A=\(\frac{\sqrt{x}+2}{x}\)B=\(\frac{\sqrt{x}}{\sqrt{x}-2}\)Tìm các giá trị của x thoả mãn \(\frac{A}{B}\)x < hoặc bằng 6\(\sqrt{x}\) -13-\(\sqrt{x-9}\)
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S
15 tháng 10 2019
\(\sqrt{9x-9}+1=13\Leftrightarrow3\sqrt{x-1}=12\Leftrightarrow\sqrt{x-1}=4\Leftrightarrow x-1=16\Leftrightarrow x=17\)
\(2.\text{bạn tự tìm đk}\)
\(A=\left(\frac{2}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-\frac{2\sqrt{x}-2}{\sqrt{x}-1}\right)\)
\(A=\frac{2\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-2\right)=\sqrt{x}\left(\sqrt{x}-2\right)< 0\Leftrightarrow x-2\sqrt{x}< 0\Leftrightarrow\left(\sqrt{x}-1\right)^2< 1\Leftrightarrow-1< \sqrt{x}-1< 1\)
\(\Leftrightarrow0< x< 4\)
15 tháng 10 2019
Câu 1:
\(\sqrt{9x-9}+1=13\)\(ĐKXĐ:x\ge1\)
\(\Leftrightarrow\sqrt{9\left(x-1\right)}=12\)
\(\Leftrightarrow3\sqrt{x-1}=12\)
\(\Leftrightarrow\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=16\)
\(\Leftrightarrow x=17\)(tm ĐKXĐ)
Câu 2
ĐKXĐ: \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(A=\left(\frac{2}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{x-\sqrt{x}}\right):\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}-\frac{2\sqrt{x}-2}{\sqrt{x}-1}\right)\)
\(=\left(\frac{2}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)
\(=\left(\frac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\sqrt{x}-2\right)\)
\(=\left(\frac{2\sqrt{x}-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\frac{1}{\sqrt{x}-2}\)
\(=\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\frac{1}{\sqrt{x}-2}\)
\(=\frac{1}{x-2\sqrt{x}}\)
b Để A có giá trị âm \(\Rightarrow\frac{1}{x-2\sqrt{x}}< 0\)
vì 1>0
\(\Rightarrow x-2\sqrt{x}< 0\)
\(\Leftrightarrow0< \sqrt{x}< 2\)
\(\Leftrightarrow0< x< 4\)
kết hợp ĐKXĐ: \(\Rightarrow1< x< 4\)
NN
11 tháng 10 2020
a) \(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne4\\x\ne9\end{cases}}\)
\(P=\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}+2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right):\left(1-\frac{3\sqrt{x}-9}{x-9}\right)\)
\(=\left[\frac{-\left(\sqrt{x}-3\right)}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{\sqrt{x}+3}+\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]:\left[1-\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]\)
\(=\left[\frac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]:\left(1-\frac{3}{\sqrt{x}+3}\right)\)
\(=\left[\frac{-x+9+x-4+x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]:\left(\frac{\sqrt{x}+3-3}{\sqrt{x}+3}\right)\)
\(=\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}:\frac{\sqrt{x}}{\sqrt{x}+3}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}+3}{\sqrt{x}}=\frac{\sqrt{x}+2}{\sqrt{x}}\)
b) Ta có: \(P=\frac{\sqrt{x}+2}{\sqrt{x}}=1+\frac{2}{\sqrt{x}}\)
Vì \(x\inℤ\)\(\Rightarrow\)Để P nguyên thì \(\frac{2}{\sqrt{x}}\inℤ\)
\(\Rightarrow2⋮\sqrt{x}\)\(\Rightarrow\sqrt{x}\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Vì \(\sqrt{x}>0\)\(\Rightarrow\sqrt{x}\in\left\{1;2\right\}\)
\(\Rightarrow x\in\left\{1;4\right\}\)
So sánh với ĐKXĐ ta thấy \(x=1\)thỏa mãn
\(\Rightarrow P=\frac{\sqrt{1}+2}{\sqrt{1}}=\frac{1+2}{1}=3\)
Vậy \(x=1\)khi đó \(P=3\)
11 tháng 10 2020
\(P=\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}+2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\div\left(1-\frac{3\sqrt{x}-9}{x-9}\right)\)
a) ĐK : \(\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)
\(=\left(\frac{3-\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\div\left(1-\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\left(\frac{\left(3-\sqrt{x}\right)\left(x+\sqrt{3}\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\div\left(1-\frac{3}{\sqrt{x}+3}\right)\)
\(=\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\div\left(\frac{\sqrt{x}+3}{\sqrt{x}+3}-\frac{3}{\sqrt{x}+3}\right)\)
\(=\left(\frac{9-x+x-4-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\div\left(\frac{\sqrt{x}}{\sqrt{x}+3}\right)\)
\(=\frac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\times\frac{\sqrt{x}+3}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+2}{\sqrt{x}}\)
b) Ta có : \(\frac{\sqrt{x}+2}{\sqrt{x}}=1+\frac{2}{\sqrt{x}}\)
Để P nguyên => \(\frac{2}{\sqrt{x}}\)nguyên
=> \(2⋮\sqrt{x}\)
=> \(\sqrt{x}\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
=> \(\sqrt{x}\in\left\{1;2\right\}\)( vì x ≥ 0 )
=> \(x\in\left\{1;4\right\}\Rightarrow x=1\)( vì x ≠ 4 )
Vậy với x = 1 thì P có giá trị nguyên
MN
4 tháng 4 2020
Bài 1 :
a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)
\(A=\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{1}{\sqrt{x}-2}\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
b) Để \(A< -1\)
\(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< -1\)
\(\Leftrightarrow\sqrt{x}-2< -\sqrt{x}-1\)
\(\Leftrightarrow2\sqrt{x}< 1\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{2}\)
\(\Leftrightarrow x< \frac{1}{4}\)
Vậy để \(A< -1\Leftrightarrow x< \frac{1}{4}\)