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1) Khi x = 36 thì A = \(\frac{\sqrt{36}+4}{\sqrt{36}+2}\Leftrightarrow\frac{5}{4}\)
Vậy khi x = 36 thì A = \(\frac{5}{4}\)
2) B = \((\frac{\sqrt{x}\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}+\frac{4\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}):\frac{x+16}{\sqrt{x}+2}\)
= \(\frac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}.\frac{\sqrt{x}+2}{x+16}=\frac{x+16}{x-16}.\frac{\sqrt{x}+2}{x+16}\)
= \(\frac{\sqrt{x}+2}{x-16}\)
Vậy B = \(\frac{\sqrt{x}+2}{x-16}\)
a/ \(P=12\)
b/ \(Q=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c/ Ta có:
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Dấu = xảy ra khi x = 3 (thỏa tất cả các điều kiện )
a. Thay x = 3 vào biểu thức P ta được :
\(p=\frac{x+3}{\sqrt{x}-2}=\frac{9+3}{\sqrt{9}-2}=12\)
b, \(Q=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{x-4}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-3\sqrt{x}+2+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c, Ta có :
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Vậy GTNN \(\frac{P}{Q}=2\sqrt{3}\) khi và chỉ khi \(x=3\)
\(a.A=\dfrac{x+\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}:\dfrac{\sqrt{x}-1}{x+1}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}.\dfrac{x+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\left(x\ge0;x\ne1\right)\)
\(b.x=4+2\sqrt{3}=3+2\sqrt{3}+1=\left(\sqrt{3}+1\right)^2\left(TM\right)\)
\(\Rightarrow\sqrt{x}=\sqrt{3}+1\)
Ta có : \(\dfrac{\sqrt{3}+1}{\sqrt{3}+1-1}=\dfrac{\sqrt{3}+1}{\sqrt{3}}\)
\(c.Để:A\in Z\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-1}=1+\dfrac{1}{\sqrt{x}-1}\in Z\)\(\Rightarrow\left(\sqrt{x}-1\right)\in\left\{\pm1\right\}\)
\(\circledast\sqrt{x}-1=1\Leftrightarrow x=4\left(TM\right)\)
\(\circledast\sqrt{x}-1=-1\Leftrightarrow x=0\left(TM\right)\)
KL.........
a) Ta có:
\(A=\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-4}{\sqrt{x}-2\sqrt{x}}\)
\(A=\frac{\sqrt{x}-3}{\sqrt{x}-2}+\frac{\sqrt{x}-4}{\sqrt{x}}\)
\(A=\frac{\left(\sqrt{x}-3\right)\sqrt{x}+\left(\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\sqrt{x}}\)
\(A=\frac{x-3\sqrt{x}+x-6\sqrt{x}+8}{\left(\sqrt{x}-2\right)\sqrt{x}}\)
\(A=\frac{2x-9\sqrt{x}+8}{\left(\sqrt{x}-2\right)\sqrt{x}}\)