Tìm x để \(\sqrt{P}< \dfrac{1}{2}\) biết \(P=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
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H9
HT.Phong (9A5)
CTVHS
20 tháng 11 2023
Ta biết: \(\sqrt{P}=\dfrac{1}{2}\Rightarrow P=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\) (1)
Với đk: \(P\ge0\)
\(\Rightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\ge0\)
\(\Leftrightarrow\sqrt{x}-2\ge0\) (vì \(\sqrt{x}+1\ge1>0\forall x\ge0\))
\(\Leftrightarrow\sqrt{x}\ge2\)
\(\Leftrightarrow x\ge4\)
\(\left(1\right)\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{1}{4}\)
\(\Leftrightarrow4\left(\sqrt{x}-2\right)=\sqrt{x}+1\)
\(\Leftrightarrow4\sqrt{x}-8=\sqrt{x}+1\)
\(\Leftrightarrow4\sqrt{x}-\sqrt{x}=1+8\)
\(\Leftrightarrow3\sqrt{x}=9\)
\(\Leftrightarrow\sqrt{x}=3\)
\(\Leftrightarrow x=3^2\)
\(\Leftrightarrow x=9\left(tm\right)\)
Vậy: ...
TN
18 tháng 5 2018
a) Ta có:
\(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left(\frac{2\sqrt{x}\left(\sqrt{x-3}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x-3}\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{3x+3}{x-9}\right):\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\left(\frac{2x-6}{x-9}+\frac{x+3\sqrt{x}}{x-9}-\frac{3x+3}{x-9}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{2x-6+x+3\sqrt{x}-3x-3}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}\)
b) \(P< \frac{-1}{2}\Rightarrow\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}< \frac{-1}{2}\)
.....Chưa nghĩ ra....
c) Ta có: \(\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-3=0\Rightarrow x=9\)
Vậy Min P = 0 khi x =9.
k - kb với tớ nhia mn!
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=\(\dfrac{1}{4}\)
10 tháng 12 2023
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)
\(B=\dfrac{x-3}{x-1}-\dfrac{2}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{x-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{x-3-2\left(\sqrt{x}-1\right)+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}-2-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(ĐặtP=\dfrac{A}{B}\)
=>\(P=\dfrac{2\sqrt{x}-2}{\sqrt{x}+1}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}-2}{\sqrt{x}+1}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{2\sqrt{x}-2}{\sqrt{x}}\)
Để P<1 thì P-1<0
=>\(\dfrac{2\sqrt{x}-2-\sqrt{x}}{\sqrt{x}}< 0\)
=>\(\sqrt{x}-2< 0\)
=>\(\sqrt{x}< 2\)
=>0<=x<4
mà x nguyên
nên \(x\in\left\{0;1;2;3\right\}\)
7 tháng 2 2021
ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
a) Ta có: \(P=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\cdot\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\left(\dfrac{1}{2\sqrt{x}}-\dfrac{x}{2\sqrt{x}}\right)^2\)
\(=\dfrac{x-2\sqrt{x}+1-\left(x+2\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(x-1\right)^2}{4x}\)
\(=\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\cdot\dfrac{\left(x-1\right)^2}{4x}\)
\(=\dfrac{-4\sqrt{x}\cdot\left(x-1\right)}{4x}\)
\(=\dfrac{-x+1}{\sqrt{x}}\)
b) Để P=2 thì \(-x+1=2\sqrt{x}\)
\(\Leftrightarrow-x+1-2\sqrt{x}=0\)
\(\Leftrightarrow x+2\sqrt{x}-1=0\)
\(\Leftrightarrow x+2\sqrt{x}+1-2=0\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=\sqrt{2}\\\sqrt{x}+1=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\sqrt{2}-1\\\sqrt{x}=-\sqrt{2}-1\left(loại\right)\end{matrix}\right.\Leftrightarrow x=3-2\sqrt{2}\)
Vậy: Để P=2 thì \(x=3-2\sqrt{2}\)
AT
13 tháng 7 2021
a) ĐKXĐ: \(x\ge0,x\ne1\)
\(P=\left(\dfrac{3}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{x-1}\right):\left(\dfrac{x+2}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)
\(=\left(\dfrac{3}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{x+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)
\(=\dfrac{3\left(\sqrt{x}+1\right)+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{1}{\sqrt{x}-1}=\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(=\dfrac{4\sqrt{x}}{\sqrt{x}+1}\)
b) \(P=\sqrt{x}-1\Rightarrow\dfrac{4\sqrt{x}}{\sqrt{x}+1}=\sqrt{x}-1\Rightarrow4\sqrt{x}=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(\Rightarrow4\sqrt{x}=x-1\Rightarrow x-4\sqrt{x}-1=0\)
