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a, dk \(x\ge0.x\ne1\)
\(\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{2\left(1-x\right)}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)=\(\left(\frac{1}{1-x}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)
=\(\left(\frac{1+x-x^2-1}{1-x^2}\right)\left(\frac{x+1}{x}\right)=\frac{x\left(1-x\right)\left(x+1\right)}{x\left(1-x\right)\left(1+x\right)}=1\)
phan b,c ban tu lam not nhe dai lam mk ko lam dau mk co vc ban rui
#)Giải :
Bài 1 :
a) \(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\left(\frac{1-x}{\sqrt{2}}\right)^2\)
\(=\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right]\frac{\left(1-x\right)^2}{2}\)
\(=\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x+1}\right)^2}{2}\)
\(=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)
\(=-\sqrt{x}\left(\sqrt{x}-1\right)\)
b) Để \(P>0\Rightarrow\hept{\begin{cases}\sqrt{x}>0\\1-\sqrt{x}>0\end{cases}\Rightarrow0< x< 1}\)
c) \(P=-x+\sqrt{x}=-\left(x-2\sqrt{x}.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Dấu ''='' xảy ra khi \(x=\frac{1}{4}\)
Giả sứ nếu x là số lớn nhất trong 3 chữ số. Ta sẽ lấy x, y để so sánh tạm nhé...
Từ đề bài ta có;
\(\sqrt{x+2011}+\sqrt{y+2012}+\sqrt{z+2013}=\sqrt{z+2011}+\sqrt{x+2012}+\)
\(\sqrt{y+2013}\)
\(\Leftrightarrow\sqrt{x+2012}-\sqrt{x+2011}+\sqrt{y+2013}-\sqrt{y+2012}=\sqrt{z+2012}-\)
\(\sqrt{z+2011}+\sqrt{z+2013}-\sqrt{z-2012}\)
\(\Leftrightarrow\frac{1}{\sqrt{x+2012}+\sqrt{x+2011}}+\frac{1}{\sqrt{x+2013}+\sqrt{x+2012}}=\)
\(\frac{1}{\sqrt{z+2012}+\sqrt{z+2011}}+\frac{1}{\sqrt{z+2013}+\sqrt{z+2012}}\)
Ta lại có
\(\frac{1}{\sqrt{x+2012}+\sqrt{x+2011}}\ge\frac{1}{\sqrt{z+2012}+\sqrt{z+2011}}\)
\(\frac{1}{\sqrt{y+2013}+\sqrt{y+2012}}\ge\frac{1}{\sqrt{z+2013}+\sqrt{z+2012}}\)
P/s; Mình ko chắc đâu nhé
Sửa lại đề nha , đề đúng nè :
\(\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{x}{x-1}\right):\)\(\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{x}{x+2\sqrt{x}+1}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\)\(\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{x}{\left(\sqrt{x}+1\right)^2}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)-x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}\left(\sqrt{x}+1\right)-x}{\left(\sqrt{x}+1\right)^2}\)
\(=\frac{x-\sqrt{x}-x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{x+\sqrt{x}-x}{\left(\sqrt{x}+1\right)^2}\)
\(=\frac{-\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\sqrt{x}}=-\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(\)