Tìm GTNN của A=\(\frac{X+4\sqrt{X}+20}{\sqrt{X}+2}\)
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PN
31 tháng 7 2017
\(A=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}+2\right)+8\sqrt{x}}{x-4}:\frac{2\left(\sqrt{x}+2\right)-2\sqrt{x}-3}{\sqrt{x}+2}\)
\(A=\frac{2x}{x-4}.\left(\sqrt{x}+2\right)=\frac{2x\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(A=\frac{2x}{\sqrt{x}-2}\)
12 tháng 1 2021
\(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}+\frac{3\sqrt{x}+1}{1-x}\)
a) ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{3\sqrt{x}+1}{x-1}\)
\(=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2x-2\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b) Với x = 4 thỏa mãn ĐKXĐ
\(A=\frac{2\sqrt{4}-1}{\sqrt{4}+1}=\frac{4-1}{2+1}=\frac{3}{3}=1\)
c) Chưa nghĩ ra :<
14 tháng 8 2020
ĐKXĐ: x \(\ge\)0; x \(\ne\)1 ; x \(\ne\)4
a) P = \(\left(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right)\)
P = \(\frac{\sqrt{x}\left(\sqrt{x}+1\right)-x-2}{\sqrt{x}+1}:\frac{\sqrt{x}\left(1-\sqrt{x}\right)-\sqrt{x}+4}{\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}\)
P = \(\frac{x+\sqrt{x}-x-2}{\sqrt{x}+1}\cdot\frac{\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-x-\sqrt{x}+4}\)
P = \(\frac{\left(1-\sqrt{x}\right)\left(\sqrt{x}-2\right)}{4-x}\)
P = \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
b) P < 0 <=> \(\frac{\sqrt{x}-1}{\sqrt{x}+2}< 0\)
Do \(\sqrt{x}+2>0\) => \(\sqrt{x}-1< 0\) => \(\sqrt{x}< 1\) => \(x< 1\)
kết hợp với đk => S = {x| \(0\le x< 1\)}
c) P = \(\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\frac{3}{\sqrt{x}+2}\ge-\frac{1}{2}\)
Do \(\sqrt{x}+2\ge2\) => \(-\frac{3}{\sqrt{x}+2}\ge-\frac{3}{2}\) => \(1-\frac{3}{\sqrt{x}+2}\ge-\frac{1}{2}\)
Dấu "=" xảy ra <=> x = 0
Vậy MinP = -1/2 khi x = 0