Cho P=\((\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}})^2\).\((\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1})\)
a)Tìm ĐKXĐ.Rút gọn P
b)Tìm x để P<0
Cho P = (\(\dfrac{1}{\sqrt{x}-1 }\) - \(\dfrac{1}{\sqrt{x}}\))(\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\) - \(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\))
a. Tìm đkxđ và rút gọn P
b. Tìm x để P = \(\dfrac{1}{4}\)
Điều kiện: x>2
P= \(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{2}+2}{\sqrt{x}-1}\right)\)
P= \(\left(\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
P= \(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)
P= \(\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b) P= \(\dfrac{1}{4}\)
⇔\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}\) =\(\dfrac{1}{4}\)
⇔\(4\sqrt{x}-8=3\sqrt{x}\)
⇔\(\sqrt{x}=8\)
⇔x=64 (TM)
Vậy X=64(TMĐK) thì P=\(\dfrac{1}{4}\)
Cho P= \((\dfrac{1}{1-\sqrt{2}}-\dfrac{1}{\sqrt{x}}):(\dfrac{2x+\sqrt{x}-1}{\sqrt{x}-x\sqrt{x}}+\dfrac{2x\sqrt{x}+x-\sqrt{x}}{\sqrt{x}+x^{2}})\)
a) Rút gọn P
b) so sánh P với \(\dfrac{3}{4}\).
c) tìm x để P=1
Cho P = (\(\dfrac{\sqrt{x}}{\sqrt{x}-1 }\) - \(\dfrac{1}{x-\sqrt{x}}\))(\(\dfrac{1}{1+\sqrt{x}}\) + \(\dfrac{2}{x-1}\))
a. Tìm đkxđ và rút gọn P
b. Tìm x để P>0
a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right)\cdot\left(\dfrac{1}{1+\sqrt{x}}+\dfrac{2}{x-1}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\cdot\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
b) Để P>0 thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}>0\)
mà \(\sqrt{x}+1>0\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}\left(\sqrt{x}-1\right)>0\)
mà \(\sqrt{x}>0\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}-1>0\)
\(\Leftrightarrow\sqrt{x}>1\)
hay x>1
Kết hợp ĐKXĐ,ta được: x>1
Vậy: Để P>0 thì x>1
cho P= (\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}\)+ \(\dfrac{\sqrt{x}}{\sqrt{x-3}}\)-\(\dfrac{3x+3}{x-9}\)) : (\(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}\)-1)
a, Rút gọn P
b, Tìm x để P < \(\dfrac{1}{2}\)
c, Tìm GTNN của P
a: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{2\sqrt{x}-2-\sqrt{x}+3}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
cho bt
P=(\(\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}\)+\(\dfrac{2\sqrt{x}+2}{\sqrt{x}+1}\)):\(\dfrac{x+1+2\sqrt{x}}{x-1}\)
a)Rút gọn P
b)Tìm x để P<0
\(a,P=\left(\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{2\sqrt{x}+2}{\sqrt{x}+1}\right):\dfrac{x+1+2\sqrt{x}}{x-1}\left(dk:x>0,x\ne1\right)\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right):\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\left(\sqrt{x}-1+2\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\left(\sqrt{x}+1\right).\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ =\sqrt{x}-1\)
\(b,P< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1\)
So với \(dk:x>0\) \(\Rightarrow S=\left\{x|0< x< 1\right\}\)
cho P= \(\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
a, tìm đkxd của P
b, rút gọn P
c, tìm x để p=\(\dfrac{1}{2}\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{9;4\right\}\end{matrix}\right.\)
b: Ta có: \(P=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
\(=\dfrac{2\sqrt{x}-9+2x-3\sqrt{x}-2-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
P=\(\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{1-\sqrt{x}}\right):\left(\dfrac{3}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}-\dfrac{2}{x-1}\right)\)
a) Rút gọn P
b) Tìm x để P = \(\sqrt{x}-1\)
a: \(P=\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}:\dfrac{3\sqrt{x}+3+\sqrt{x}-1-2}{x-1}\)
\(=\dfrac{-4\sqrt{x}}{3\sqrt{x}}=\dfrac{-4}{3}\)
b: Để \(P=\sqrt{x}-1\) thì \(\sqrt{x}=-\dfrac{1}{3}\)(vô lý)
13. P= \(\dfrac{x+2}{x\sqrt{x}+1}+\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}-\dfrac{\sqrt{x}-1}{x-1}\)
a. Rút gọn P
b. Tìm x để \(\left|P\right|\)=\(\dfrac{2}{3}\)
a: \(P=\dfrac{x+2+x-1}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{2x+1-x+\sqrt{x}-1}{x\sqrt{x}+1}=\dfrac{x+\sqrt{x}}{x\sqrt{x}+1}=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)
b: |P|=2/3
mà P>=0
nên P=2/3
=>căn x/(x-căn x+1)=2/3
=>2x-2căn x+2=3căn x
=>2x-5*căn x+2=0
=>(căn x-2)(2căn x-1)=0
=>x=4 hoặc x=1/4
(P): (\(\dfrac{1}{x-\sqrt{x}}\)+\(\dfrac{1}{\sqrt{x}-1}\)):\(\dfrac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
a)rút gọn bt P
b) tìm x để P=-1
a: \(P=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
b: Để P=-1 thì \(\sqrt{x}-1=-\sqrt{x}\)
=>x=1/4(nhận)
Cho biểu thức:
P=\(\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\left(\dfrac{1-x}{\sqrt{2}}\right)^2\)
a) Rút gọn P
b) Tìm x để P>0