tìm giá trị nhỏ nhất của \(\frac{x+8}{\sqrt{x}+1}\)với x > 0
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22 tháng 2 2019
Áp dụng bất đẳng thức Cô-si ta có :
\(P=\frac{x}{\sqrt{1-x}}+\frac{y}{\sqrt{1-y}}=\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\)
\(=\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{xy}}=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{xy}}\)
\(\ge\frac{2\sqrt{\sqrt{x}.\sqrt{y}}\left(x+y-\frac{x+y}{2}\right)}{\sqrt{xy}}\)
\(=\frac{x+y}{\sqrt[4]{xy}}\ge\frac{x+y}{\sqrt{\frac{x+y}{2}}}=\frac{1}{\sqrt{\frac{1}{2}}}=\sqrt{2}\)
Dấu "=" khi x = y = 1/2
29 tháng 7 2021
I) Đk: x > 0 và x \(\ne\)9
\(D=\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(D=\frac{x+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(D=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+3}\)
=> \(\frac{1}{D}=\frac{\sqrt{x}+3}{\sqrt{x}+1}=\frac{\sqrt{x}+1+2}{\sqrt{x}+1}=1+\frac{2}{\sqrt{x}+1}\)
Để 1/D nguyên <=> \(\frac{2}{\sqrt{x}+1}\in Z\)
<=> \(2⋮\left(\sqrt{x}+1\right)\) <=> \(\sqrt{x}+1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
Do \(x>0\) => \(\sqrt{x}+1>1\) => \(\sqrt{x}+1=2\)
<=> \(\sqrt{x}=1\) <=> x = 1 (tm)
29 tháng 7 2021
\(E=\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right)\cdot\frac{4\sqrt{x}}{3}\)
\(E=\frac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}\)
\(E=\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}=\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
b) Với x\(\ge\)0; ta có:
\(E=\frac{8}{9}\) <=> \(\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\frac{8}{9}\)
<=> \(3\sqrt{x}=2x-2\sqrt{x}+2\)
<=> \(2x-4\sqrt{x}-\sqrt{x}+2=0\)
<=> \(\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
<=> \(\orbr{\begin{cases}x=\frac{1}{4}\left(tm\right)\\x=4\left(tm\right)\end{cases}}\)
e) Ta có: \(E=\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\ge0\forall x\in R\) (vì \(x-\sqrt{x}+1=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\))
Dấu "=" xảy ra<=> x = 0
Vậy MinE = 0 <=> x = 0
Lại có: \(\frac{1}{E}=\frac{3\left(x-\sqrt{x}+1\right)}{4\sqrt{x}}=\frac{3}{4}\left(\sqrt{x}-1+\frac{1}{\sqrt{x}}\right)\ge\frac{3}{4}\left(2\sqrt{\sqrt{x}\cdot\frac{1}{\sqrt{x}}}-1\right)\)(bđt cosi)
=> \(\frac{1}{E}\ge\frac{3}{2}.\left(2-1\right)=\frac{3}{2}\)=> \(E\le\frac{2}{3}\)
Dấu "=" xảy ra<=> \(\sqrt{x}=\frac{1}{\sqrt{x}}\) <=> x = 1
Vậy MaxE = 2/3 <=> x = 1
I
1 tháng 8 2020
a) Thay x=4 zô là đc . ra kết quả \(\frac{7}{6}\)là dúng
b) \(B=\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\)
\(=\frac{3x+3\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\)
\(=>P=A.B=\frac{3\sqrt{x}+1}{x+\sqrt{x}}.\frac{3\left(x+\sqrt{x}\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}=\frac{3}{3\sqrt{x}-1}\)
c) xét \(\frac{1}{P}=\frac{3\sqrt{x}-1}{3}\)
do \(\sqrt{x}\ge0=>3\sqrt{x}-1\ge-1\)\(=>\frac{3\sqrt{x}-1}{3}\ge-\frac{1}{3}\)
\(=>\frac{1}{P}\ge-\frac{1}{3}\)
dấu = xảy ra khi x=0
zậy ..
15 tháng 7 2018
\(E=\frac{x+\sqrt{x}}{x-2\sqrt{x}+1}:\left(\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{1}{1-\sqrt{x}}+\frac{2-x}{x-\sqrt{x}}\right)\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\) \(\left[\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\sqrt{x}}+\frac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\)\(\left[\frac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(E=\frac{x}{\sqrt{x}-1}\)
15 tháng 7 2018
b) \(E>1\Leftrightarrow\frac{x}{\sqrt{x}-1}>1\)
\(\Leftrightarrow\frac{x}{\sqrt{x}-1}-1>0\)
\(\Leftrightarrow\frac{x}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\frac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\frac{x-2\sqrt{x}+1+\sqrt{x}}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\frac{\left(\sqrt{x}-1\right)^2+\sqrt{x}}{\sqrt{x}-1}>0\)
\(\Rightarrow\sqrt{x}-1>0\) vì tử của phân số luôn \(\ge0\forall x\ge0\)
\(\Rightarrow x>1\)
kết hợp với ĐKXĐ \(x\ge0\Rightarrow x>1\)
vậy \(x>1\) thì \(E>1\)
22 tháng 10 2017
Ta có:
\(3=x+y+z\ge3\sqrt[3]{xyz}\)
\(\Leftrightarrow xyz\le1\)
Ta lại có:
\(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\ge\frac{3}{\sqrt[6]{xyz}}\ge\frac{3}{1}=3\)
20 tháng 8 2021
\(A=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}}{1-x}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(A=\left(\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)\(\div\left(\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(A=\left(\frac{x+2\sqrt{x}+1+x-\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\frac{2x+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{4\sqrt{x}}\)
\(A=\frac{2x+1}{4\sqrt{x}}\)
c, \(A=\frac{2x+1}{4\sqrt{x}}=\frac{\sqrt{x}}{2}+\frac{1}{4\sqrt{x}}\)
ap dụng cô si ta có \(\frac{\sqrt{x}}{2}+\frac{1}{4\sqrt{x}}\ge2\sqrt{\frac{\sqrt{x}}{2}\cdot\frac{1}{4\sqrt{x}}}=\frac{\sqrt{2}}{2}\)
dấu = xảy ra khi \(\frac{\sqrt{x}}{2}=\frac{1}{4\sqrt{x}}\Leftrightarrow x=\frac{1}{2}\) (tm)
\(\frac{x+8}{\sqrt{x}+1}=\frac{x-1+9}{\sqrt{x}+1}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+9}{\sqrt{x}+1}=\sqrt{x}-1+\frac{9}{\sqrt{x}+1}\)
\(=\sqrt{x}+1+\frac{9}{\sqrt{x}+1}-2\ge2\sqrt{\left(\sqrt{x}+1\right).\frac{9}{\sqrt{x}+1}}-2=2.3-2=4\)
Vậy: GTNN là 4 \(\Leftrightarrow\sqrt{x}+1=\frac{9}{\sqrt{x}+1}\Leftrightarrow x=4\)