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24 tháng 7 2017
a. ĐKXĐ \(x\ge0\)và \(x\ne9\)
Ta có \(K=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{3x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(x-2\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
b. Để \(K< -1\Rightarrow\frac{3\sqrt{x}-9+\sqrt{x}+3}{\sqrt{x}+3}< 0\Rightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\Rightarrow4\sqrt{x}-6< 0\)vì \(\sqrt{x}+3\ge3\)
\(\Rightarrow0\le x< \frac{9}{4}\left(tm\right)\)
Vậy với \(0\le x< \frac{9}{4}\)thì K<-1
c. \(K=\frac{3\sqrt{x}-9}{\sqrt{x}+3}=3+\frac{-18}{\sqrt{x}+3}\)
Ta có \(\sqrt{x}+3\ge3\Rightarrow\frac{1}{\sqrt{x}+3}\le\frac{1}{3}\Rightarrow-\frac{18}{\sqrt{x}+3}\ge-6\Rightarrow3+\frac{-18}{\sqrt{x}+3}\ge-3\)
\(\Rightarrow K\ge-3\)
Vậy \(MinK=-3\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)
N
20 tháng 9 2019
a.\(DK:x\ge0\)
\(A=\frac{x-2\sqrt{x}+1}{x+1}.\frac{\left(x+1\right)\left(\sqrt{x}+1\right)}{x-2\sqrt{x}+1}=\sqrt{x}+1\)
b.Dat \(P=\frac{1}{A}\left(x+3\right)=\frac{x+3}{\sqrt{x}+1}\left(P>0\right)\)
\(\Rightarrow P\sqrt{x}+P=x+3\)
\(\Leftrightarrow x-P\sqrt{x}+3-P=0\)
Dat \(t=\sqrt{x}\left(t\ge0\right)\)
Ta co:
\(\Delta\ge0\)
\(\Leftrightarrow P^2-4\left(3-P\right)\ge0\)
\(\Leftrightarrow P^2+4P-12\ge0\)
\(\Leftrightarrow\left(P-2\right)\left(P+6\right)\ge0\)
TH1:
\(\hept{\begin{cases}P-2\ge0\\P+6\ge0\end{cases}\Leftrightarrow P\ge2}\)
TH2:
\(\hept{\begin{cases}P-2\le0\\P+6\le0\end{cases}\Leftrightarrow P\le2\left(P>0\right)}\)
Vi la de bai tim min nen lay TH1 thoi
Dau '=' xay ra khi \(x=\frac{P}{2}=1\)
Vay \(P_{min}=2\)khi \(x=1\)
11 tháng 10 2016
a) \(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right):\left(\frac{2}{x}-\frac{2-x}{x\sqrt{x}+x}\right)\left(ĐK:x>0;x\ne1\right)\)
\(=\left[\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\left[\frac{2}{x}-\frac{2-x}{x\left(\sqrt{x}+1\right)}\right]\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{2\left(\sqrt{x}+1\right)-2+x}{x\left(\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{x\left(\sqrt{x}+1\right)}{2\sqrt{x}+2-2+x}\)
\(=\frac{x+2\sqrt{x}}{\sqrt{x}-1}\cdot\frac{x}{2\sqrt{x}+x}=\frac{x}{\sqrt{x}-1}\)
b)Để P>2
\(\Leftrightarrow\frac{x}{\sqrt{x}-1}>2\)
\(\Leftrightarrow\frac{x}{\sqrt{x}-1}-2>0\)
\(\Leftrightarrow\frac{x-2\sqrt{x}+2}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\frac{\left(\sqrt{x}-1\right)^2+1}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\sqrt{x}-1>0\Leftrightarrow x>1\left(tm\right)\)
Vậy x>1 thì P>2
19 tháng 6 2017
a/ Căn xác định với \(2\le x< 3\) ta có \(\frac{\left(x-2\right)^2}{3-x}+\frac{x^2+1}{x-3}=0\)
<=> \(\frac{\left(x-2\right)^2}{3-x}-\frac{x^2+1}{3-x}=0\)<=> \(^{x^2-4x+4-x^2-1=0}\)<=> x = 3/4 ( Không TM ) Vậy PTVN
TN
19 tháng 6 2017
Bài 2:
*)GTNN: Áp dụng BĐT \(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\) ta có:
\(A=\sqrt{x+3}+\sqrt{5-x}\)
\(\ge\sqrt{x+3+5-x}=\sqrt{8}\)
Đẳng thức xảy ra khi \(-3\le x\le5\)
*)GTLN:Áp dụng BĐT Cauchy-Schwarz ta có:
\(A^2=\left(\sqrt{x+3}+\sqrt{5-x}\right)^2\)
\(\le\left(1+1\right)\left(x+3+5-x\right)\)
\(=2\cdot8=16\)
\(\Rightarrow A^2\le16\Rightarrow A\le4\)
Đẳng thức xảy ra khi \(x=1\)
6 tháng 2 2022
a: \(P=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{x+1+\sqrt{x}}{x+1}\)
\(=\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{x+1}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
KT
14 tháng 7 2018
a) ĐKXĐ: \(x\ge0;x\ne9\)
mk chỉnh lại đề bài nhé, chắc có lẽ bn ghi nhầm:
\(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left(\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-\frac{\sqrt{x}-3}{\sqrt{x}-3}\right)\)
\(=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\frac{2\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{-3}{\sqrt{x}+3}\)