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ĐK: 0 =< 1 # 0
a) \(\text{P}=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}1}\right).\frac{\left(1-x\right)^2}{2}\)
\(\text{P}=\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{\left(1-x\right)^2}{2}\)
\(\text{P}=\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}-1\right)^3}{2}\)
\(\text{P}=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)
\(\text{P}=-\sqrt{x}\left(1-\sqrt{x}\right)\)
b) \(\text{P}=\sqrt{x}\left(\sqrt{x}-1\right)\)
Để P > 0 thì \(\hept{\begin{cases}\sqrt{x}>0\\1-\sqrt{x}>0\end{cases}\Rightarrow0< x< 1}\)
c) \(\text{P}=-x+\sqrt{x}=-\left(x-2\sqrt{x}.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
\(\Rightarrow MAX_P=\frac{1}{4}\text{ khi }x=\frac{1}{4}\)
Answer:
a. \(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\left(\frac{1-x}{\sqrt{2}}\right)^2\) ĐK: \(x\ge0;x\ne1\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(1-x\right)^2}{2}\)
\(=\frac{-2\sqrt{x}}{\sqrt{x}+1}.\frac{x-1}{2}\)
\(=\frac{\sqrt{x}\left(1-x\right)}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\sqrt{x}\left(1-\sqrt{x}\right)\)
b. Vì \(0< x< 1\Rightarrow\hept{\begin{cases}\sqrt{x}\ge0\\1-\sqrt{x}>0\end{cases}}\Rightarrow\sqrt{x}\left(1-\sqrt{x}\right)>0\)
Do vậy \(\sqrt{x}\left(1-\sqrt{x}\right)>0\)
c. \(P=\sqrt{x}\left(1-\sqrt{x}\right)\)
\(=-\left(\sqrt{x}\right)^2+\sqrt{x}\)
\(=-\left(x-2\sqrt{x}.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}\)
\(=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x\)
Dấu "=" xảy ra khi \(\sqrt{x}-\frac{1}{2}=0\Rightarrow x=\frac{1}{4}\)
a: \(P=\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x-1-x+4}\)
\(=\dfrac{1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-2}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b: P=1/4
=>\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\)
=>\(4\left(\sqrt{x}-2\right)=3\sqrt{x}\)
=>\(4\sqrt{x}-8-3\sqrt{x}=0\)
=>\(\sqrt{x}=8\)
=>x=64
c: Khi \(x=4+2\sqrt{3}\) thì \(P=\dfrac{\sqrt{4+2\sqrt{3}}-2}{3\cdot\sqrt{4+2\sqrt{3}}}\)
\(=\dfrac{\sqrt{3}+1-2}{3\left(\sqrt{3}+1\right)}=\dfrac{\sqrt{3}-1}{3\sqrt{3}+3}=\dfrac{2-\sqrt{3}}{3}\)
a/ \(P=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\left(\frac{1-x}{\sqrt{2}}\right)^2\)
=> \(P=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right)\left(\frac{1-x}{\sqrt{2}}\right)^2\)
\(P=\left(\frac{x+\sqrt{x}-2\sqrt{x}-2-x+\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right)\left(\frac{1-x}{\sqrt{2}}\right)^2\)
\(P=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(1-\sqrt{x}\right)^2\left(1+\sqrt{x}\right)^2}{2}\)
=> \(P=-\sqrt{x}\left(\sqrt{x}-1\right)\)
b/ Nếu 0<x<1 => \(\sqrt{x}-1< 0\); và \(\sqrt{x}>0\)
=> \(P=-\sqrt{x}\left(\sqrt{x}-1\right)>0\)
c/ \(P=-\sqrt{x}\left(\sqrt{x}-1\right)=-x+\sqrt{x}=-x+2.\frac{1}{2}\sqrt{x}-\frac{1}{4}+\frac{1}{4}\)
=> \(P=\frac{1}{4}-\left(\sqrt{x}-\frac{1}{2}\right)^2\le\frac{1}{4}\)
=> \(P_{max}=\frac{1}{4}\)
Đạt được khi x=1/4
1: Khi x=9 thì \(A=\dfrac{3+1}{3-1}=\dfrac{4}{2}=2\)
2:
a: \(P=\left(\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: \(2P=2\sqrt{x}+5\)
=>\(P=\sqrt{x}+\dfrac{5}{2}\)
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\dfrac{5}{2}=\dfrac{2\sqrt{x}+5}{2}\)
=>\(\sqrt{x}\left(2\sqrt{x}+5\right)=2\sqrt{x}+2\)
=>\(2x+3\sqrt{x}-2=0\)
=>\(\left(\sqrt{x}+2\right)\left(2\sqrt{x}-1\right)=0\)
=>\(2\sqrt{x}-1=0\)
=>x=1/4
Bạn có thể làm hộ mình câu c được không?Nếu được thì mình cảm ơn bạn nhiều!
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