ĐKXĐ: x>=1

\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{1}{2}\left(x+3\right)\)

\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=\dfrac{1}{2}\left(x+3\right)\)

=>\(\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\dfrac{1}{2}\left(x+3\right)\)

=>\(\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=\dfrac{1}{2}\left(x+3\right)\)

TH1: \(x>=2\)

PT sẽ tương đương với \(\sqrt{x-1}+1+\sqrt{x-1}-1=\dfrac{1}{2}\left(x+3\right)\)

=>\(2\sqrt{x-1}=\dfrac{1}{2}\left(x+3\right)\)

=>\(4\sqrt{x-1}=x+3\)

=>\(\sqrt{16x-16}=x+3\)

=>x>=-3 và (x+3)^2=16x-16

=>x>=-3 và x^2+6x+9-16x+16=0

=>x>=-3 và x^2-7x+25=0

=>Loại

TH2: 1<=x<2

PT sẽ là \(\sqrt{x-1}+1+1-\sqrt{x-1}=\dfrac{1}{2}\left(x+3\right)\)

=>1/2(x+3)=2

=>x+3=4

=>x=1(nhận)

1) \(\sqrt{x^2+1}=\sqrt{5}\)

\(\Leftrightarrow x^2+1=5\)

\(\Leftrightarrow x^2=5-1\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x^2=2^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

2) \(\sqrt{2x-1}=\sqrt{3}\) (ĐK: \(x\ge\dfrac{1}{2}\)) 

\(\Leftrightarrow2x-1=3\)

\(\Leftrightarrow2x=3+1\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=\dfrac{4}{2}\)

\(\Leftrightarrow x=2\left(tm\right)\)

3) \(\sqrt{43-x}=x-1\) (ĐK: \(x\le43\))

\(\Leftrightarrow43-x=\left(x-1\right)^2\)

\(\Leftrightarrow x^2-2x+1=43-x\)

\(\Leftrightarrow x^2-x-42=0\)

\(\Leftrightarrow\left(x-7\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(tm\right)\\x=-6\left(tm\right)\end{matrix}\right.\)

4) \(x-\sqrt{4x-3}=2\) (ĐK: \(x\ge\dfrac{3}{4}\))

\(\Leftrightarrow\sqrt{4x-3}=x-2\)

\(\Leftrightarrow4x-3=\left(x-2\right)^2\)

\(\Leftrightarrow x^2-4x+4=4x-3\)

\(\Leftrightarrow x^2-8x+7=0\)

\(\Leftrightarrow\left(x-7\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

5) \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{1}{2}\) (ĐK: \(x\ge0\))

\(\Leftrightarrow\sqrt{x}+3=2\sqrt{x}+2\)

\(\Leftrightarrow2\sqrt{x}-\sqrt{x}=3-2\)

\(\Leftrightarrow\sqrt{x}=1\)

\(\Leftrightarrow x=1^2\)

\(\Leftrightarrow x=1\left(tm\right)\)

1)

\(\sqrt{x^2+1}=\sqrt{5}\\ \Leftrightarrow x^2+1=5\\ \Leftrightarrow x^2=5-1=4\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy PT có nghiệm `x=2` hoặc `x=-2`

2)

ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\sqrt{2x-1}=\sqrt{3}\\ \Leftrightarrow2x-1=3\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\left(tm\right)\)

Vậy PT có nghiệm `x=2`

3)

\(ĐKXĐ:x\le43\)

PT trở thành:

\(43-x=\left(x-1\right)^2=x^2-2x+1\\ \Leftrightarrow43-x-x^2+2x-1=0\\ \Leftrightarrow-x^2+x+42=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\left(tm\right)\\x=7\left(tm\right)\end{matrix}\right.\)

Vậy PT có nghiệm `x=-6` hoặc `x=7`

4)

ĐKXĐ: \(x\ge\dfrac{3}{4}\)

PT trở thành:

\(\sqrt{4x-3}=x-2\\ \Leftrightarrow4x-3=\left(x-2\right)^2=x^2-4x+4\\ \Leftrightarrow4x-3-x^2+4x-4=0\\ \Leftrightarrow-x^2+8x-7=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=7\left(tm\right)\end{matrix}\right.\)

Vậy PT có nghiệm \(x=1\) hoặc \(x=7\)

5) 

ĐKXĐ: \(x\ge0\)

PT trở thành:

\(\sqrt{x+3}=2\sqrt{x}+2\\ \Leftrightarrow x+3=\left(2\sqrt{x}+2\right)^2=4x+8\sqrt{x}+4\\ \Leftrightarrow x+3-4x-8\sqrt{x}-4=0\\ \Leftrightarrow-3x-8\sqrt{x}-1=0\left(1\right)\)

Đặt \(\sqrt{x}=t\left(t\ge0\right)\)

Khi đó:

(1)\(\Leftrightarrow3t^2+8t+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-4+\sqrt{13}}{3}\left(loại\right)\\t=\dfrac{-4-\sqrt{13}}{3}\left(loại\right)\end{matrix}\right.\)

Vậy PT vô nghiệm.

