left(\frac{1}{1-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\) đkxđ:\(a>0;a\ne1\)
\(A=\left(\frac{1}{1-\sqrt{a}}-\frac{1}{1-\sqrt{a}}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}}\)\
\(A=0\)
\(P=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\cdot\left(\sqrt{a}-1\right)}:\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+4}\)
\(=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
ĐK: \(a>0;a\ne1\)
\(S=\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a-\sqrt{a}+1}{\sqrt{a}}\)
\(=\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{a-\sqrt{a}+1}{\sqrt{a}}\)
\(=\frac{a+\sqrt{a}+1}{\sqrt{a}}-\frac{a-\sqrt{a}+1}{\sqrt{a}}\)
\(=\frac{2\sqrt{a}}{\sqrt{a}}=2\)
a)
\(=\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2+4\sqrt{a}\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}.\frac{1}{2a\sqrt{a}}\)
\(=\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4a\sqrt{a}-4\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}.\frac{1}{2a\sqrt{a}}\)
\(=\frac{4a\sqrt{a}}{a-1}.\frac{1}{2a\sqrt{a}}=\frac{2}{a-1}\)
b) \(\frac{2}{a-1}=a\Rightarrow a^2-a-2=0\)
Ta có: 1+1+(-2)=0, nên pt có 2 nghiệm a1=-1<0 (không thỏa mãn đk)=> loại
a2=2(thỏa mãn đk)=> chọn
Vậy a=2 thì P=a
\(C=\left(\frac{a+\sqrt{a}}{\sqrt{a}+1}-\frac{\sqrt{a}-1}{a-\sqrt{a}}\right):\frac{\sqrt{a}-1}{a}\)
\(=\left(\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\frac{\sqrt{a}-1}{a}\)
\(=\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right):\frac{\sqrt{a}-1}{a}=\frac{a-1}{\sqrt{a}}:\frac{\sqrt{a}-1}{a}=\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}.\frac{a}{\sqrt{a}-1}\)
\(=\left(\sqrt{a}+1\right)\sqrt{a}\)
Với a > 0 , a khác 1 ta có :
\(C=\left[\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]\cdot\frac{a}{\sqrt{a}-1}\)
\(=\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\cdot\frac{a}{\sqrt{a}-1}=\frac{a-1}{\sqrt{a}}\cdot\frac{a}{\sqrt{a}-1}=\sqrt{a}\left(\sqrt{a}+1\right)=a+\sqrt{a}\)
*bài này làm sao ấy tìm Min không ra:v*
a) \(A=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\left(a>0;a\ne1\right)\)
\(=\left[\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a}-1}\right]:\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
\(=\frac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(=\frac{\sqrt{a}-1}{\sqrt{a}}\)
b) Để \(A=\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{a}-1}{\sqrt{a}}=\frac{1}{2}\)
\(\Leftrightarrow2\sqrt{a}-2=\sqrt{a}\)
\(\Leftrightarrow\sqrt{a}=2\Leftrightarrow a=4\left(tm\right)\)
1)))))))
\(\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{2}{\sqrt{ab}}:\frac{\left(\sqrt{b}-\sqrt{a}\right)^2}{\left(\sqrt{ab}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{2}{\sqrt{ab}}.\frac{\left(\sqrt{ab}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{2\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{2\sqrt{ab}-a-b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{-\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}=-1\)
\(\text{VT}=\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)=\left(1+\frac{\sqrt{x}.\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\left(1-\frac{\sqrt{x}.\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)
\(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=\text{VP(điều phải chứng minh)}\)
Ý tưởng : tử và mẫu có thể đặt nhân tử chung dc, ta rút gọn tử và mẫu cho nha, sau đó làm tiếp...
\(B=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{1}{\sqrt{a}}\)
\(=\left(\frac{\sqrt{a}^3-1}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{a^3+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right):\frac{1}{\sqrt{a}}\)
\(=\left(\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\)\(:\frac{1}{\sqrt{a}}\)
\(=\left(\frac{\sqrt{a}-1}{\sqrt{a}}-\frac{\sqrt{a}+1}{\sqrt{a}}\right):\frac{1}{\sqrt{a}}\)
\(=\frac{\sqrt{a}-1-\sqrt{a}-1}{\sqrt{a}}:\frac{1}{\sqrt{a}}=\frac{-2\sqrt{a}}{\sqrt{a}}=-2\)