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\(a,PT\Leftrightarrow\sqrt{x-1-2\sqrt{x-1}+1}=3\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-1\right)^2}=3\)
\(\Leftrightarrow\sqrt{x-1}=4\Leftrightarrow x-1=16\Leftrightarrow x=17\)
Vậy............................................
\(b,PT\Leftrightarrow\sqrt{\left(x^2-1\right)^2}=x-1\)
\(\Leftrightarrow x^2-1=x-1\Leftrightarrow x^2=x\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy...............................................
a) đkxđ: \(\begin{cases}\sqrt{x^2-4}\ge0\\\sqrt{x^2}+4x+4\ge0\end{cases}\) \(\Leftrightarrow\begin{cases}\begin{cases}x-2\ge0\\x+2\ge0\end{cases}\\x+2\ge0\end{cases}\) \(\Leftrightarrow\begin{cases}x\ge2\\x\le-2\end{cases}\) \(\Leftrightarrow-2\ge x\ge2\)
\(\sqrt{x^2-4}+\sqrt{x^2+4x+4}=0\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x+2\right)}+\sqrt{\left(x+2\right)^2}=0\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x+2\right)}=x+2\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=\left(x+2\right)^2\)
\(\Leftrightarrow\left(x+2\right)\left(x-2-x+2\right)=0\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
S={-2}
b) đkxđ: \(\begin{cases}\sqrt{1-x^2}\ge0\\\sqrt{x+1}\ge0\end{cases}\) \(\Leftrightarrow\begin{cases}1-x^2\ge0\\x+1\ge0\end{cases}\) \(\Leftrightarrow\begin{cases}x^2\le1\\x\ge-1\end{cases}\) \(\Leftrightarrow\begin{cases}\begin{cases}x\le1\\x\ge-1\end{cases}\\x\ge-1\end{cases}\) \(\Leftrightarrow-1\le x\le1\)
\(\sqrt{1-x^2}+\sqrt{x+1}=0\)
\(\Leftrightarrow\sqrt{1-x^2}=-\sqrt{x+1}\)
\(\Leftrightarrow1-x^2=x+1\)
\(\Leftrightarrow-x-x^2=0\)
\(\Leftrightarrow-x\left(1+x\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}-x=0\\1+x=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\left(N\right)\\x=-1\left(N\right)\end{array}\right.\)
S={-1;0}
a, \(x-3\sqrt{x}+2=0\Leftrightarrow x-2\sqrt{x}-\sqrt{x}+2=0\)đk : x >= 0
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)=0\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\Leftrightarrow x=1;x=4\)
b, \(\sqrt{x^2-1}-\sqrt{x+1}=0\Leftrightarrow\sqrt{\left(x-1\right)\left(x+1\right)}-\sqrt{x+1}=0\)đk : \(x\ge1\)
\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{x-1}-1\right)=0\)
TH1 : \(x=-1\)( loại )
TH2 : \(\sqrt{x-1}=1\Leftrightarrow x-1=1\Leftrightarrow x=2\)
c, \(x^2+4x+4-\sqrt{2x+1}-\left(x-1\right)^2=0\)đk : x>= -1/2
\(\Leftrightarrow\left(x+2\right)^2-\left(x-1\right)^2-\sqrt{2x+1}=0\)
\(\Leftrightarrow3\left(2x+1\right)-\sqrt{2x+1}=0\Leftrightarrow\sqrt{2x+1}\left(3\sqrt{2x+1}-1\right)=0\)
TH1 : \(x=-\frac{1}{2}\)
TH2 : \(\sqrt{2x+1}=\frac{1}{3}\Leftrightarrow2x+1=\frac{1}{9}\Leftrightarrow x=\frac{\frac{1}{9}-1}{2}=\frac{-\frac{8}{9}}{2}=-\frac{4}{9}\)
a) ĐK : x \(\ge0\)
