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1,
\(A=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{a+2}{a-2}\left(đk:a\ne0;1;2;a\ge0\right)\)
\(=\frac{\left(a\sqrt{a}-1\right)\left(a+\sqrt{a}\right)-\left(a\sqrt{a}+1\right)\left(a-\sqrt{a}\right)}{a^2-a}.\frac{a-2}{a+2}\)
\(=\frac{a^2\sqrt{a}+a^2-a-\sqrt{a}-\left(a^2\sqrt{a}-a^2+a-\sqrt{a}\right)}{a\left(a-1\right)}.\frac{a-2}{a+2}\)
\(=\frac{2a\left(a-1\right)\left(a-2\right)}{a\left(a-1\right)\left(a+2\right)}=\frac{2\left(a-2\right)}{a+2}\)
Để \(A=1\)\(=>\frac{2a-4}{a+2}=1< =>2a-4-a-2=0< =>a=6\)
2,
a, Điều kiện xác định của phương trình là \(x\ne4;x\ge0\)
b, Ta có : \(B=\frac{2\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)
\(=\frac{2\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}-\frac{\sqrt{x}-2}{x-4}\)
\(=\frac{2\sqrt{x}+2+2}{x-4}=\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{2}{\sqrt{x}-2}\)
c, Với \(x=3+2\sqrt{3}\)thì \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)
1/
a/ ĐKXĐ: \(x\ge0\) và \(x\ne\frac{1}{9}\)
b/ \(P=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\frac{3\sqrt{x}+1}{3}\)
\(=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}.\frac{1}{3}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c/ \(P=\frac{6}{5}\Rightarrow\frac{x+\sqrt{x}}{3\sqrt{x}-1}=\frac{6}{5}\Rightarrow6\left(3\sqrt{x}-1\right)=5\left(x+\sqrt{x}\right)\)
\(\Rightarrow5x-13\sqrt{x}+6=0\Rightarrow\left(5\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{3}{5}\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}}\)
Vậy x = 9/25 , x = 4
1) a) ĐKXĐ : \(0\le x\ne\frac{1}{9}\)
b) \(P=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
\(=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}+\frac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]:\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}=\frac{3x+3\sqrt{x}}{3\left(3\sqrt{x}-1\right)}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c) \(P=\frac{6}{5}\Leftrightarrow18\sqrt{x}-6=5x+5\sqrt{x}\Leftrightarrow5x-13\sqrt{x}+6=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}\)
Bài 1
a) \(P=\frac{3a+\sqrt{9a}-3}{a+\sqrt{a}-2}-\frac{\sqrt{a}+1}{\sqrt{a}+2}+\frac{\sqrt{a}-2}{1-\sqrt{a}}\) (ĐK : x\(\ge0\) ; x\(\ne\) 1)
\(=\frac{3a+\sqrt{9a}-3}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}+1}{\sqrt{a}+2}-\frac{\sqrt{a}-2}{\sqrt{a}-1}\)
\(=\frac{3a+\sqrt{9a}-3-\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{3a+\sqrt{9a}-3-a+1-a+4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\sqrt{a}+1}{\sqrt{a}-1}\)
b) \(P=\frac{\sqrt{a}+1}{\sqrt{a}-1}=\frac{\sqrt{a}-1+2}{\sqrt{a}-1}=1+\frac{2}{\sqrt{a}-1}\)
Vậy để P là số nguyên thì: \(\sqrt{a}-1\inƯ\left(2\right)\)
Mà Ư(2)={-1;1;2;-1}
=> \(\sqrt{a}-1\in\left\{1;-1;2;-2\right\}\)
Ta có bảng sau:
| \(\sqrt{a}-1\) | 1 | -1 | 2 | -2 |
| a | 4 | 0 | 9 | \(\sqrt{a}=-1\) (ktm) |
vậy a={0;4;9} thì P nguyên
Bài 2
\(P=\frac{\sqrt{a+4\sqrt{a-4}}+\sqrt{a-4\sqrt{a-4}}}{\sqrt{1-\frac{8}{a}+\frac{16}{a^2}}}\)(ĐK:a\(\ge\)8)
\(=\frac{\sqrt{\left(a-4\right)+4\sqrt{a-4}+4}+\sqrt{\left(a-4\right)-4\sqrt{a-4}+4}}{\sqrt{\left(1-\frac{4}{a}\right)^2}}\)
\(=\frac{\sqrt{\left(\sqrt{a-4}+2\right)^2}+\sqrt{\left(\sqrt{a-4}-2\right)^2}}{1-\frac{4}{a}}\)
\(=\sqrt{a-4}+2+\sqrt{a-4}-2:\frac{a-4}{a}\)
\(=2\sqrt{a-4}\cdot\frac{a}{a-4}\)
\(=\frac{2a}{\sqrt{a-4}}\)
a/ Điều kiện \(\hept{\begin{cases}a\ge0\\a\ne\frac{1}{9}\end{cases}}\) \(\Rightarrow0\le a\ne\frac{1}{9}\)
