Cho biểu thức A = \(\left\{\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+2}{x\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right\}:\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
1. Rút gọn biểu thức
2. Chứng minh rằng 0<A<2
GIÚP MÌNH VỚI Ạ !
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3 tháng 4 2017
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 :)
NV
13 tháng 7 2016
ĐKXĐ: \(x\ge4\)
a/ \(A=\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left[\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right]\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left(\frac{x-4-x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(-3\right)}\)
\(=\frac{\sqrt{x}-2}{-3\sqrt{x}}\)
b/ A = 0 \(\Rightarrow\frac{\sqrt{x}-2}{-3\sqrt{x}}=0\Rightarrow\sqrt{x}-2=0\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
MT
17 tháng 8 2015
1)))))))
\(\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{2}{\sqrt{ab}}:\frac{\left(\sqrt{b}-\sqrt{a}\right)^2}{\left(\sqrt{ab}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{2}{\sqrt{ab}}.\frac{\left(\sqrt{ab}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{2\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{2\sqrt{ab}-a-b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{-\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}=-1\)
MT
17 tháng 8 2015
\(\text{VT}=\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)=\left(1+\frac{\sqrt{x}.\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\left(1-\frac{\sqrt{x}.\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)
\(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=\text{VP(điều phải chứng minh)}\)
26 tháng 6 2019
\(đkxđ\Leftrightarrow\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(a,A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{1+\sqrt{x}}+\frac{2}{x-1}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x\left(\sqrt{x}-1\right)}\right):\left(\frac{1-\sqrt{x}}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}-\frac{2}{1-x}\right)\)
\(=\left(\frac{x.\sqrt{x}}{x.\left(\sqrt{x}-1\right)}-\frac{1}{x\left(\sqrt{x}-1\right)}\right):\left(\frac{1-\sqrt{x}}{1-x}-\frac{2}{1-x}\right)\)
\(=\frac{x.\sqrt{x}-1}{x\left(\sqrt{x}-1\right)}.\frac{1-x}{-\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(x.\sqrt{x}-1\right)\left(1-x\right)}{x\left(1-x\right)}=\frac{\sqrt{x^3}-1}{x}\)
\(b,\)\(A=\frac{\sqrt{x}^3-1}{x}=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x}\)
Để A > 0 \(\Rightarrow\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x}>0\)
Mà \(x>0\)và \(x+\sqrt{x}+1>0\)( do x lớn hơn 0 )
\(\Rightarrow\sqrt{x}-1>0\)
\(\Rightarrow\sqrt{x}>1\Leftrightarrow\sqrt{x}>\sqrt{1}\Leftrightarrow x>1\)
YN
5 tháng 2 2022
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 \(\sqrt{x}-\frac{1}{2}=0\Rightarrow x=\frac{1}{4}\)