B=\(\frac{\sqrt{x}+2}{\sqrt{x}-3}\)-\(\frac{\sqrt{x}+1}{\sqrt{x}-2}\)-\(3\frac{\sqrt{x}-1}{x-5\sqrt{x}+6}\).
rút gọn B, tìm x để B<-1. tìm x thuộc Z sao cho 2B thuộc Z
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a ) \(ĐKXĐ\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)
\(A=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(2+\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x-3}\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
b ) \(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}< 1\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-3}-1< 0\)
\(\Leftrightarrow\frac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}< 0\)
\(\Leftrightarrow\frac{4}{\sqrt{x}-3}< 0\)
\(\sqrt{x}-3< 0\)
\(\Leftrightarrow x< 9\)
Vậy với \(0\le x\le9;x\ne4\) thì ...
Chúc bạn học tốt !!!
a: \(A=\dfrac{1}{\sqrt{x}+1}:\left(\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{1}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
b: Để A<0 thì \(\sqrt{x}-2< 0\)
hay 0<x<4
Để M có nghĩa thì \(\hept{\begin{cases}\sqrt{x}-3\ne0\\2-\sqrt{x}\ne0\\x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}}\)
ta có \(M=\frac{2\sqrt{x}-9+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(M=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
b.\(M=5=\frac{\sqrt{x}+1}{\sqrt{x}-3}\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\)
ĐK \(\hept{\begin{cases}x\ge0\\x\ne4;x\ne9\end{cases}}\)
a. Ta có \(A=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
b. Để \(A< 1\Rightarrow\frac{\sqrt{x}+1}{\sqrt{x}-3}-1< 0\Rightarrow\frac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}< 0\Rightarrow\frac{4}{\sqrt{x}-3}< 0\)
\(\Rightarrow\sqrt{x}-3< 0\Rightarrow0\le x< 9\)
Kết hợp đk thì \(0\le x< 9\)và \(x\ne4\)thì \(A< 1\)
\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{x-2}-\frac{2\sqrt{x}+1}{3\sqrt{x}}\)
a) A= (\(\left(\frac{1+\sqrt{x}}{1+\sqrt{x}}-\frac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x-2}\right)}+\frac{\sqrt{x}+2}{x-2\sqrt{x}-3\sqrt{x}+6}\right)\)
A=\(\left(\frac{1+\sqrt{x}-\sqrt{x}}{1+\sqrt{x}}\right):\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}\right)\)
A= \(\left(\frac{1}{1+\sqrt{x}}\right):\left(\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{x-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
A=\(\left(\frac{1}{1+\sqrt{x}}\right):\left(\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
A=\(\left(\frac{1}{1+\sqrt{x}}\right):\left(\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
A=\(\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
\(B=\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{3\left(\sqrt{x}-1\right)}{x-5\sqrt{x}+6}\left(ĐKXĐ:x\ne4;x\ne9;x\ge0\right)\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-4-\left(x-2\sqrt{x}-3\right)-3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{2-\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{1}{3-\sqrt{x}}\)
\(B< -1\)\(\Leftrightarrow\) \(\frac{1}{3-\sqrt{x}}< -1\)\(\Rightarrow\sqrt{x}-3< 1\Leftrightarrow x< 16\)
Mặt khác : Vì \(B< -1< 0\)nên \(3-\sqrt{x}< 0\Rightarrow x>9\)
Vậy để \(B< -1\)thì \(9< x< 16\)
\(2B\in Z\Leftrightarrow B\in Z\)
\(\Leftrightarrow\frac{1}{3-\sqrt{x}}\in Z\)=> \(3-\sqrt{x}\inƯ\left(1\right)\)
\(\Rightarrow3-\sqrt{x}\in\left\{-1;1\right\}\)\(\Rightarrow x\in\left\{16\right\}\)( Loại x = 4 vì không thoả mãn điều kiện)
Xin lỗi vì để bài mình ghi lộn :))
Còn lại thì ổn rồi :))