b/ \(M=\left(\frac{2\sqrt{a}}{3\sqrt{a}+1}+\frac{\sqrt{a}-2}{1-3\sqrt{a}}-\frac{5\sqrt{a}+3}{9a-1}\right):\left(a-\frac{2\sqrt{a}-6}{3\sqrt{a}-1}\right)\)
\(=\frac{2\sqrt{a}\left(1-3\sqrt{a}\right)+\left(\sqrt{a}-2\right)\left(1+3\sqrt{a}\right)+5\sqrt{a}+3}{\left(1-3\sqrt{a}\right)\left(1+3\sqrt{a}\right)}:\left(\frac{3a\sqrt{a}-2\sqrt{a}+6-a}{3\sqrt{a}-1}\right)\)
\(=\frac{2\sqrt{a}-6a+\sqrt{a}+3a-2-6\sqrt{a}+5\sqrt{a}+3}{\left(1-3\sqrt{a}\right)\left(1+3\sqrt{a}\right)}.\left(\frac{3\sqrt{a}-1}{3a\sqrt{a}-2\sqrt{a}+6-a}\right)\)
\(=\frac{3a-2\sqrt{a}-1}{1+3\sqrt{a}}.\frac{1}{3a\sqrt{a}-2\sqrt{a}+6-a}\)
\(=\frac{\left(3\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{1+3\sqrt{a}}.\frac{1}{3a\sqrt{a}-2\sqrt{a}+6-a}\)
\(=\frac{\sqrt{a}-1}{3a\sqrt{a}-2\sqrt{a}+6-a}\)
Hình như đề sai rồi bạn :(
a/ Điều kiện xác định : \(\hept{\begin{cases}a\ge0\\a\ne9\end{cases}\Leftrightarrow}0\le a\ne9\)
b/ \(M=\left(\frac{2\sqrt{a}}{3\sqrt{a}+1}+\frac{\sqrt{a}-2}{1-3\sqrt{a}}-\frac{5\sqrt{a}+3}{9a-1}\right):\left(1-\frac{2\sqrt{a}-6}{3\sqrt{a}-1}\right)\)
\(=\frac{2\sqrt{a}\left(3\sqrt{a}-1\right)+\left(2-\sqrt{a}\right)\left(3\sqrt{a}+1\right)-5\sqrt{a}-3}{\left(3\sqrt{a}+1\right)\left(3\sqrt{a}-1\right)}:\frac{\sqrt{a}+5}{3\sqrt{a}-1}\)
\(=\frac{6a-2\sqrt{a}+6\sqrt{a}+2-3a-\sqrt{a}-5\sqrt{a}-3}{\left(3\sqrt{a}+1\right)\left(3\sqrt{a}-1\right)}.\frac{3\sqrt{a}-1}{\sqrt{a}+5}\)
\(=\frac{3a-2\sqrt{a}-1}{3\sqrt{a}+1}.\frac{1}{\sqrt{a}+5}\)
\(=\frac{\left(3\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(3\sqrt{a}+1\right)\left(\sqrt{a}+5\right)}=\frac{\sqrt{a}-1}{\sqrt{a}+5}\)
c/ \(a=9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\) thay vào M được
\(\frac{\sqrt{5}-2-1}{\sqrt{5}-2+5}=\frac{\sqrt{5}-3}{\sqrt{5}+3}=\frac{-7+3\sqrt{5}}{2}\)
d/ \(M=\frac{\sqrt{a}-1}{\sqrt{a}+5}=\frac{\sqrt{a}+5-6}{\sqrt{a}+5}=1-\frac{6}{\sqrt{a}+5}\)
Với mọi \(0\le a\ne9\) thì ta luôn có \(\sqrt{a}+5\ge5\Leftrightarrow\frac{6}{\sqrt{a}+5}\le\frac{6}{5}\Leftrightarrow-\frac{6}{\sqrt{a}+5}\ge-\frac{6}{5}\Leftrightarrow1-\frac{6}{\sqrt{a}+5}\ge1-\frac{6}{5}\)
\(\Rightarrow M\ge-\frac{1}{5}\)
Đẳng thức xảy ra khi a = 0
Vậy giá trị nhỏ nhất của M bằng \(-\frac{1}{5}\) khi a = 0
\(đkxđ\Leftrightarrow x\ge0;x\ne1;x\ne4\)
\(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right).\)
\(=\left(\frac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\)\(\left(\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\left(\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3\sqrt{a}\left(\sqrt{a}-1\right)}=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
\(A< \frac{1}{6}\Rightarrow\frac{\sqrt{a}-2}{3\sqrt{a}}>\frac{1}{6}\Rightarrow\frac{\sqrt{a}-2}{3\sqrt{a}}-\frac{1}{6}>0\)
\(\Rightarrow\frac{2\left(\sqrt{a}-2\right)}{6\sqrt{a}}-\frac{\sqrt{a}}{6\sqrt{a}}>0\Rightarrow\frac{\sqrt{a}-4}{6\sqrt{a}}>0\)
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
Kiểm tra đề lại nhé b. Sao rút gọn rồi mà còn phức tạp thế
a/ Để P có nghĩa thì
\(\hept{\begin{cases}a>0\\\sqrt{a}-1\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}a>0\\a\ne1\end{cases}}\)
b/ \(P=\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}:\frac{\sqrt{a}+2}{\sqrt{a}-1}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\)
\(=\frac{1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}+2\right)}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\)
\(=\frac{2\sqrt{a}+a-a+2\sqrt{a}-1-a\sqrt{a}-4a-4\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
\(=-\frac{1+4a+a\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)