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a) P= √x+1 √x−1 + x+2 x√x−1 - √x+1 x+√x+1 \(\Leftrightarrow0\)
b)\(\sqrt{x}\left(2x+2\right)+2x+abp^2-2\)
ĐK ; \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
a, \(Q=\frac{\sqrt{x}\left(\sqrt{x}+1\right)-3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x-8\sqrt{x}+7}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-7\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-7}{\sqrt{x}+1}\)
b. \(Q< \frac{1}{2}\Rightarrow\frac{\sqrt{x}-7}{\sqrt{x}+1}-\frac{1}{2}< 0\Rightarrow\frac{\sqrt{x}-15}{2\left(\sqrt{x}+1\right)}< 0\Rightarrow\sqrt{x}-15< 0\)
\(\Rightarrow0\le x< 225\)và \(x\ne4\)
c. \(Q=\frac{\sqrt{x}-7}{\sqrt{x}+1}=1-\frac{8}{\sqrt{x}+1}\)
Ta thấy \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow\frac{-8}{\sqrt{x}+1}\ge-8\Rightarrow1-\frac{8}{\sqrt{x}+1}\ge-7\)
\(\Rightarrow Q\ge-7\)
Vậy \(MinQ=-7\). Dấu bằng xảy ra \(\Rightarrow x=0\)
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{2\left(\sqrt{x}+1\right)-2+x}{x\left(\sqrt{x}+1\right)}\right)\)
\(\Leftrightarrow P=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{x\left(\sqrt{x}+1\right)}=\frac{x}{\sqrt{x}-1}\)
b. ta có \(x=\frac{8-4\sqrt{3}}{2-\sqrt{3}}=4\)
vậy \(P=\frac{4}{\sqrt{4}-1}=4\)
c.\(P=\frac{x}{\sqrt{x}-1}=\sqrt{x}-1+\frac{1}{\sqrt{x}-1}+2\ge2+2=4\)
vậy \(\sqrt{P}\ge2\)
ĐKXĐ: \(x\ge0,x\ne9\)
a) \(P=\frac{3\sqrt{x}+2}{\sqrt{x}+1}+\frac{2\sqrt{x}+3}{\sqrt{x}-3}-\frac{3\left(3\sqrt{x}-5\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x-3}\right)}\)
\(=\frac{\left(3\sqrt{x}+2\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)+3\left(3\sqrt{x}-5\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{3x-9\sqrt{x}+2\sqrt{x}-6+2x+2\sqrt{x}-3\sqrt{x}-3-9\sqrt{x}+15}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{5x-17\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{5x-15\sqrt{x}-2\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(5\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{5\sqrt{x}-2}{\sqrt{x}+1}\)
b) Ta có: \(x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}+1\)
Do đó: \(P=\frac{5\left(\sqrt{3}+1\right)-2}{\left(\sqrt{3}+1\right)+1}=\frac{5\sqrt{3}+3}{\sqrt{3}+2}=\frac{\left(5\sqrt{3}+3\right)\left(2-\sqrt{3}\right)}{\left(\sqrt{3}+2\right)\left(2-\sqrt{3}\right)}=7\sqrt{3}-9\)
c) Ta có \(P=\frac{5\sqrt{x}-2}{\sqrt{x}+1}=\frac{5\sqrt{x}+5-7}{\sqrt{x}+1}\)
\(P=5-\frac{7}{\sqrt{x}+1}\)
Vì \(\frac{7}{\sqrt{x}+1}>0\)nên \(P\)có giá trị nhỏ nhất khi và chỉ khi \(\frac{7}{\sqrt{x}+1}\)lớn nhất
\(\Leftrightarrow\sqrt{x}+1\)nhỏ nhất \(\Leftrightarrow x=0\)
Khi đó minP=5-7=-2