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a) \(ĐK:x\ge0,x\ne9\)
Với\(x\ge0,x\ne9\)thì \(B=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]:\left[\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right]\)\(=\left[\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]:\left[\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right]\)\(=\left[\frac{2x-6\sqrt{x}}{x-9}+\frac{x+3\sqrt{x}}{x-9}-\frac{3\sqrt{x}+9}{x-9}\right]:\left[\frac{\sqrt{x}+1}{\sqrt{x}-3}\right]\)\(=\left[\frac{3x-6\sqrt{x}-9}{x-9}\right].\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{\left(\sqrt{x}+1\right)\left(3\sqrt{x}-9\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}=\frac{3\sqrt{x}-9}{\sqrt{x}+3}\)
b) \(B< -1\Leftrightarrow\frac{3\sqrt{x}-9}{\sqrt{x}+3}< -1\Leftrightarrow\frac{3\sqrt{x}-9}{\sqrt{x}+3}+1< 0\Leftrightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\)
Mà \(\sqrt{x}+3>0\)nên \(4\sqrt{x}-6< 0\Leftrightarrow\sqrt{x}< \frac{3}{2}\Leftrightarrow x< \frac{9}{4}\)
Vậy với \(0\le x< \frac{9}{4}\)thì B < -1
c) \(B=\frac{4\sqrt{x}-6}{\sqrt{x}+3}=\frac{4\left(\sqrt{x}+3\right)-18}{\sqrt{x}+3}=4-\frac{18}{\sqrt{x}+3}\)
Ta có: \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+3\ge3\Leftrightarrow\frac{18}{\sqrt{x}+3}\le6\Leftrightarrow-\frac{18}{\sqrt{x}+3}\ge-6\Leftrightarrow4-\frac{18}{\sqrt{x}+3}\ge-2\)
Vậy \(MinB=-2\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)
Nhìn nhầm câu c)
\(B=\frac{3\sqrt{x}-9}{\sqrt{x}+3}\)làm tương tự
a) đk: \(x\ge0;x\ne9\)
Ta có:
\(B=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]\div\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(B=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+3\right)\sqrt{x}-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(B=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(B=\frac{3x-6\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(B=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(B=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}=\frac{3\sqrt{x}-9}{\sqrt{x}+3}\)
b) \(B< -1\Leftrightarrow\frac{3\sqrt{x}-9}{\sqrt{x}+3}+1< 0\)
\(\Leftrightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\) , mà \(\sqrt{x}+3\ge3>0\left(\forall x\right)\)
=> \(4\sqrt{x}-6< 0\)
\(\Leftrightarrow4\sqrt{x}< 6\)
\(\Rightarrow\sqrt{x}< \frac{3}{2}\)
\(\Rightarrow x< \frac{9}{4}\)
Vậy \(0\le x< \frac{9}{4}\)
c) Ta có: \(B=\frac{3\sqrt{x}-9}{\sqrt{x}+3}=\frac{3\left(\sqrt{x}+3\right)-18}{\sqrt{x}+3}=3-\frac{18}{\sqrt{x}+3}\)
Vì \(\sqrt{x}+3\ge3\Rightarrow\frac{18}{\sqrt{x}+3}\le6\)
\(\Leftrightarrow3-\frac{18}{\sqrt{x}+3}\ge-3\)
\(\Rightarrow A\ge-3\)
Dấu "=" xảy ra khi: \(\sqrt{x}+3=3\Rightarrow x=0\)
Vậy \(Min_A=-3\Leftrightarrow x=0\)
từ dòng cuối là sai rồi bạn à
Bạn bỏ dòng cuối đi còn lại đúng rồi
Ở tử đặt nhân tử chung căn x chung rồi lại đặt căn x +1 chung
Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra
rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
a) ĐK: x > 1
\(P=\left(\frac{\sqrt{x-1}}{3+\sqrt{x-1}}+\frac{x+8}{9-\left(x-1\right)}\right):\left(\frac{3\sqrt{x-1}+1}{\left(x-1\right)-3\sqrt{x-1}}-\frac{1}{\sqrt{x-1}}\right)\)
\(P=\frac{\sqrt{x-1}\left(3-\sqrt{x-1}\right)+x+8}{9-\left(x-1\right)}:\frac{3\sqrt{x-1}+1-\left(\sqrt{x-1}-3\right)}{\sqrt{x-1}\left(\sqrt{x-1}-3\right)}\)
\(P=\frac{3\sqrt{x-1}-x+1+x+8}{10-x}:\frac{2\sqrt{x-1}+4}{\sqrt{x-1}\left(\sqrt{x-1}-3\right)}\)
\(P=\frac{3\left(\sqrt{x-1}+3\right)}{10-x}.\frac{\sqrt{x-1}\left(\sqrt{x-1}-3\right)}{2\sqrt{x-1}+4}\)
\(P=\frac{-3\sqrt{x-1}}{2\sqrt{x-1}+4}\)
b) \(x=\sqrt[4]{\frac{17+12\sqrt{2}}{1}}-\sqrt[4]{\frac{17-12\sqrt{2}}{1}}=1+\sqrt{2}-\left(\sqrt{2}-1\right)=2\)
Vậy \(P=\frac{-3\sqrt{2-1}}{2\sqrt{2-1}+4}=-\frac{1}{2}\)
cô Hoàng Thị Thu Huyền làm rõ cho em ý b đc ko ạ chỗ biến đổi x
ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)
\(=\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right)\div\frac{1}{2\left(\sqrt{x}-2\right)}\)
\(=\left(\frac{x-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\frac{x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right)\times\frac{2\left(\sqrt{x}-2\right)}{1}\)
\(=\left(\frac{x-4-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right)\times\frac{2\left(\sqrt{x}-2\right)}{1}\)
\(=\frac{5}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\times\frac{2\left(\sqrt{x}-2\right)}{1}\)
\(=\frac{5\times2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{10}{\sqrt{x}-3}\)
\(\left(\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}+3}{\sqrt{x}-2}\right):\frac{1}{2\sqrt{x}-4}\)
\(=\left(\frac{x-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\frac{x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right):\frac{1}{2\sqrt{x}-4}\)
\(=\frac{5}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}.\frac{2\left(\sqrt{x}-2\right)}{1}\)
\(=\frac{10\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{10}{\sqrt{x}-3}\)