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a: \(M=\dfrac{x+6\sqrt{x}-3\sqrt{x}+18-x}{x-36}\)
\(=\dfrac{3\left(\sqrt{x}+6\right)}{x-36}=\dfrac{3}{\sqrt{x}-6}\)
b: \(N=\dfrac{x^2}{y}\cdot\sqrt{xy\cdot\dfrac{y}{x}}-x^2\)
\(=\dfrac{x^2}{y}\cdot y-x^2=0\)
a) \(2\sqrt{x^2}=2.\left|x\right|=-2x\)(vì x<0)
b) \(\frac{1}{2}\sqrt{x^{10}}=\frac{1}{2}\sqrt{\left(x^5\right)^2}\frac{1}{2}\left|x^5\right|=-\frac{1}{2}x^5\)(vì x>0)
c) \(x-4+\sqrt{x^2-8x+16}=x-4+\sqrt{\left(x-4\right)^2}=x-4+\left|x-4\right|=x-4+4-x=0\)(vì x<4 nên x-4<0)
d) \(\frac{3-\sqrt{x}}{x-9}=\frac{-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{-1}{\sqrt{x}+3}\)
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 \(x=\frac{1}{4}\)
Trả lời:
a, \(P=\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\left(ĐK:x>0;x\ne1\right)\)
\(=\left(\frac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}+2}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\left(x+\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(x+2\sqrt{x}-\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left[\sqrt{x}\left(\sqrt{x}+2\right)-\left(\sqrt{x}+2\right)\right]\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\) (đpcm)
b, \(2P=2\sqrt{x}+5\Leftrightarrow\frac{2\left(\sqrt{x}+1\right)}{\sqrt{x}}=2\sqrt{x}+5\) \(\left(ĐK:x>0\right)\)
\(\Leftrightarrow\frac{2\sqrt{x}+2}{\sqrt{x}}=2\sqrt{x}+5\)
\(\Leftrightarrow\frac{2\sqrt{x}+2}{\sqrt{x}}=\frac{2x}{\sqrt{x}}+\frac{5\sqrt{x}}{\sqrt{x}}\)
\(\Rightarrow2\sqrt{x}+2=2x+5\sqrt{x}\)
\(\Leftrightarrow2x+3\sqrt{x}-2=0\)
\(\Leftrightarrow2x+4\sqrt{x}-\sqrt{x}-2=0\)
\(\Leftrightarrow2\sqrt{x}\left(\sqrt{x}+2\right)-\left(\sqrt{x}+2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}+2\right)\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}+2=0\\2\sqrt{x}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=-2\left(voli\right)\\2\sqrt{x}=1\end{cases}\Leftrightarrow}\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\left(tm\right)}\)
Vậy x = 1/4 là giá trị cần tìm.
b) \(B=\dfrac{x-\sqrt{x}}{1-\sqrt{x}}-\dfrac{x\sqrt{x}}{\sqrt{x}}=\dfrac{\sqrt{x}\left(x-\sqrt{x}\right)-x\sqrt{x}\left(1-\sqrt{x}\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}\) = \(\dfrac{x\sqrt{x}-x-x\sqrt{x}+x^2}{\sqrt{x}-x}=\dfrac{x^2-x}{\sqrt{x}-x}\)
c) \(C=\dfrac{x+2\sqrt{x}}{\sqrt{x}-x}-\dfrac{x\sqrt{x}}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)\left(x+2\sqrt{x}\right)-x\sqrt{x}\left(\sqrt{x}-x\right)}{\left(\sqrt{x}-x\right)\left(\sqrt{x}+1\right)}=x+2\sqrt{x}-x\sqrt{x}\)
\(d,D=\dfrac{x+2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}=\dfrac{x+2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\) \(\dfrac{\left(x+2\sqrt{x}\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{x+7\sqrt{x}-2}{\sqrt{x}+2}\)
e) \(E=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}-24}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\) = \(\dfrac{2\sqrt{x}-24}{\sqrt{x}+3}\)
F) F = \(\dfrac{3}{\sqrt{x}+5}+\dfrac{20-2\sqrt{x}}{x-25}=\dfrac{3\left(\sqrt{x}-5\right)+20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\dfrac{23-2\sqrt{x}}{\sqrt{x}+5}\)
a) P < 1; P < 2
b) P < 1 --> P(1 - P) > 0
--> P > P^2