\(x-3\sqrt{x}+2=0\)
<=> \(\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
<=> \(\orbr{\begin{cases}\sqrt{x}-1=0\\\sqrt{x}-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=4\end{cases}}\)(tm)
b) ĐK \(\hept{\begin{cases}x\ge-1\\x\notin\left\{x\in R|-1< x< 0\right\}\end{cases}}\)
\(\sqrt{x^2-1}-\sqrt{x+1}=0\)
<=> \(\sqrt{x-1}\sqrt{x+1}-\sqrt{x+1}=0\)
<=> \(\sqrt{x-1}\left(\sqrt{x+1}-1\right)=0\)
<=> \(\orbr{\begin{cases}\sqrt{x+1}=0\\\sqrt{x-1}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-1=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)(tm)
c) ĐK : \(x\ge-\frac{1}{2}\)
\(x^2+4x+4-\sqrt{2x+1}-\left(x-1\right)^2=0\)
<=> \(6x+3-\sqrt{2x+1}=0\)
<=> \(\sqrt{2x+1}\left(3\sqrt{2x+1}-1\right)=0\)
<=> \(\orbr{\begin{cases}\sqrt{2x+1}=0\\3\sqrt{2x+1}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=-\frac{4}{9}\end{cases}}\)(tm)
a)\(\sqrt{4x+20}\) +\(\sqrt{x-5}\) -\(\dfrac{1}{3}\)\(\sqrt{9x-45}\)=4 ; ĐKXĐ : x ≥_+ 5
⇔ \(\sqrt{2^2x+2^2.5}\) +\(\sqrt{x-5}\) -\(\dfrac{1}{3}\)\(\sqrt{3^2x-3^2.5}\) =4
⇔ 2\(\sqrt{x+5}\) +\(\sqrt{x-5}\) -\(\dfrac{1}{3}\)3\(\sqrt{x-5}\) =4 ⇔ 2\(\sqrt{x+5}\) +\(\sqrt{x-5}\) -\(\sqrt{x-5}\) =4⇔2\(\sqrt{x+5}\)=4(tm)
⇔\(\sqrt{x+5}\)=2⇔x+5=4 ⇔x=-1
Vậy x=-1
b) \(\sqrt{x^2-36}\) - \(\sqrt{x-6}\) =0 ; ĐKXĐ: x≥_+6
⇔ \(\sqrt{\left(x-6\right)\left(x+6\right)}\) - \(\sqrt{x-6}\) =0 ⇔ \(\sqrt{x-6}\).\(\sqrt{x+6}\) - \(\sqrt{x-6}\) =0
⇔ \(\sqrt{x-6}\)(\(\sqrt{x+6}\) -1 )=0 ⇔\([\) \(\begin{matrix}\sqrt{x-6}&=0\\\sqrt{x+6}-1&=0\end{matrix}\) ⇔ \([\) \(\begin{matrix}x-6&=0\\x+6-1&=0\end{matrix}\) ⇔\([\) \(\begin{matrix}x&=6\left(ktm\right)\\x&=-5\left(tm\right)\end{matrix}\)
Vậy x=-5
c) \(\sqrt{4-x^2}\) -x +2 =0 ; ĐKXĐ: -2≤x≤2
⇔ \(\sqrt{\left(2-x\right)\left(2+x\right)}\) -x+2 =0 ⇔ \(\sqrt{\left(2-x\right)\left(2+x\right)}\) -(x-2)=0
⇔ \(\sqrt{\left(2-x\right)\left(2+x\right)}\) =(x-2) ⇔ (2-x)(2+x)=(x-2)2 ⇔ 4-x2 = x2-4x+4 ⇔ -x2-x2+4x=4-4
⇔-2x2+4x=0 ⇔ -2x(x-2)=0 ⇔ \([\) \(\begin{matrix}-2x&=0\\x-2&=0\end{matrix}\) ⇔\([\) \(\begin{matrix}x&=0\left(tm\right)\\x&=2\left(tm\right)\end{matrix}\)
Vậy S=\(\left\{0;2\right\}\)
d) \(\sqrt{\left(2x-3\right)\left(x-1\right)}-\sqrt{x-1}=0\) ; ĐKXĐ: x≥\(\dfrac{3}{2}\);x ≥ 1
⇔\(\sqrt{2x-3}.\sqrt{x-1}-\sqrt{x-1}=0\) ⇔ \(\sqrt{x-1}.\left(\sqrt{2x-3}-1\right)=0\)
⇔ \(\left[{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{2x-3}-1=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x-1=0\\2x-3-1=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=1\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
Vậy s=\(\left\{1:2\right\}\)
a/\(\sqrt{\left(x-2\right)^2}+\sqrt{\left(x+2\right)^2}=0\Leftrightarrow x-2+x+2=0\Rightarrow x=0\)
\(x^2-4=\left(x-2\right)^2\) à chắc bn thông minh lắm mới sáng chế bđt mới đc đó