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MN
3 tháng 3 2020
a) \(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne9\\x\ne4\end{cases}}\)
\(P=\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\right):\frac{\sqrt{x}-3}{2\sqrt{x}-x}\)
\(\Leftrightarrow P=\frac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\frac{\sqrt{x}-3}{\sqrt{x}\left(2-\sqrt{x}\right)}\)
\(\Leftrightarrow P=\frac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\cdot\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\)
\(\Leftrightarrow P=\frac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\cdot\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\)
\(\Leftrightarrow P=\frac{4x\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow P=\frac{4x}{\sqrt{x}-3}\)
b) Để P < 0
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-3< 0\Leftrightarrow4x>0\\\sqrt{x}-3>0\Leftrightarrow4x< 0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}< 3\Leftrightarrow x>0\\\sqrt{x}>3\Leftrightarrow x< 0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x< 9\Leftrightarrow x>0\left(ktm\right)\\x>9\Leftrightarrow x< 0\left(ktm\right)\end{cases}}\)
Vậy để \(P< 0\Leftrightarrow x\in\varnothing\)
Để P > 0
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-3>0\Leftrightarrow4x>0\\\sqrt{x}-3< 0\Leftrightarrow4x< 0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}>3\Leftrightarrow x>0\left(tm\right)\\\sqrt{x}< 3\Leftrightarrow x< 0\left(ktm\right)\end{cases}}\)
\(\Leftrightarrow x>9\Leftrightarrow x>0\left(tm\right)\)
Vậy để \(P>0\Leftrightarrow x>9\)
c) Để \(\left|P\right|=1\)
\(\Leftrightarrow\orbr{\begin{cases}P=1\left(tm\right)\\P=-1\left(ktm\right)\end{cases}}\)
\(\Leftrightarrow\frac{4x}{\sqrt{x}-3}=1\)
\(\Leftrightarrow4x=\sqrt{x}-3\)
\(\Leftrightarrow4x-\sqrt{x}+3=0\)
\(\Leftrightarrow\left(2\sqrt{x}-\frac{1}{4}\right)^2+\frac{47}{48}=0\left(ktm\right)\)
Vậy để \(\left|P\right|=1\Leftrightarrow x\in\varnothing\)
29 tháng 8 2018
Vì x>0; y>0
Nên áp dụng BĐT Cô-si ta có: \(x+y\ge2\sqrt{xy}\)
\(\Rightarrow\)\(\frac{1}{x}+\frac{1}{y}\ge2\sqrt{\frac{1}{x}.\frac{1}{y}}=2\sqrt{\frac{1}{xy}}\)
Mà \(\frac{1}{x}+\frac{1}{y}=\frac{1}{2}\)
Nên \(\frac{1}{2}\ge2.\frac{1}{\sqrt{xy}}\Rightarrow\frac{1}{4}\ge\frac{1}{\sqrt{xy}}\)
\(\Rightarrow4\le\sqrt{xy}\) (C)
Ta có: \(\sqrt{x}+\sqrt{y}\ge2\sqrt{\sqrt{xy}}\)
Thế (C) vào ta được: \(\sqrt{x}+\sqrt{y}\ge2\sqrt{4}=4\)
Dấu "=" xảy ra <=> x = y
Vậy AMin = 4 khi và chỉ khi x = y
29 tháng 8 2018
\(\frac{1}{x}+\frac{1}{y}>=\frac{4}{x+y}\Rightarrow\frac{1}{2}>=\frac{4}{x+y}\Rightarrow x+y>=8\left(1\right)\)(bđt svacxo)
\(\frac{1}{x}+\frac{1}{y}>=2\sqrt{\frac{1}{x}\cdot\frac{1}{y}}=\frac{2}{\sqrt{xy}}\Rightarrow\frac{1}{2}>=\frac{2}{\sqrt{xy}}\Rightarrow\sqrt{xy}>=4\Rightarrow2\sqrt{xy}>=8\left(2\right)\)(bđt cosi)
từ \(\left(1\right);\left(2\right)\Rightarrow x+2\sqrt{xy}+y>=8+8=16\Rightarrow\left(\sqrt{x}+\sqrt{y}\right)^2>=16\)
mà \(\sqrt{x}>0;\sqrt{y}>0\Rightarrow\sqrt{x}+\sqrt{y}>=4\)
dấu = xảy ra khi x=y=4
vậy min A là 4 khi x=y=4
24 tháng 7 2019
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{4\sqrt{x}-3}{2\sqrt{x}-x}\right):\)\(\left(\frac{\sqrt{x}+2}{\sqrt{x}}-\frac{\sqrt{x}-4}{\sqrt{x}-2}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)\(:\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{x-4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}:\frac{x-4-x+4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}-3}{4}\)
\(b,\)Để \(P>0\Rightarrow\frac{\sqrt{x}-3}{4}>0\)
Mà \(4>0\Rightarrow\sqrt{x}-3>0\Rightarrow\sqrt{x}>3\Rightarrow x>9\)
\(c,\sqrt{P}_{min}=0\Rightarrow\frac{\sqrt{x}-3}{4}=0\)
\(\Leftrightarrow\sqrt{x}-3=0\Rightarrow\sqrt{x}=3\Rightarrow x=9\)
NK
1
12 tháng 6 2015
\(\frac{x+8}{\sqrt{x}+1}=\frac{x-1+9}{\sqrt{x}+1}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+9}{\sqrt{x}+1}=\sqrt{x}-1+\frac{9}{\sqrt{x}+1}\)
\(=\sqrt{x}+1+\frac{9}{\sqrt{x}+1}-2\ge2\sqrt{\left(\sqrt{x}+1\right).\frac{9}{\sqrt{x}+1}}-2=2.3-2=4\)
Vậy: GTNN là 4 \(\Leftrightarrow\sqrt{x}+1=\frac{9}{\sqrt{x}+1}\Leftrightarrow x=4\)
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 ..
\(P=x-\sqrt{x}\)
\(P=x-2\cdot\sqrt{x}\cdot\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\)
\(P=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge\frac{-1}{4}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}\)( thỏa )
Vậy \(minP=\frac{-1}{4}\Leftrightarrow x=\frac{1}{4}\)