\(\Delta=\left(-4\right)^2-4.\left(-1\right)=20\Rightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{4-2\sqrt{5}}{2}=2-\sqrt{5}\\\sqrt{x}=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{4+2\sqrt{5}}{2}=2+\sqrt{5}\end{matrix}\right.\)
mà \(\sqrt{x}\ge0\Rightarrow\sqrt{x}=2+\sqrt{5}\Rightarrow x=9+4\sqrt{5}\)
c) \(P=\dfrac{4\sqrt{x}}{\sqrt{x}+1}=\dfrac{4\left(\sqrt{x}+1\right)-4}{\sqrt{x}+1}=4-\dfrac{4}{\sqrt{x}+1}\)
Để \(P\in Z\Rightarrow4⋮\sqrt{x}+1\Rightarrow\sqrt{x}+1\in\left\{1;2;4\right\}\left(\sqrt{x}+1\ge1\right)\)
\(\Rightarrow x\in\left\{0;1;9\right\}\) mà \(x\ne1\Rightarrow x\in\left\{0;9\right\}\)
AT
13 tháng 7 2021
Từ khúc có \(x-4\sqrt{x}-1=0\)
Ta có: \(\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=4-5=-1\)
Thế vào \(\Rightarrow x-4\sqrt{x}+\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=0\)
\(\Rightarrow x-\sqrt{x}\left(2-\sqrt{5}+2+\sqrt{5}\right)+\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=0\)
\(\Rightarrow x-\left(2-\sqrt{5}\right)\sqrt{x}-\left(2+\sqrt{5}\right)\sqrt{x}+\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=0\)
\(\Rightarrow\sqrt{x}\left(\sqrt{x}-\left(2-\sqrt{5}\right)\right)-\left(2+\sqrt{5}\right)\left(\sqrt{x}-\left(2-\sqrt{5}\right)\right)=0\)
\(\Rightarrow\left(\sqrt{x}-\left(2-\sqrt{5}\right)\right)\left(\sqrt{x}-\left(2+\sqrt{5}\right)\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=2-\sqrt{5}\\\sqrt{x}=2+\sqrt{5}\end{matrix}\right.\) rồi khúc sau như trên
AT
13 tháng 8 2018
Rút gọn: đkxđ: x >=0; x khác 9; x khác 4
A = \(\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
\(=\dfrac{1+\sqrt{x}-\sqrt{x}}{1+\sqrt{x}}:\left[\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
\(=\dfrac{1}{1+\sqrt{x}}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{1+\sqrt{x}}:\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{1+\sqrt{x}}\cdot\left(\sqrt{x}-2\right)=\dfrac{\sqrt{x}-2}{1+\sqrt{x}}\)
Ta thấy: \(1+\sqrt{x}\ge1>0\forall xTMĐKXĐ\)
=> A < 0 <=> \(\sqrt{x}-2< 0\)
\(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
kết hợp với đkxđ => 0 ≤ x < 4
Y
8 tháng 7 2023
\(a,P\) xác định \(\Leftrightarrow\left[{}\begin{matrix}x>0\\x\ne1\\x\ne4\end{matrix}\right.\)
\(b,P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\\ =\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x-1-x+4}\\ =\dfrac{1}{\sqrt{x}}.\dfrac{\sqrt{x}-2}{3}\\ =\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
\(c,P=\dfrac{1}{4}\Leftrightarrow\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\\ \Leftrightarrow\dfrac{4\left(\sqrt{x}-2\right)-3\sqrt{x}}{12\sqrt{x}}=0\\ \Leftrightarrow4\sqrt{x}-8-3\sqrt{x}=0\\ \Leftrightarrow\sqrt{x}=8\\ \Leftrightarrow x=64\left(tmdk\right)\)
Vậy \(x=64\) thì \(P=\dfrac{1}{4}\)
12 tháng 8 2021
a: Ta có: \(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b: Thay \(x=\dfrac{1}{4}\) vào P, ta được:
\(P=\left(\dfrac{1}{2}-1\right):\left(\dfrac{1}{2}+1\right)=\dfrac{-1}{2}:\dfrac{3}{2}=-\dfrac{1}{3}\)
c: Ta có: \(P< \dfrac{1}{2}\)
\(\Leftrightarrow P-\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}-2-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\sqrt{x}< 3\)
hay x<9
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)
Ta có: \(\sqrt{P}< \dfrac{1}{2}\Rightarrow P< \left(\dfrac{1}{2}\right)^2\Leftrightarrow P< \dfrac{1}{4}\) (1)
Với đk: \(P\ge0\)
\(\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\ge0\)
\(\Leftrightarrow\sqrt{x}-2\ge0\) (vì \(\sqrt{x}+1>0\forall x\ge0\))
\(\Leftrightarrow\sqrt{x}\ge2\)
\(\Leftrightarrow x\ge4\)
\(\left(1\right)\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< \dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}-\dfrac{1}{4}< 0\)
\(\Leftrightarrow\dfrac{4\sqrt{x}-8}{4\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{4\left(\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\dfrac{4\sqrt{x}-8-\sqrt{x}-1}{4\left(\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\dfrac{3\sqrt{x}-9}{4\left(\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow3\sqrt{x}-9< 0\)
\(\Leftrightarrow\sqrt{x}< 3\)
\(\Leftrightarrow x< 9\)
Kết hợp với đk: \(4\le x< 9\)
sai dấu r bn ơi \(\sqrt{x}+1\\ \) mà bn bn nhìn lại đi