Bài 1:

$\sqrt{x^2+1}=\sqrt{5}$

$\Leftrightarrow x^2+1=5$

$\Leftrightarrow x^2-4=0$

$\Leftrightarrow (x-2)(x+2)=0$

$\Leftrightarrow x-2=0$ hoặc $x+2=0$

$\Leftrightarrow x=\pm 2$ (đều tm)

2. ĐKXĐ: $x\geq \frac{1}{2}$

PT $\Leftrightarrow 2x-1=3$

$\Leftrightarrow 2x=4$

$\Leftrightarrow x=2$ (tm) 

3. ĐKXĐ: $x\leq 43$

PT \(\Rightarrow \left\{\begin{matrix} x-1\geq 0\\ 43-x=(x-1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ x^2-x-42=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ (x+6)(x-7)=0\end{matrix}\right.\)

$\Rightarrow x=7$ (tm) 

ĐKXĐ: \(x\ge1\)

Ta có:

\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+3}{2}\\ \Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\dfrac{x+3}{2}\\ \Leftrightarrow\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=\dfrac{x+3}{2}\\ \Leftrightarrow\sqrt{x-1}+\left|\sqrt{x-1}-1\right|=\dfrac{x+1}{2}\left(1\right)\)

Ta xét 2 trường hợp sau:

TH1: \(x\ge2\)

Khi đó:

\(\left(1\right)\Leftrightarrow2\sqrt{x-1}-1=\dfrac{x+1}{2}\\ \Leftrightarrow2\sqrt{x-1}=\dfrac{x+3}{2}\\ \Leftrightarrow16\left(x-1\right)=x^2+6x+9\\ \Leftrightarrow x^2-10x+25=0\\ \Leftrightarrow\left(x-5\right)^2=0\\ \Leftrightarrow x=5\left(TMĐK\right)\)

TH2: \(1\le x< 2\)

Khi đó:

\(\left(1\right)\Leftrightarrow1=\dfrac{x+1}{2}\Leftrightarrow x=1\left(TMĐK\right)\)

Vậy x=1 hoặc x=5

\(1,\sqrt{x+2+4\sqrt{x-2}}=5\left(x\ge2\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-2}+4\right)^2}=5\\ \Leftrightarrow\sqrt{x-2}+4=5\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\Leftrightarrow x=3\\ 2,\sqrt{x+3+4\sqrt{x-1}}=2\left(x\ge1\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-1}+4\right)^2}=2\\ \Leftrightarrow\sqrt{x-1}+4=2\\ \Leftrightarrow\sqrt{x-1}=-2\\ \Leftrightarrow x\in\varnothing\left(\sqrt{x-1}\ge0\right)\)

\(3,\sqrt{x+\sqrt{2x-1}}=\sqrt{2}\left(x\ge\dfrac{1}{2};x\ne1\right)\\ \Leftrightarrow x+\sqrt{2x-1}=2\\ \Leftrightarrow x-2=-\sqrt{2x-1}\\ \Leftrightarrow x^2-4x+4=2x-1\\ \Leftrightarrow x^2-6x+5=0\\ \Leftrightarrow\left(x-5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=1\left(loại\right)\end{matrix}\right.\)

\(4,\sqrt{x-2+\sqrt{2x-5}}=3\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-4+2\sqrt{2x-5}}=6\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}=6\\ \Leftrightarrow\sqrt{2x-5}+1=6\\ \Leftrightarrow\sqrt{2x-5}=5\\ \Leftrightarrow2x-5=25\Leftrightarrow x=15\left(TM\right)\)

4) Ta có: \(\left(x+3\right)\cdot\sqrt{10-x^2}=x^2-x-12\)

