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\(P=\frac{1}{x^2-\sqrt{x}}:\frac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}\)\(=\frac{1}{\sqrt{x}\left(\sqrt{x^3}-1^3\right)}:\frac{\sqrt{x}+1}{\sqrt{x}\left(x+\sqrt{x}+1\right)}\)
\(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\sqrt{x}+1}\)\(\frac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{1}{x-1}\)
2) De 3P=1+x
=> \(3\left(\frac{1}{x-1}\right)=x+1\)<=>\(\frac{3}{x-1}=x+1\)<=>\(\left(x+1\right)\left(x-1\right)=3\)
<=> \(x^2-1=3\)<=> \(x^2=4\)<=> \(x=2\)hoac \(x=-2\)
Vay voi x =2 va x=-2 ta co 3P=x+1
a) Ta có: \(A=\frac{x+2\sqrt{x}+1}{\sqrt{x}+1}.\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\frac{x+2\sqrt{x}+1}{\sqrt{x}+1}.\frac{\sqrt{2x}-x-1}{\sqrt{x}-1}\)
\(=\frac{x+2\sqrt{x}+1}{\sqrt{x}+1}.\frac{1-2\sqrt{x}+x}{1-\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}.\frac{\left(1-\sqrt{x}\right)^2}{1-\sqrt{x}}\)
\(=\left(\sqrt{x}+1\right)\left(1-\sqrt{x}\right)\)
\(=1^2-\left(\sqrt{x}\right)^2=1-x\).
Vậy \(A=1-x\).
b) Ta có: \(A=1-x\)
Để \(A>0\)\(\Rightarrow1-x>0\Rightarrow1-0>x\Rightarrow1>x\Rightarrow x< 1.\)
Vậy để A > 0 thì x < 1.
Chúc bn hc tốt!
a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\left(\frac{1-x}{\sqrt{2}}\right)^2\)
\(=\left[\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right].\frac{\left(1-x\right)^2}{2}\)
\(=\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{\left(x-1\right)^2}{2}\)
\(=\left[\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{x+\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{\left(\sqrt{x}-1\right)^2.\left(\sqrt{x}+1\right)^2}{2}\)
\(=\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2.\left(\sqrt{x}+1\right)^2}{2}\)
\(=\frac{-2\sqrt{x}.\left(\sqrt{x}-1\right)}{2}=-\sqrt{x}\left(\sqrt{x}-1\right)=-x+\sqrt{x}\)
b) Với \(0< x< 1\)\(\Rightarrow0< \sqrt{x}< 1\)
\(\Rightarrow\sqrt{x}-1< 0\)
mà \(\sqrt{x}>0\)\(\Rightarrow\sqrt{x}.\left(\sqrt{x}-1\right)< 0\)
\(\Rightarrow-\sqrt{x}.\left(\sqrt{x}-1\right)>0\)\(\Rightarrow P>0\)( đpcm )
c) \(P=-x+\sqrt{x}=-x+\sqrt{x}-\frac{1}{4}+\frac{1}{4}\)
\(=-\left(x-\sqrt{x}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\)
Vì \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\)\(\Rightarrow-\left(\sqrt{x}-\frac{1}{2}\right)^2\le0\)
\(\Rightarrow-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Dấu " = " xảy ra \(\Leftrightarrow\sqrt{x}-\frac{1}{2}=0\)\(\Leftrightarrow\sqrt{x}=\frac{1}{2}\)\(\Leftrightarrow x=\frac{1}{4}\)( thỏa mãn ĐKXĐ )
Vậy \(maxP=\frac{1}{4}\)\(\Leftrightarrow x=\frac{1}{4}\)
ĐKXĐ \(\hept{\begin{cases}x\ne1\\x\ge0\end{cases}}\)
a, Ta có \(P=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\left(\frac{\left(1-\sqrt{x}\right).\left(1+\sqrt{x}\right)}{\sqrt{2}}\right)^2\)
\(P=\left(\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\left(\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{\sqrt{2}}\right)^2\)
\(P=\left(\frac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\left(\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{\sqrt{2}}\right)^2\)
\(P=\frac{2\sqrt{x}-2x}{\sqrt{2}}\)
\(P=\sqrt{2x}-\sqrt{2}x\)
\(P=\sqrt{2x}\left(1-\sqrt{x}\right)\)
b, Vì \(0< x< 1\Rightarrow\sqrt{x}< 1\Rightarrow1-\sqrt{x}< 1\)
\(\Rightarrow\sqrt{2x}\left(1-\sqrt{x}\right)>0\)
c, Ta có \(P=-\sqrt{2}\left(x-\sqrt{x}\right)\)
\(P=-\sqrt{2}\left(x-\frac{1}{2}.2.\sqrt{x}+\frac{1}{4}-\frac{1}{4}\right)\)
\(P=-\sqrt{2x}\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{\sqrt{8}}\le\frac{1}{\sqrt{8}}\)
Dấu = xảy ra \(\Leftrightarrow\)\(\sqrt{x}-\frac{1}{2}=0\)
\(\Rightarrow x=\frac{1}{4}\)
vậy GTLN của P là \(\frac{1}{\sqrt{8}}\)với x=\(\frac{1}{4}\)
#)Giải :
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right)\div\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)
\(P=\left(\frac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\div\left(\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\div\frac{1}{\sqrt{x}-1}=\frac{x-1}{\sqrt{x}}\)
Thấy : \(\sqrt{x}\ge0\)
\(\Rightarrow P=\dfrac{\sqrt{x}+2}{2\sqrt{x}+1}>0\)
\(\Rightarrow\left|P\right|=P\)
Ta có : \(\left|P\right|=P\ge P\)
=> P = P .
Vậy \(\forall x>0\) TMYC đè bài
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