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b) Với \(x\ge0;x\ne1\)Thì M=\(-\sqrt{x}\left(\sqrt{x}-1\right)\)
Ta có : 0<x<1 ( Thỏa mãn ĐKXĐ )
=> \(0< \sqrt{x}< 1\)
=>\(-1< \sqrt{x}-1< 0\)
=>\(\sqrt{x}>-\sqrt{x}\left(\sqrt{x}-1\right)>0\)
Vậy với 0<x<1 Thì M>0
ĐKXĐ:
\(\left\{{}\begin{matrix}2-\sqrt{x}\\2+\sqrt{x}\\x-4\end{matrix}\right.\ne0\Leftrightarrow x\ne4\)
P=\(\dfrac{\left(2+\sqrt{x}\right)\left(2+\sqrt{x}\right)-\left(2-\sqrt{x}\right)\left(2-\sqrt{x}\right)+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
P=\(\dfrac{\left(4+4\sqrt{x}+x\right)-\left(4-4\sqrt{x}+x\right)+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
P=\(\dfrac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
P=\(\dfrac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
P=\(\dfrac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ P=\dfrac{4\sqrt{x}}{2-\sqrt{x}}\)
b) Th P>0
<=> \(\dfrac{4\sqrt{x}}{2-\sqrt{x}}\)>0<=>\(4\sqrt{x}\)>0 <=> x>0(x\(\ne\)4)
TH P < 0
<=>\(\dfrac{4\sqrt{x}}{2-\sqrt{x}}\)<0 <=>\(4\sqrt{x}\)<0<=> \(\sqrt{x}< 0\)(vô lý)
c) |P|=1
=>P=1 hoặc P=-1
TH P=1
=>\(\dfrac{4\sqrt{x}}{2-\sqrt{x}}\)=1 <=> \(4\sqrt{x}\)=\(2-\sqrt{x}\) <=> x=\(\dfrac{4}{25}\)
TH P= -1
=>\(\dfrac{4\sqrt{x}}{2-\sqrt{x}}\)=-1<=> \(4\sqrt{x}\)=\(\sqrt{x}-2\)<=> \(\sqrt{x}=-\dfrac{2}{3}\)(vô lý)
d/ Ta có:
\(A=\left(-x+\sqrt{x}-\dfrac{1}{4}\right)+\dfrac{1}{4}\)
\(=\dfrac{1}{4}-\left(\sqrt{x}-\dfrac{1}{2}\right)^2\le\dfrac{1}{4}\)
Vậy GTLN là \(A=\dfrac{1}{4}\) đạt được tại \(x=\dfrac{1}{4}\)
b/ \(\sqrt{1x}-x\)
c/ Ta có:
x < 1
\(\Rightarrow\sqrt{x}< 1\)
\(\Rightarrow1-\sqrt{x}>0\)
Ta lại có: x > 0
\(\Rightarrow A=\sqrt{x}-x=\sqrt{x}\left(1-\sqrt{x}\right)>0\)
a: ĐKXĐ: x>0; x<>1
\(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\dfrac{1}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Để P<0 thì \(\sqrt{x}-1< 0\)
=>0<x<1
c: Để P là số nguyên thì \(\sqrt{x}-1+2⋮\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}-1\in\left\{1;-1;2\right\}\)
hay \(x\in\left\{4;0;9\right\}\)
điều kiện xác định : \(x\ge0;x\ne1\)
a) ta có : \(G=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{x^2-2x+1}{2}\)
\(\Leftrightarrow G=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(x-1\right)^2}{2}\)
\(\Leftrightarrow G=\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\dfrac{\left(x-1\right)^2}{2}\) \(\Leftrightarrow G=\left(\dfrac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\dfrac{\left(x-1\right)^2}{2}=\sqrt{x}-x\)
b) thay \(x=0,16\) vào \(G\) ta có : \(G=\sqrt{0,16}-0,16=0,24\)
c) ta có : \(G=-\left(x-\sqrt{x}+\dfrac{1}{4}\right)-\dfrac{1}{4}=-\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge\dfrac{-1}{4}\)
\(\Rightarrow G_{max}=\dfrac{-1}{4}\) khi \(\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\)
d) ta có : \(G=\sqrt{x}-x\) \(\Rightarrow\) để \(G\in Z\) \(\Rightarrow x=a^2\ne1\)
e) ta có : \(G>0\Leftrightarrow\sqrt{x}-x>0\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>0\\1-\sqrt{x}>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 0\\1-\sqrt{x}< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0< x< 1\\x\in\varnothing\end{matrix}\right.\) \(\Rightarrow\left(đpcm\right)\)
f) để \(G< 0\Leftrightarrow\sqrt{x}-x< 0\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>0\\1-\sqrt{x}< 0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 0\\1-\sqrt{x}>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>1\\x\in\varnothing\end{matrix}\right.\) vậy \(x>1\)
bạn có thể làm chi tiết dòng thứ tư phần rút gọn đc ko ?
a: \(P=\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)-8x}{x-4}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4x-8\sqrt{x}-8x}{x-4}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}\)
\(=\dfrac{-4x-8\sqrt{x}}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}}{-\sqrt{x}+3}\)
\(=\dfrac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\cdot\dfrac{-\sqrt{x}}{\sqrt{x}-3}=\dfrac{4x}{\sqrt{x}-3}\)
b để P=-1 thì \(\dfrac{4x}{\sqrt{x}-3}=-1\)
\(\Leftrightarrow4x=-\sqrt{x}+3\)
\(\Leftrightarrow4x+\sqrt{x}-3=0\)
\(\Leftrightarrow4x+4\sqrt{x}-3\sqrt{x}-3=0\)
\(\Leftrightarrow4\sqrt{x}-3=0\)
hay x=9/16
c: Để P<0 thì \(\sqrt{x}-3< 0\)
hay 0<x<9
a. ĐKXĐ: x\(\ne1\) x, \(\ne-1\)
b. \(\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{x^2-2x+1}{2}\)
=\(\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}\)
=\(\left(\dfrac{\left(\sqrt{x}-2\right).\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)^2}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}\)
\(\left(\dfrac{\left(\sqrt{x}-2\right).\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}=\left(\dfrac{\left(\sqrt{x}-2\right)-\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}=\left(\dfrac{\sqrt{x}-2-\sqrt{x}-2}{\sqrt{x}+1}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}=\dfrac{4}{\sqrt{x}-1}.\dfrac{\sqrt{x}-1^2}{2}=2\left(\sqrt{x}-1\right)=2\sqrt{x}-2\)
c. khi x=0,16 thì G=\(2\sqrt{x}-2=2\sqrt{0,16}-2=2.0,4-2=0,8-2=-1,2\)
Bài 1:
\(a,E=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(b,E>0\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}>0\)
Mà: \(\sqrt{x}>0\\ \Rightarrow\sqrt{x}-1>0\\ \Leftrightarrow\sqrt{x}>1\\ \Leftrightarrow x>1\)
Bài 2:
\(a,G=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{1}{1-\sqrt{x}}-\dfrac{2\sqrt{x}}{x-1}\right)\left(\sqrt{x}+1\right)\\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\left(\sqrt{x}+1\right)\\ =\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}+1-2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\left(\sqrt{x}+1\right)\\ =\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\left(\sqrt{x}+1\right)\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\\ =\sqrt{x}-1\)
câu này đâu khó bn,suy nghĩ kỉ lm là đc mak
hỏi rk mà cx hỏi!