\(\Leftrightarrow\left(x+3\right)\cdot\sqrt{10-x^2}-\left(x-4\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(\sqrt{10-x^2}-x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\\sqrt{10-x^2}=x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\10-x^2=x^2-8x+16\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x^2-8x+16-10+x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\2x^2-8x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\2\left(x^2-4x+3\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\\left(x-1\right)\left(x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\\x=3\end{matrix}\right.\)

Đặt căn x=a; căn 1-x=b

Theo đề, ta có: a+b=1+2/3ab

=>3a+3b=3+2ab

=>3a+3b-2ab=3

=>a(3-2b)+3b-4,5=-1,5

=>-a(2b-3)+3(b-1,5)=-1,5

=>-2a(b-1,5)+3(b-1,5)=-1,5

=>(-2a+3)(b-1,5)=-1,5

=>(2a-3)(b-1,5)=1,5

=>(2a-3)(2b-3)=3

=>(2a-3;2b-3) thuộc {(1;3); (3;1);(-1;-3); (-3;-1)}

=>(a,b) thuộc {(2;3); (3;2); (1;0); (0;1)}

TH1: a=2; b=3

=>căn x=2 và căn 1-x=3

=>x=4 và 1-x=9

=>Loại

TH2: a=3 và b=2

=>căn x=3 và căn 1-x=2

=>x=9 và 1-x=4(loại)

TH3: a=1 và b=0

=>x=1 và 1-x=0

=>x=1

TH4: a=0 và b=1

=>x=0 và 1-x=1

=>x=0

ĐKXĐ: \(\left\{{}\begin{matrix}-1\le x\le3\\x\ne1\end{matrix}\right.\)

\(\dfrac{\sqrt{x+1}\left(\sqrt{x+1}+\sqrt{3-x}\right)}{2\left(x-1\right)}>x-\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{x+1+\sqrt{-x^2+2x+3}}{x-1}>2x-1\)

- TH1: Với \(x>1\) BPT tương đương:

\(x+1+\sqrt{-x^2+2x+3}>\left(2x-1\right)\left(x-1\right)\)

\(\Leftrightarrow\sqrt{-x^2+2x+3}>2x^2-4x\)

Đặt \(\sqrt{-x^2+2x+3}=t\ge0\Rightarrow2x^2-4x=-2t^2+6\)

BPt trở thành: \(t>-2t^2+6\Leftrightarrow2t^2+t-6>0\)

\(\Rightarrow t>\dfrac{3}{2}\Rightarrow-x^2+2x+3>\dfrac{9}{4}\Rightarrow1< x< \dfrac{2+\sqrt{7}}{2}\)

TH2: với \(x< 1\) BPT tương đương:

\(x+1+\sqrt{-x^2+2x+3}< \left(2x-1\right)\left(x-1\right)\)

\(\Leftrightarrow\sqrt{-x^2+2x+3}< 2x^2-4x\)

Tương tự như trên, đặt  \(t=\sqrt{-x^2+2x+3}\ge0\) ta được \(0\le t< \dfrac{3}{2}\)

\(\Rightarrow-x^2+2x+3< \dfrac{9}{4}\) \(\Rightarrow-1\le x< \dfrac{2-\sqrt{7}}{2}\)

Vậy nghiệm của BPT là: \(\left[{}\begin{matrix}-1\le x< \dfrac{2-\sqrt{7}}{2}\\1< x< \dfrac{2+\sqrt{7}}{2}\end{matrix}\right.\)

1) \(\dfrac{x+2\sqrt[]{x}}{\sqrt[]{x}-1}=8\left(1\right)\)

Điều kiện \(\left\{{}\begin{matrix}x\ge0\\\sqrt[]{x}-1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x+2\sqrt[]{x}=8\left(\sqrt[]{x}-1\right)\)

\(\Leftrightarrow x-6\sqrt[]{x}+8=0\left(2\right)\)

Đặt \(t^2=x\Leftrightarrow t=\sqrt[]{x}\)

\(\left(2\right)\Leftrightarrow t^2-6t+8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt[]{x}=2\\\sqrt[]{x}=4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=16\end{matrix}\right.\) (thỏa điều kiện)

2) \(\sqrt[]{\dfrac{2x-3}{x-1}}=2\left(1\right)\)

Điều kiện \(\dfrac{2x-3}{x-1}\ge0\Leftrightarrow\left[{}\begin{matrix}x< 1\\x\ge\dfrac{3}{2}\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\dfrac{2x-3}{x-1}=4\)

\(\Leftrightarrow2x-3=4\left(x-1\right)\)

\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\) (thỏa điều kiện)