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\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}-1}\)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 ....
1. \(VT=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=2+\sqrt{3}-2+\sqrt{3}=VP\)
Bài 1.
Ta có : \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{3+4\sqrt{3}+4}-\sqrt{3-4\sqrt{3}+4}\)
\(=\sqrt{\left(\sqrt{3}+2\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\left|\sqrt{3}+2\right|-\left|\sqrt{3}-2\right|\)
\(=\sqrt{3}+2-\left(2-\sqrt{3}\right)\)
\(=\sqrt{3}+2-2+\sqrt{3}=2\sqrt{3}\left(đpcm\right)\)
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: \(P=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
c: Để \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\) là số nguyên thì \(\sqrt{x}+1-2⋮\sqrt{x}+1\)
=>\(\sqrt{x}+1\in\left\{1;2\right\}\)
=>x=0
a: \(A=\dfrac{\sqrt{x}-1+\sqrt{x}}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left(\dfrac{-\left(2x+\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)+\left(2x\sqrt{x}+x-\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\right)\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}:\dfrac{-2x^2+x\sqrt{x}-2\sqrt{x}+1+2x^2-x\sqrt{x}-2x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{-\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}{-2x-\sqrt{x}+1}\)
\(=\dfrac{-\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{-\sqrt{x}\left(2x+\sqrt{x}-1\right)}\)
\(=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\)
\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
b: Thay \(x=17-12\sqrt{2}=\left(3-2\sqrt{2}\right)^2\) vào A, ta được:
\(A=\dfrac{17-12\sqrt{2}-\sqrt{2}+1+1}{3-2\sqrt{2}}=\dfrac{19-13\sqrt{2}}{3-2\sqrt{2}}=5-\sqrt{2}\)
ĐKXĐ: x>=0
\(P-1=\dfrac{\sqrt{x}-2}{2\sqrt{x}+1}-1\)
\(=\dfrac{\sqrt{x}-2-2\sqrt{x}-1}{2\sqrt{x}+1}=\dfrac{-\sqrt{x}-3}{2\sqrt{x}+1}\)
\(=-\dfrac{\left(\sqrt{x}+3\right)}{2\sqrt{x}+1}< 0\)
=>P<1