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ĐK:x\(\ge2\)\(\sqrt{x-1+2\sqrt{x-2}}-\sqrt{x-1-2\sqrt{x-2}}=1\Leftrightarrow\sqrt{x-2+2\sqrt{x-2}+1}-\sqrt{x-2-2\sqrt{x}-2+1}=1\Leftrightarrow\sqrt{\left(\sqrt{x-2}+1\right)^2}-\sqrt{\left(\sqrt{x-2}-1\right)^2}=1\Leftrightarrow\left|\sqrt{x-2}+1\right|-\left|\sqrt{x-2}-1\right|=1\Leftrightarrow\sqrt{x-2}+1-\left|\sqrt{x-2}-1\right|=1\)(1)
TH1: nếu \(\sqrt{x-2}< 1\Leftrightarrow x-2< 1\Leftrightarrow x< 3\) và x>2 thì
(1)⇔\(\sqrt{x-2}+1-1+\sqrt{x-2}=1\Leftrightarrow2\sqrt{x-2}=1\Leftrightarrow\sqrt{x-2}=\dfrac{1}{2}\Leftrightarrow x-2=\dfrac{1}{4}\Leftrightarrow x=\dfrac{9}{4}\left(tm\right)\)TH2: nếu \(\sqrt{x-2}\ge1\Leftrightarrow x\ge3\) thì
(1)\(\Leftrightarrow\sqrt{x-2}+1-\sqrt{x-2}+1=1\Leftrightarrow2=1\left(ktm\right)\)
Vậy S={\(\dfrac{9}{4}\)}
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a/ ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1}=\sqrt{5x-1}+\sqrt{3x-2}\)
\(\Leftrightarrow x-1=8x-3+2\sqrt{\left(5x-1\right)\left(3x-2\right)}\)
\(\Leftrightarrow2-7x=2\sqrt{\left(5x-1\right)\left(3x-2\right)}\)
Do \(x\ge1\Rightarrow2-7x< 0\Rightarrow\left\{{}\begin{matrix}VP\ge0\\VT< 0\end{matrix}\right.\)
Phương trình vô nghiệm
b/ ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|+\left|1-\sqrt{x-1}\right|=2\)
Mà \(\left|\sqrt{x-1}+1\right|+\left|1-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}+1+1-\sqrt{x-1}\right|=2\)
Dấu "=" xảy ra khi và chỉ khi \(1-\sqrt{x-1}\ge0\Rightarrow x\le2\Rightarrow1\le x\le2\)
Vậy nghiệm của pt là \(1\le x\le2\)
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a) \(\sqrt{x+3}-\sqrt{x-1}=\sqrt{2x+2}\)
Điều kiện: \(\hept{\begin{cases}x+3\ge0\\x-1\ge0\\2x+2\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge-3\\x\ge1\\x\ge-1\end{cases}\Leftrightarrow x\ge1}\)
\(\Leftrightarrow\left(\sqrt{x+3}-\sqrt{x-1}\right)^2=\left(\sqrt{2x+2}\right)^2\)
\(\Leftrightarrow x+3-2\sqrt{\left(x+3\right)\left(x-1\right)}+x-1=2x+2\)
\(\Leftrightarrow2x+2-2\sqrt{\left(x+3\right)\left(x-1\right)}=2x+2\)
\(\Leftrightarrow-2\sqrt{\left(x+3\right)\left(x-1\right)}=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\left(l\right)\\x=1\left(n\right)\end{cases}}\)
Vậy \(S=\left\{1\right\}\)
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Điều kiện xác định : \(\hept{\begin{cases}2\ge\frac{1}{\sqrt{2-x}}\\x< 2\\x\ge0\end{cases}}\) \(\Leftrightarrow0\le x\le\frac{7}{4}\)
Ta có : \(\sqrt{2-\frac{1}{\sqrt{2-x}}}=x\)
\(\Rightarrow2-\frac{1}{\sqrt{2-x}}=x^2\)
\(\Leftrightarrow x^2\sqrt{2-x}-2\sqrt{2-x}+1=0\)
Đặt \(t=\sqrt{2-x},t\ge0\Rightarrow x=2-t^2\)
Ta có : \(\left(2-t^2\right)^2.t-2t+1=0\)
\(\Leftrightarrow t\left[\left(2-t^2\right)^2-1\right]-\left(t-1\right)=0\)
\(\Leftrightarrow t\left(2-t^2-1\right)\left(2-t^2+1\right)-\left(t-1\right)=0\)
\(\Leftrightarrow t\left(t-1\right)\left(t+1\right)\left(t^2-3\right)-\left(t-1\right)=0\)
\(\Leftrightarrow\left(t-1\right)\left[t\left(t+1\right)\left(t^2-3\right)-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}t-1=0\\t\left(t+1\right)\left(t^2-3\right)-1=0\end{cases}}\)
- Nếu t - 1 = 0 => t = 1 ta có \(x=2-1^2=1\)(tmđk)
- Nếu \(t\left(t+1\right)\left(t^2-3\right)-1=0\) , từ điều kiện \(0\le x\le\frac{7}{4}\)ta có \(t\left(t+1\right)\left(t^2-3\right)-1\le-\frac{179}{256}< 0\)=> pt này vô nghiệm.
Vậy pt có nghiệm x = 1
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