\(đkxđ\Leftrightarrow\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)

\(A=\)\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\)\(\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)

\(=\left(\frac{\sqrt{a}.\sqrt{a}}{2\sqrt{a}}-\frac{1}{2\sqrt{a}}\right)^2\)\(\left(\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

\(=\left(\frac{a-1}{2\sqrt{a}}\right)^2\left(\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)

\(=\frac{\left(a-1\right)^2}{\left(2\sqrt{a}\right)^2}\left(\frac{a-2\sqrt{a}+1-a-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

\(=\frac{\left(a-1\right)^2.-4\sqrt{a}}{4a\left(a-1\right)}=\frac{a-1}{\sqrt{a}}\)

\(b,A< 0\Rightarrow\frac{a-1}{\sqrt{a}}< 0\)

Mà \(\sqrt{a}\ge0\Rightarrow a-1\le0\Rightarrow a\le1\)

\(A=2\Rightarrow\frac{a-1}{\sqrt{a}}=2\)

\(\Rightarrow a-1=2\sqrt{a}\Rightarrow a-2\sqrt{a}-1=0\)

\(\Rightarrow a-2\sqrt{a}+1-2=0\)

\(\Rightarrow\left(\sqrt{a}-1\right)^2-\sqrt{2}^2=0\)

\(\Rightarrow\left(\sqrt{a}-1-\sqrt{2}\right)\left(\sqrt{a}-1+\sqrt{2}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{a}=1+\sqrt{2}\\\sqrt{a}=1-\sqrt{2}\end{cases}\Rightarrow\orbr{\begin{cases}a=\left(1+\sqrt{2}\right)^2=3+2\sqrt{2}\\a=\left(1-\sqrt{2}\right)^2=3-2\sqrt{2}\end{cases}}}\)

\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)

\(=\left(\frac{a-1}{2\sqrt{a}}\right)^2\left(\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

\(=\frac{\left(a-1\right)^2}{4a}.\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)

\(=\frac{\left(a-1\right)^2}{4a}.\frac{\left(\sqrt{a}-1+\sqrt{a}+1\right)\left(\sqrt{a}-1-\sqrt{a}-1\right)}{a-1}\)

\(=\frac{a-1}{4a}.\frac{2\sqrt{a}.\left(-2\right)}{1}\)

\(=\frac{a-1}{4a}.\frac{-4\sqrt{a}.}{1}\)

\(=\frac{1-a}{\sqrt{a}}\)

\(đkxđ\Leftrightarrow x\ge0;x\ne1;x\ne4\)

\(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right).\)

\(=\left(\frac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\)\(\left(\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)

\(=\left(\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)

\(=\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3\sqrt{a}\left(\sqrt{a}-1\right)}=\frac{\sqrt{a}-2}{3\sqrt{a}}\)

\(A< \frac{1}{6}\Rightarrow\frac{\sqrt{a}-2}{3\sqrt{a}}>\frac{1}{6}\Rightarrow\frac{\sqrt{a}-2}{3\sqrt{a}}-\frac{1}{6}>0\)

\(\Rightarrow\frac{2\left(\sqrt{a}-2\right)}{6\sqrt{a}}-\frac{\sqrt{a}}{6\sqrt{a}}>0\Rightarrow\frac{\sqrt{a}-4}{6\sqrt{a}}>0\)

Vì \(6\sqrt{a}>0\Rightarrow\sqrt{a}-4>0\Rightarrow\sqrt{a}>4\Rightarrow a>16\)

Vậy \(P>\frac{1}{6}\Leftrightarrow a>16\)

Tự làm đi easy quá mà :)))) không biết quy đồng mà rút gọn hay sao

M ngon m làm đi nói nhiều 

a, Để A xác định =>  \(\hept{\begin{cases}\sqrt{a}+1\ne0\\\sqrt{a}-1\ne0\\\sqrt{a}>0\end{cases}}\)

=>\(\hept{\begin{cases}a\ne1\\a>0\end{cases}}\)

\(A=\left(\frac{a}{2\sqrt{a}}-\frac{1}{2\sqrt{a}}\right)^2.\left(\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{a-1}\right)\)

=\(\frac{a-1}{2\sqrt{a}}.\frac{\left(-4\sqrt{a}\right)}{a-1}\)

=  -2

b, Vì giá trị của A không phụ thuộc vào giá trị của a=>với mọi giá trị của biến a thì A=-2 luôn bé hơn 0

\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)

\(=\left(\frac{a-1}{2\sqrt{a}}\right)^2\left(\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

\(=\frac{\left(a-1\right)^2}{4a}\left(\frac{\left(\sqrt{a}-1-\sqrt{a}-1\right)\left(\sqrt{a}-1+\sqrt{a}+1\right)}{a-1}\right)\)

\(=\frac{\left(a-1\right)\left(-2\right)2\sqrt{a}}{4a}=-\frac{\left(a-1\right)}{\sqrt{a}}\)

h di roi t se trl

Pain Thiên Đạo ý là sao

\(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)

\(=\frac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\sqrt{a}}:\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}:\frac{a-1-\left(a-4\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}:\frac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}.\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)

\(=\frac{\sqrt{a}-2}{3\sqrt{a}}\)

\(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right).\)

\(A=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(A=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)\)

\(A=\frac{\sqrt{a}-2}{\sqrt{a}}\)

a,

ĐK :a>0    ;    a  khác 1 , khác 4

\(Q=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right)\div\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\\ \)

\(Q=\left(\frac{\sqrt{a}-\sqrt{a+1}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)

\(Q=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(Q=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\times\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)

\(Q=\frac{\sqrt{a}-2}{3\sqrt{a}}\)

b,

để Q đạt Giá Trị dương

\(\Rightarrow Q>0\Leftrightarrow\frac{\sqrt{a}-2}{3\sqrt{a}}>0\)

có \(a>0\Leftrightarrow\sqrt{a}>0\Leftrightarrow3\sqrt{a}>0\)

Suy Ra : để Q dương thì \(\sqrt{a}-2>0\)

\(\Leftrightarrow a>4\)   Thỏa mãn ĐK :  a > 0   ;a  khác 1 , khác 4

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:

\(1P=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)

\(=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)

\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)\)

\(=\frac{\sqrt{a}-2}{\sqrt{a}}\)

không hiểu nhan