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a
\(ĐKXĐ:a\ne0;a\ne1;a\ne\sqrt{2}\)
\(Q=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(Q=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(Q=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(Q=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{1}\)
\(Q=\frac{\sqrt{a}-2}{\sqrt{a}}\)
b
\(Q>0\Leftrightarrow\frac{\sqrt{a}-2}{\sqrt{a}}>0\Leftrightarrow\sqrt{a}-2>0\Leftrightarrow\sqrt{a}>2\Leftrightarrow a>\sqrt{2}\)
Điều kiện: \(\hept{\begin{cases}a>0\\\sqrt{a}-1\ne0\\\sqrt{a}-2\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}a>0\\a\ne1\\a\ne4\end{cases}}\)
Ta có:
\(1P=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)\)
\(=\frac{\sqrt{a}-2}{\sqrt{a}}\)
I don't now
sorry
...................
nha
\(a,ĐKXĐ:\hept{\begin{cases}a>0\\a\ne1\end{cases}}\)
\(b,A=\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)\left(\frac{a-\sqrt{a}}{\sqrt{a}+1}-\frac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\frac{a-1}{2\sqrt{a}}.\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}+1}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}-1}\right)\)
\(=\frac{a-1}{2\sqrt{a}}.\frac{\sqrt{a}.\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{a-1}\)
\(=\frac{\sqrt{a}\left(\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2\right)}{2\sqrt{a}}\)
\(=\frac{\sqrt{a}.\left(\sqrt{a}-1-\sqrt{a}-1\right).\left(\sqrt{a}-1+\sqrt{a}+1\right)}{2\sqrt{a}}\)
\(=\frac{\sqrt{a}.\left(-2\right).2\sqrt{a}}{2\sqrt{a}}\)
\(=-2\sqrt{a}\)
\(c,\)Để A= -4 thì
\(-2\sqrt{a}=-4\Leftrightarrow\sqrt{a}=2\Leftrightarrow a=4\)
Kết bạn với mình nha ....
a) ĐKXĐ: \(a\ne1;a\ne0\))
\(A=\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{\sqrt{a+1}}{a}\)
\(=\left(\frac{\sqrt{a}.\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{\sqrt{a}.\left(\sqrt{a}+1\right)}\right):\frac{\sqrt{a+1}}{a}\)
\(=\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right):\frac{\sqrt{a+1}}{a}\)
\(=\frac{a-1}{\sqrt{a}}.\frac{a}{\sqrt{a+1}}=\frac{\sqrt{a}\left(a-1\right)}{\sqrt{a+1}}\)
a) Điều kiện xác định : \(a>0\); \(a\ne1\)
b) Ta có :
\(A=\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{\sqrt{a}+1}{a}=\left(\frac{\sqrt{a}.\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{\sqrt{a}.\left(\sqrt{a}+1\right)}\right).\frac{a}{\sqrt{a}+1}\)
\(=\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right).\frac{a}{\sqrt{a}+1}=\frac{a-1}{\sqrt{a}}.\frac{a}{\sqrt{a}+1}=\frac{a.\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}{\sqrt{a}.\left(\sqrt{a}+1\right)}\)
\(=\sqrt{a}.\left(\sqrt{a}-1\right)=a-\sqrt{a}\)
c)
Ta có : \(A=a-\sqrt{a}=\left(a-2.\frac{1}{2}.\sqrt{a}+\frac{1}{4}\right)-\frac{1}{4}=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\)
Vì \(a>0\)và \(a\ne1\)nên \(\left(\sqrt{a}-\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow\) \(A=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
Vậy \(Min_A=-\frac{1}{4}\) khi và chỉ khi \(\sqrt{a}-\frac{1}{2}=0\Rightarrow\sqrt{a}=\frac{1}{2}\Rightarrow a=\frac{1}{4}\)
a) ĐKXĐ: \(a>0;a\ne1\)
b) ta có:
\(P=\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{\sqrt{a}+1}{a}\)
\(=\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right):\frac{\sqrt{a}+1}{a}\)
\(=\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right):\frac{\sqrt{a}+1}{a}=\frac{a-1}{\sqrt{a}}.\frac{a}{\sqrt{a}+1}\)
\(=\sqrt{a}\left(\sqrt{a}-1\right)\)
c) ta có:
\(P=\sqrt{a}\left(\sqrt{a}-1\right)=a-\sqrt{a}=a-\sqrt{a}+\frac{1}{4}-\frac{1}{4}\)
\(=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
Dấu "=" xảy ra khi : \(a=\frac{1}{4}\)
Vậy min P =-1/4 khi a=1/4
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