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12 tháng 8 2019
a) A= \(\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{2}\right)\) (x ≥ 0; x ≠ 4)
= \(\left(\frac{x+2}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)\cdot\sqrt{x}}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right):\frac{\sqrt{x}-1}{2}\)
=\(\left(\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right)\cdot\frac{2}{\sqrt{x}-1}\)
=\(\left(\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right)\cdot\frac{2}{\sqrt{x}-1}\)
= \(\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\cdot\frac{2}{\sqrt{x}-1}\)
=\(\frac{2}{x+\sqrt{x}+1}\)
b) Ta có: x ≥ 0 ⇒ \(\sqrt{x}\) ≥ 0
⇒x+\(\sqrt{x}\)+1 ≥ 1 > 0
mà 2 > 0
⇒ A > 0 (1)
Ta có:
\(x+\sqrt{x}+1\) ≥ 1
⇒ \(\frac{1}{x+\sqrt{x}+1}\) ≤ 1
⇒\(\frac{2}{x+\sqrt{x}+1}\) ≤ 2
⇒A ≤ 2 (2)
Từ (1) và (2) => 0 < A ≤ 2
1 tháng 8 2019
a) ĐKXĐ: x\(\ge0,x\ne1\)
A = \(\frac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\frac{\sqrt{x}-1}{2}\)
= \(\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x +\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)
= \(\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)
= \(\frac{2}{x+\sqrt{x}+1}\)
b) Ta có x\(\ge0,x\ne1\) =>\(x+\sqrt{x}+1>0\Rightarrow\frac{2}{x+\sqrt{x}+1}>0\)
=> A>0 (1)
Mặt khác \(x\ge0,x\ne1\Rightarrow x+\sqrt{x}+1\ge1\)
\(\Rightarrow\frac{2}{x+\sqrt{x}+1}\le2\) \(\Rightarrow A\ge2\) (2)
Từ (1) và (2) => \(0< A\le2\)
21 tháng 9 2020
a)\(G=\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\frac{\sqrt{x}-1}{2}\)
\(=\frac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)
\(=\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)
\(=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)
\(=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)
\(=\frac{2}{x+\sqrt{x}+1}\)
b) \(x+\sqrt{x}+1>0\Rightarrow G>0\)
\(x+\sqrt{x}+1>0+0+1=1\)
\(\Rightarrow\frac{2}{x+\sqrt{x}+1}< \frac{2}{1}=2\Rightarrow G< 2\)
\(\Rightarrow O< G< 2\)
HT
27 tháng 8 2016
Ta có:
\(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\left(\frac{1-x}{\sqrt{2}}\right)^2\)
\(P=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\frac{\left(1-x\right)^2}{2}\)
\(P=\left(\frac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\frac{\left(x-1\right)^2}{2}\)
\(P=\left(\frac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\frac{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)^2}{2}\)
\(P=\left(-\sqrt{x}\right)\left(\sqrt{x}-1\right)\)
\(P=\sqrt{x}-x\)
b) Để \(P>0\) thì \(\sqrt{x}-x>0\)
- \(\sqrt{x}-x>0\)
\(\Rightarrow\sqrt{x}\left(1-\sqrt{x}\right)>0\)
Suy ra: TH1: \(\sqrt{x}< 0\) và \(1-\sqrt{x}< 0\) (Loại) vì \(\sqrt{x}\ge0\)
TH2:\(\sqrt{x}>0\) và \(1-\sqrt{x}>0\) (Nhận)
Ta có \(\sqrt{x}>0\) và \(1-\sqrt{x}>0\) để \(P>0\)
- \(\sqrt{x}>0\) \(\Rightarrow x>0\)
- \(1-\sqrt{x}>0\) \(\Rightarrow\sqrt{x}< 1\) \(\Rightarrow x< 1\)
Vậy để \(P>0\) thì \(0< x< 1\)
c)\(P=\sqrt{x}-x\)
\(P=-\left(x-\sqrt{x}\right)\)
\(P=-\left(\left(\sqrt{x}\right)^2-2.\frac{1}{2}.\sqrt{x}+\frac{1}{4}-\frac{1}{4}\right)\)
\(P=-\left(\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\right)\)
\(P=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\)
Vì \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow-\left(\sqrt{x}-\frac{1}{2}\right)^2\le0\)
Nên \(-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Dấu "=" xảy ra khi \(\sqrt{x}-\frac{1}{2}=0\) \(\Rightarrow x=\frac{1}{4}\)
Vậy GTLN của \(P\) là \(\frac{1}{4}\) khi \(x=\frac{1}{4}\)
10 tháng 7 2019
\(3,\)Áp dụng bđt Mincopski \(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\)hai lần có
\(VT\ge\sqrt{\left(\sqrt{x}+\sqrt{y}\right)^2+\left(\sqrt{yz}+\sqrt{zx}\right)^2}+\sqrt{z+xy}\)
\(\ge\sqrt{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2+\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}\)
\(=\sqrt{x+y+z+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)+\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}\)
\(=\sqrt{1+2t+t^2}\left(t=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)
\(=\sqrt{\left(t+1\right)^2}=t+1=VP\left(Đpcm\right)\)
10 tháng 7 2019
\(2,\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\frac{2\sqrt{ab}}{2\sqrt{\sqrt{a}.\sqrt{b}}}=\sqrt{\sqrt{ab}}\left(đpcm\right)\)