cho biểu thức \(P=\dfrac{x+1-2\sqrt{x}}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\)
a)Tìm x để biết P có nghĩa và rút gọn P
b)với giá trị nào của x thì P<1
K
Khách
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6 tháng 12 2021
a: \(P=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
22 tháng 2 2021
Sửa đề: \(P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}\)
a) Ta có: \(P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}\)
\(=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
\(=\dfrac{x-1}{x}\)
b) Sửa đề: \(2\sqrt{x+1}=5\)
Ta có: \(2\sqrt{x+1}=5\)
\(\Leftrightarrow\sqrt{x+1}=\dfrac{5}{2}\)
\(\Leftrightarrow x+1=\dfrac{25}{4}\)
hay \(x=\dfrac{21}{4}\)(thỏa ĐK)
Thay \(x=\dfrac{21}{4}\) vào biểu thức \(P=\dfrac{x-1}{x}\), ta được:
\(P=\left(\dfrac{21}{4}-1\right):\dfrac{21}{4}=\dfrac{17}{4}\cdot\dfrac{4}{21}=\dfrac{17}{21}\)
Vậy: Khi \(2\sqrt{x+1}=5\) thì \(P=\dfrac{17}{21}\)
c) Để \(P>\dfrac{1}{2}\) thì \(P-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{2\left(x-1\right)}{2x}-\dfrac{x-1}{2x}>0\)
mà \(2x>0\forall x\) thỏa mãn ĐKXĐ
nen \(2\left(x-1\right)-x+1>0\)
\(\Leftrightarrow2x-2-x+1>0\)
\(\Leftrightarrow x-1>0\)
hay x>1
Kết hợp ĐKXĐ, ta được: x>1
Vậy: Để \(P>\dfrac{1}{2}\) thì x>1
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 12 2022
ĐKXĐ: \(x>0;x\ne9\)
\(P=\left(\dfrac{x+7}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}-\dfrac{4\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\right)\left(\dfrac{\sqrt{x}+6}{\sqrt{x}}\right)\)
\(=\left(\dfrac{x+7-4\sqrt{x}-4+\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\right)\left(\dfrac{\sqrt{x}+6}{\sqrt{x}}\right)\)
\(=\left(\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\right).\left(\dfrac{\sqrt{x}+6}{\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}.\dfrac{\left(\sqrt{x}+6\right)}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+6}{\sqrt{x}+1}\)
b.
Ta có \(P=\dfrac{\sqrt{x}+1+5}{\sqrt{x}+1}=1+\dfrac{5}{\sqrt{x}+1}\)
Do \(\sqrt{x}+1>0\Rightarrow\dfrac{5}{\sqrt{x}+1}>0\Rightarrow P>1\)
\(P=\dfrac{6\left(\sqrt{x}+1\right)-5\sqrt{x}}{\sqrt{x}+1}=6-\dfrac{5\sqrt{x}}{\sqrt{x}+1}\)
Do \(\left\{{}\begin{matrix}5\sqrt{x}>0\\\sqrt{x}+1>0\end{matrix}\right.\) ;\(\forall x>0\Rightarrow\dfrac{5\sqrt{x}}{\sqrt{x}+1}>0\)
\(\Rightarrow P< 6\Rightarrow1< P< 6\)
Mà P nguyên \(\Rightarrow P=\left\{2;3;4;5\right\}\)
- Để \(P=2\Rightarrow\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=2\Rightarrow\sqrt{x}+6=2\sqrt{x}+2\Rightarrow x=16\)
- Để \(P=3\Rightarrow\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=3\Rightarrow\sqrt{x}+6=3\sqrt{x}+3\Rightarrow\sqrt{x}=\dfrac{3}{2}\Rightarrow x=\dfrac{9}{4}\)
- Để \(P=4\Rightarrow\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=4\Rightarrow\sqrt{x}+6=4\sqrt{x}+4\Rightarrow\sqrt{x}=\dfrac{2}{3}\Rightarrow x=\dfrac{4}{9}\)
- Để \(P=5\Rightarrow\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=5\Rightarrow\sqrt{x}+6=5\sqrt{x}+5\Rightarrow\sqrt{x}=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{16}\)
15 tháng 5 2022
a: \(P=\dfrac{x+\sqrt{x}}{x-\sqrt{x}}\cdot\dfrac{3}{\sqrt{x}+1}=\dfrac{3}{\sqrt{x}-1}\)
b: Để P=1 thì \(\sqrt{x}-1=3\)
hay x=16
15 tháng 5 2022
\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{3}\)
\(P=\left(\dfrac{x+\sqrt{x}}{x\left(\sqrt{x}-1\right)}\right):\dfrac{\sqrt{x}+1}{3}\)
\(P=\left(\dfrac{x\left(\sqrt{x}+1\right)}{x\left(\sqrt{x}-1\right)}\right).\dfrac{3}{\sqrt{x}+1}\)
\(P=\dfrac{3}{\sqrt{x}-1}\)
\(P=1\)
\(\Leftrightarrow1=\dfrac{3}{\sqrt{x}-1}\)
\(\Leftrightarrow\sqrt{x}-1=3\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\left(tm\right)\)
28 tháng 6 2023
a: \(P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
b: Khi x=9 thì P=9-3+1=7
c: P=3
=>x-căn x-2=0
=>(căn x-2)(căn x+1)=0
=>x=4
AT
29 tháng 5 2021
a) ĐKXĐ: \(x>0,x\ne1\)
\(P=\dfrac{x-2\sqrt{x}}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}+\dfrac{1+2x-2\sqrt{x}}{x^2-\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}}{\left(\sqrt{x}\right)^3-1}+\dfrac{\sqrt{x}+1}{\sqrt{x}\left(x+\sqrt{x}+1\right)}+\dfrac{1+2x-2\sqrt{x}}{\sqrt{x}\left(\left(\sqrt{x}\right)^3-1\right)}\)
\(=\dfrac{x-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{\sqrt{x}\left(x+\sqrt{x}+1\right)}+\dfrac{1+2x-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\left(x-2\sqrt{x}\right)\sqrt{x}+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+1+2x-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x\sqrt{x}+x-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(x+\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
b) Ta có: \(\left\{{}\begin{matrix}\sqrt{x}+2>0\\x+\sqrt{x}+1>0\end{matrix}\right.\Rightarrow P>0\)
Vì \(x>0\Rightarrow2x+\sqrt{x}>0\Rightarrow2x+2\sqrt{x}+2-\left(\sqrt{x}+2\right)>0\)
\(\Rightarrow2\left(x+\sqrt{x}+1\right)>\sqrt{x}+2\Rightarrow\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}< 2\)
mà P nguyên \(\Rightarrow\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}=1\Rightarrow\sqrt{x}+2=x+\sqrt{x}+1\)
\(\Rightarrow x-1=0\Rightarrow x=1\) mà \(x\ne1\Rightarrow\) không có x để P nguyên
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có: \(P=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\)
\(=\sqrt{x}-1+\sqrt{x}\)
\(=2\sqrt{x}-1\)
b) Để P<1 thì \(2\sqrt{x}< 2\)
\(\Leftrightarrow x< 1\)
Kết hợp ĐKXĐ, ta được: \(0\le x< 1\)
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