Cho \(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
a, Rút gọn P
b, Tìm GTNN của P
K
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24 tháng 8 2021
a: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{2\sqrt{x}-2-\sqrt{x}+3}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
AT
11 tháng 6 2021
a) ĐKXĐ: \(x\ge0,x\ne1\)
\(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) Để \(P< \dfrac{1}{2}\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}< \dfrac{1}{2}\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{1}{2}< 0\)
\(\Rightarrow\dfrac{2\sqrt{x}-2-\sqrt{x}-1}{2\sqrt{x}+2}< 0\Rightarrow\dfrac{\sqrt{x}-3}{2\sqrt{x}+2}< 0\)
mà \(2\sqrt{x}+2>0\Rightarrow\sqrt{x}-3< 0\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\)
\(\Rightarrow0\le x< 9\left(x\ne1\right)\)
H9
HT.Phong (9A5)
CTVHS
24 tháng 9 2023
a) \(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{x-4}\right):\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]\)
\(P=\left[\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\left[\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{-4x-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{-\left(\sqrt{x}-3\right)}\)
\(P=\dfrac{-4\sqrt{x}\cdot\sqrt{x}}{-\left(\sqrt{x}-3\right)}\)
\(P=\dfrac{4x}{\sqrt{x}-3}\)
b) \(P=\dfrac{4x}{\sqrt{x}-3}\)
\(P=4\left(\sqrt{x}-3\right)+\dfrac{36}{\sqrt{x}-3}+24\)
Theo BĐT côsi ta có:
\(P\ge\sqrt{\dfrac{4\left(\sqrt{x}-3\right)\cdot36}{\sqrt{x}-3}}+24=36\)
Vậy: \(P_{min}=36\Leftrightarrow x=36\)
3 tháng 2 2021
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=\(\dfrac{1}{4}\)
21 tháng 7 2023
a: \(P=\dfrac{x-1}{\sqrt{x}}:\dfrac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
b: \(x=\dfrac{2}{2+\sqrt{3}}=2\left(2-\sqrt{3}\right)=4-2\sqrt{3}\)
Khi x=4-2căn 3 thì \(P=\dfrac{\left(\sqrt{3}-1+1\right)^2}{\sqrt{3}-1}=\dfrac{3}{\sqrt{3}-1}=\dfrac{3\sqrt{3}+3}{2}\)
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 \(P=5\Rightarrow\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=5\Rightarrow\sqrt{x}+6=5\sqrt{x}+5\Rightarrow\sqrt{x}=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{16}\)
16 tháng 5 2023
1: \(B=\dfrac{2\sqrt{x}-x-2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\dfrac{-x}{\left(\sqrt{x}-2\right)\cdot\sqrt{x}}\)
\(=\dfrac{-\sqrt{x}}{\sqrt{x}-2}\)
6 tháng 7 2021
ĐK:\(x\ge0;x\ne9\)
a) \(P=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\dfrac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+x-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)
b)\(P=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}=1+\dfrac{2}{\sqrt{x}+2}\le1+\dfrac{2}{0+2}=2\)
Dấu "=" xảy ra khi \(x=0\)
Vậy \(P_{max}=2\)
ĐKXĐ: \(x\ge0;x\ne1\)
\(P=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b.
\(P=\dfrac{2\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}=-1+\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)
Do \(\left\{{}\begin{matrix}2\sqrt{x}\ge0\\\sqrt{x}+1>0\end{matrix}\right.\) \(\Rightarrow\dfrac{2\sqrt{x}}{\sqrt{x}+1}\ge0\)
\(\Rightarrow P\ge-1\)
\(P_{min}=-1\) khi \(x=0\)
a) Ta có: \(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)