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\(A=\frac{15\sqrt{x}-11}{x-\sqrt{x}+3\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{45\sqrt{x}-11}{\left(\sqrt{x}+3\right)(\sqrt{x}-1)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{45\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{37\sqrt{x}-5x-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
Sửa đề chút nhé
Đk: x khác 25, x lớn bằng 0
\(A=\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{10\sqrt{x}}{x-25}-\frac{5}{\sqrt{x}+5}=\frac{\sqrt{x}\left(\sqrt{x}+5\right)}{x-25}-\frac{10\sqrt{x}}{x-25}-\frac{5\left(\sqrt{x}-5\right)}{x-25}\)
=\(\frac{x-10\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\sqrt{x}-5}{\sqrt{x}+5}\)
b) Em tự làm
c) với đk trên
\(\frac{\sqrt{x}-5}{\sqrt{x}+5}< \frac{1}{3}\Leftrightarrow3\sqrt{x}-15< \sqrt{x}+5\Leftrightarrow2\sqrt{x}< 20\Leftrightarrow x< 100\)
Vậy \(0\le x\le100,x\ne25\)
a) ĐKXĐ: \(\hept{\begin{cases}x-9\ne0\\\sqrt{x}\ge0\\\sqrt{x}\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne9\\x\ge0\\x\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne9\\x>0\end{cases}}}\)
\(A=\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(\Leftrightarrow A=\left(\frac{x+3}{x-9}+\frac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(\Leftrightarrow A=\left(\frac{x+3}{x-9}+\frac{\sqrt{x}-3}{x-9}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(\Leftrightarrow A=\frac{x+\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(\Leftrightarrow A=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+3}.\frac{1}{\sqrt{x}}=\frac{\sqrt{x}+1}{\sqrt{x}+3}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{x-9}\)
b) \(x=\sqrt{6+4\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)
\(\Leftrightarrow x=\sqrt{4+4\sqrt{2}+2}-\sqrt{2+2\sqrt{2}+1}\)
\(\Leftrightarrow x=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(\Leftrightarrow x=\left|2+\sqrt{2}\right|-\left|\sqrt{2}+1\right|\)
\(\Leftrightarrow x=2+\sqrt{2}-\sqrt{2}-1=1\left(TM\right)\)
Vậy với x= 1 thì giá trị của biểu thức \(A=\frac{\left(1+1\right)\left(1-3\right)}{1-9}=\frac{2.\left(-2\right)}{-8}=\frac{-4}{-8}=\frac{1}{2}\)
c)
Ta có :
\(\frac{x-9}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}+1}=1+\frac{2}{\sqrt{x}+1}\)
+) \(\frac{1}{A}\)nguyên
\(\Leftrightarrow1+\frac{2}{\sqrt{x}+1}\)nguyên
\(\Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)\)
\(\Leftrightarrow x=1\)
Vậy ..............
a, ĐKXĐ: \(x\ge0;x\ne9\)
\(A=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}-3-\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x-\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+2}{\sqrt{x}+3}\)
b, \(x=5+2\sqrt{6}=2+3+2\sqrt{3}.\sqrt{2}=\left(\sqrt{3}+\sqrt{2}\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{3}+\sqrt{2}\)
\(\Rightarrow A=\frac{\sqrt{x}+2}{\sqrt{x}+3}=\frac{\sqrt{3}+\sqrt{2}+2}{\sqrt{3}+\sqrt{2}+3}\)
c, \(A=\frac{\sqrt{x}+2}{\sqrt{x}+3}=\frac{3}{5}\Leftrightarrow5\sqrt{x}+10=3\sqrt{x}+9\)
\(\Leftrightarrow2\sqrt{x}=-1\Rightarrow\) không tồn tại giá trị \(x\) thỏa mãn
d, \(A=\frac{\sqrt{x}+2}{\sqrt{x}+3}\Leftrightarrow\sqrt{x}.A+3A=\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}\left(A-1\right)=2-3A\)
\(\Leftrightarrow\frac{2-3A}{A-1}=\sqrt{x}\ge0\Rightarrow\frac{2-3A}{A-1}\ge0\)
Do \(A=\frac{\sqrt{x}+2}{\sqrt{x}+3}< 1\Rightarrow A-1< 0\) nên \(2-3A\le0\Leftrightarrow A\ge\frac{2}{3}\)
\(\Rightarrow MinA=\frac{2}{3}\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}+3}=\frac{2}{3}\Leftrightarrow x=0\)