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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 \(x=3\)
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}-1}\)ĐK x>=0 x khác -1
=\(\frac{\sqrt{x}+1}{x+1}.\frac{x+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b/ x =\(\frac{2+\sqrt{3}}{2}=\frac{4+2\sqrt{3}}{4}=\frac{3+2\sqrt{3}+1}{4}=\frac{\left(\sqrt{3}+1\right)^2}{4}\)
\(\Rightarrow\sqrt{x}=\frac{\sqrt{3}+1}{2}\)
Em thay vào tính nhé!
c) với x>1
A=\(\frac{\sqrt{x}+1}{\sqrt{x}-1}.\sqrt{x}=\frac{x+\sqrt{x}}{\sqrt{x}-1}=\sqrt{x}+2+\frac{2}{\sqrt{x}-1}=\sqrt{x}-1+\frac{2}{\sqrt{x}-1}+3\)
Áp dụng bất đẳng thức Cosi
A\(\ge2\sqrt{2}+3\)
Xét dấu bằng xảy ra ....
Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
có phải/....
1) \(A=\dfrac{x+3}{\sqrt{x}-2}\)
\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-2}{x-4}\) hay \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\left(\sqrt{x}-2\right)}{x-4}\)
2) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
Bài 1:
a: \(B=\dfrac{\sqrt{x}+x+\sqrt{x}-x}{1-x}\cdot\dfrac{x-1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)
b: Để B=-1 thì \(2\sqrt{x}=-\sqrt{x}+3\)
=>3 căn x=3
=>căn x=1
hay x=1(loại)
a: \(P=\left(\dfrac{\sqrt{x}}{x+1}+\dfrac{1}{x+1}\right)\cdot\dfrac{x+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Khi \(x=\dfrac{2+\sqrt{3}}{2}=\dfrac{4+2\sqrt{3}}{4}\) thì
\(P=\left(\dfrac{\sqrt{3}+1}{2}+1\right):\left(\dfrac{\sqrt{3}+1}{2}-1\right)\)
\(=\dfrac{\sqrt{3}+3}{2}:\dfrac{\sqrt{3}-1}{2}=\dfrac{3+\sqrt{3}}{\sqrt{3}-1}\)
1: Khi x=9 thì \(A=\dfrac{9+3}{3-2}=12\)
2: \(Q=\dfrac{x-3\sqrt{x}+2-5\sqrt{x}+2}{x-4}\)
\(=\dfrac{x-8\sqrt{x}+4}{x-4}\)
a: \(P=\dfrac{2x+2}{\sqrt{x}}+\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x+2+x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{3x+\sqrt{x}+3-x+\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
b: Thay \(x=3-2\sqrt{2}\) vào P, ta được:
\(P=\dfrac{2\cdot\left(3-2\sqrt{2}\right)+2\left(\sqrt{2}-1\right)+2}{\sqrt{2}-1}\)
\(=\dfrac{6-4\sqrt{2}+2\sqrt{2}-2+2}{\sqrt{2}-1}=\dfrac{6-2\sqrt{2}}{\sqrt{2}-1}=4\sqrt{2}+2\)
\(A=P:Q=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}:\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+4}=1+\dfrac{-5}{\sqrt{x}+4}\)
Điều kiện : \(x\ge4\Rightarrow\sqrt{x}+4\ge4\Rightarrow-\dfrac{5}{\sqrt{x}+4}\le-\dfrac{5}{4}\Rightarrow\dfrac{5}{\sqrt{x}+4}\ge\dfrac{5}{4}\)
Dấu ''='' xảy ra \(\Leftrightarrow x=0\)
Vậy \(min_A=\dfrac{5}{4}\Leftrightarrow x=0\)