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Bởi vì đúng đề như vầy thì min A bằng 0
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{4\sqrt{x}-3}{2\sqrt{x}-x}\right):\)\(\left(\frac{\sqrt{x}+2}{\sqrt{x}}-\frac{\sqrt{x}-4}{\sqrt{x}-2}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)\(:\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{x-4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}:\frac{x-4-x+4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}-3}{4}\)
\(b,\)Để \(P>0\Rightarrow\frac{\sqrt{x}-3}{4}>0\)
Mà \(4>0\Rightarrow\sqrt{x}-3>0\Rightarrow\sqrt{x}>3\Rightarrow x>9\)
\(c,\sqrt{P}_{min}=0\Rightarrow\frac{\sqrt{x}-3}{4}=0\)
\(\Leftrightarrow\sqrt{x}-3=0\Rightarrow\sqrt{x}=3\Rightarrow x=9\)
a) A=\(\left(2\sqrt{32}-2\sqrt{18}-\sqrt{50}\right):\sqrt{2}=\left(2\sqrt{16.2}-2\sqrt{9.2}-\sqrt{25.2}\right):\sqrt{2}=\left(8\sqrt{2}-6\sqrt{2}-5\sqrt{2}\right):\sqrt{2}=-3\sqrt{2}:\sqrt{2}=-3\)
B=\(\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{5\sqrt{x}+2}{4-x}=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}-\frac{5\sqrt{x}+2}{x-4}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{5\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{2x-4\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{5\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
b) Để A>B thì \(\frac{3\sqrt{x}}{\sqrt{x}+2}< -3\Leftrightarrow-\sqrt{x}>\sqrt{x}+2\Leftrightarrow2\sqrt{x}< -2\Leftrightarrow\sqrt{x}< -1\left(ktm\right)\)Vậy không có giá trị của x để giá trị biểu thức A lớn hơn giá trị của biểu thức B
\(a,A=\sqrt{27}+\frac{2}{\sqrt{3}-2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
\(=3\sqrt{3}+\frac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\left(\sqrt{3}-1\right)\)
\(=3\sqrt{3}+\frac{2\sqrt{3}+4}{3-4}-\sqrt{3}+1\)
\(=3\sqrt{3}-2\sqrt{3}-4-\sqrt{3}+1\)
\(=-3\)
\(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
b, Ta có \(B< A\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}< -3\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}+3< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-1+3\sqrt{x}}{\sqrt{x}}< 0\)
\(\Leftrightarrow\frac{4\sqrt{x}-1}{\sqrt{x}}< 0\)
\(\Leftrightarrow4\sqrt{x}-1< 0\left(Do\sqrt{x}>0\right)\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{4}\)
\(\Leftrightarrow0< x< \frac{1}{2}\)(Kết hợp ĐKXĐ)
Vậy ...
ĐKXĐ: ...
Đặt \(\left(x+\frac{2}{x}\right)^2=a\ge8\Rightarrow x^2+\frac{4}{x^2}=a-4\)
\(\Rightarrow A=\sqrt{\left(a-4\right)^2-8a+50}\)
\(A=\sqrt{a^2-16a+66}\)
\(A=\sqrt{\left(a-8\right)^2+2}\ge\sqrt{2}\)
\(A_{min}=\sqrt{2}\) (khi \(a=8\) hay \(x=\pm\sqrt{2}\))