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ĐKXĐ : \(x\ge1\)
\(\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}=\sqrt{\left(\sqrt{x-1}+1\right)^2}-\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
\(=\sqrt{x-1}+1-\left|\sqrt{x-1}-1\right|\)
Xét các trường hợp :
1. Nếu \(1\le x\le2\)thì \(\sqrt{x-1}+1-\left|\sqrt{x-1}-1\right|=\sqrt{x-1}+1-\left(1-\sqrt{x-1}\right)=2\sqrt{x-1}\le2\)
2. Nếu \(x>2\) thì
\(\sqrt{x-1}+1-\left|\sqrt{x-1}-1\right|=\sqrt{x-1}+1-\sqrt{x-1}+1=2\)
Gộp hai trường hợp có đpcm.
a, ĐK \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(P=\frac{x-1}{\sqrt{x}}:\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
Ta thấy \(P=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}>0\forall x>0,x\ne1\)
b, P=\(\frac{x+2\sqrt{x}+1}{\sqrt{x}-1}=\frac{\frac{2}{2+\sqrt{3}}+2\sqrt{\frac{2}{2+\sqrt{3}}}+1}{\sqrt{\frac{2}{2+\sqrt{3}}}-1}\)
=\(\frac{\frac{4}{\left(\sqrt{3}+1\right)^2}+2.\sqrt{\left(\frac{2}{\left(\sqrt{3}+1\right)^2}\right)}+1}{\sqrt{\left(\frac{2}{2+\sqrt{3}}\right)^2}-1}=\frac{\frac{4}{\left(\sqrt{3}+1\right)^2}+2.\frac{2}{\sqrt{3}+1}+1}{\frac{2}{\sqrt{3}+1}-1}\)
\(=\frac{12+6\sqrt{3}}{1-3}=-6-3\sqrt{3}\)
a. rút gọn b. Tính giá trị A khi x =\(\sqrt{3+\sqrt{8}}\)
c. Tìm x=\(\sqrt{5}\)
Trả lời:
a, \(P=\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\left(ĐK:x>0;x\ne1\right)\)
\(=\left(\frac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}+2}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\left(x+\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(x+2\sqrt{x}-\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left[\sqrt{x}\left(\sqrt{x}+2\right)-\left(\sqrt{x}+2\right)\right]\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\) (đpcm)
b, \(2P=2\sqrt{x}+5\Leftrightarrow\frac{2\left(\sqrt{x}+1\right)}{\sqrt{x}}=2\sqrt{x}+5\) \(\left(ĐK:x>0\right)\)
\(\Leftrightarrow\frac{2\sqrt{x}+2}{\sqrt{x}}=2\sqrt{x}+5\)
\(\Leftrightarrow\frac{2\sqrt{x}+2}{\sqrt{x}}=\frac{2x}{\sqrt{x}}+\frac{5\sqrt{x}}{\sqrt{x}}\)
\(\Rightarrow2\sqrt{x}+2=2x+5\sqrt{x}\)
\(\Leftrightarrow2x+3\sqrt{x}-2=0\)
\(\Leftrightarrow2x+4\sqrt{x}-\sqrt{x}-2=0\)
\(\Leftrightarrow2\sqrt{x}\left(\sqrt{x}+2\right)-\left(\sqrt{x}+2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}+2\right)\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}+2=0\\2\sqrt{x}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=-2\left(voli\right)\\2\sqrt{x}=1\end{cases}\Leftrightarrow}\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\left(tm\right)}\)
Vậy x = 1/4 là giá trị cần tìm.
a) ta có : \(P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\left(\dfrac{1-x}{\sqrt{2}}\right)^2\)
\(\Leftrightarrow P=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\left(\dfrac{1-x}{\sqrt{2}}\right)^2\)
\(\Leftrightarrow P=\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\left(\dfrac{1-x}{\sqrt{2}}\right)^2\)
\(\Leftrightarrow P=\left(\dfrac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\left(\dfrac{1-x}{\sqrt{2}}\right)^2\) \(\Leftrightarrow P=\left(\dfrac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\) \(\Leftrightarrow P=\sqrt{x}-x\)b) ta có : \(x< 1\Leftrightarrow x-1< 0\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)< 0\)
\(\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow x-\sqrt{x}< 0\Leftrightarrow\sqrt{x}-x>0\)
\(\Leftrightarrow P>0\left(đpcm\right)\)
\(A=\sqrt{x-2\sqrt{x}-1}+\sqrt{x-2\sqrt{x}-1}vớix\ge2\)
\(=\sqrt{x-2\sqrt{x}+1-2}+\sqrt{x-2\sqrt{x}+1-2}\)
\(=\sqrt{\left(\sqrt{x}-1\right)^2-2}+\sqrt{\left(\sqrt{x}-1\right)^2-2}\)
\(=\sqrt{x-1}-2+\sqrt{x-1}-2\) (Do \(x\ge2\Rightarrow\)x dương)
\(=2\sqrt{x-1}-4\)
K biết đúng hay sai nữa,sai thì t xin lỗi nha
Xin người.
:)