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mình giải thế này
a)\(P=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right)\frac{\left(1-x\right)^2}{2}\)
\(P=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x+1}\right)^2}{2}\)
\(P=-\sqrt{x}.\left(\sqrt{x}-1\right)=-x+\sqrt{x}\)
b)\(0< x< 1\Rightarrow\sqrt{x}< 1\Rightarrow\sqrt{x}-1< 0\)
\(\Rightarrow-x\left(\sqrt{x}-1\right)>0\)vì \(x>0\)
xong rồi nhé :)
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
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) \(x+\sqrt{x}+1>0\Rightarrow G>0\)
\(x+\sqrt{x}+1>0+0+1=1\)
\(\Rightarrow\frac{2}{x+\sqrt{x}+1}< \frac{2}{1}=2\Rightarrow G< 2\)
\(\Rightarrow O< G< 2\)
ĐKXĐ : \(0\le x\ne1\)
a) \(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\frac{\left(1-x\right)^2}{2}\)
\(=\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{\left(1-x\right)^2}{2}\)
\(=\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)
\(=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)
\(=-\sqrt{x}\left(\sqrt{x}-1\right)\)
b) \(P=\sqrt{x}\left(1-\sqrt{x}\right)\)
Để P > 0 thì \(\hept{\begin{cases}\sqrt{x}>0\\1-\sqrt{x}>0\end{cases}\Rightarrow}0< x< 1\)
c) \(P=-x+\sqrt{x}=-\left(x-2\sqrt{x}.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Vậy max P = 1/4 khi x = 1/4
Trả lời:
a, \(B=\left(\frac{x+\sqrt{x}-1}{x\sqrt{x}-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\right):\frac{1}{\sqrt{x}-1}\left(ĐK:x>0;x\ne1\right)\)
\(=\left(\frac{x+\sqrt{x}-1}{\left(\sqrt{x}\right)^3-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\right):\frac{1}{\sqrt{x}-1}\)
\(=\left(\frac{x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\right).\left(\sqrt{x}-1\right)\)
\(=\left(\frac{x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\left(\sqrt{x}-1\right)\)
\(=\frac{x+\sqrt{x}-1-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(=\frac{x+\sqrt{x}-1-\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(=\frac{x+\sqrt{x}-1-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(=\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
b, \(B< \frac{1}{3}\Leftrightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}< \frac{1}{3}\)
\(\Leftrightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{3}< 0\)
\(\Leftrightarrow\frac{3\sqrt{x}}{3\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{3\left(x+\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\frac{3\sqrt{x}-x-\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\frac{-x+2\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\frac{-\left(x-2\sqrt{x}+1\right)}{3\left(x+\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\frac{-\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}< 0\)
Vì \(-\left(\sqrt{x}-1\right)^2< 0\) với mọi \(x>0;x\ne1\)
\(3\left(x+\sqrt{x}+1\right)>0\) với mọi \(x>0;x\ne1\)
\(\Rightarrow\frac{-\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}< 0\) luôn đúng với mọi \(x>0;x\ne1\)
Vậy \(B< \frac{1}{3}\)
c, \(B=\frac{1}{2\sqrt{x}+1}\Leftrightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}=\frac{1}{2\sqrt{x}+1}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)=x+\sqrt{x}+1\)
\(\Leftrightarrow2x+\sqrt{x}=x+\sqrt{x}+1\)
\(\Leftrightarrow x=1\) (tm)
Vậy x = 1 là giá trị cần tìm
a) đk: \(x\ge0\)
\(P=\frac{1}{\sqrt{x}+1}-\frac{3}{x\sqrt{x}+1}+\frac{2}{x-\sqrt{x}+1}\)
\(P=\frac{x-\sqrt{x}+1-3+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(P=\frac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(P=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)
b) Ta thấy \(\hept{\begin{cases}\sqrt{x}\ge0\\x-\sqrt{x}+1>0\end{cases}\left(\forall x\right)\Rightarrow}\frac{\sqrt{x}}{x-\sqrt{x}+1}\ge0\) (1)
Mặt khác ta thấy: \(1-\frac{\sqrt{x}}{x-\sqrt{x}+1}=\frac{x-2\sqrt{x}+1}{x-\sqrt{x}+1}=\frac{\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+1}\ge0\left(\forall x\right)\)
=> \(1\ge\frac{\sqrt{x}}{x-\sqrt{x}+1}\) (2)
Từ (1) và (2) => \(0\le\frac{\sqrt{x}}{x-\sqrt{x}+1}\le0\)
=> \(0\le P\le1\)