@Akai Haruma mai em thi rồi huhu

ĐKXĐ: \(x\ge-1\)

\(\sqrt{x+1+2\sqrt{x+1}+1}+\sqrt{x+1-6\sqrt{x+1}+9}=2\sqrt{x+1-2\sqrt{x+1}+1}\)

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

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

Ta có:

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

Dấu "=" xảy ra khi và chỉ khi:

\(\sqrt{x+1}-3\ge0\Rightarrow x\ge8\)

Vậy nghiệm của pt là \(x\ge8\)

ĐKXĐ: \(x\ge1\)

Do \(\sqrt{x-\sqrt{x^2-1}}.\sqrt{x+\sqrt{x^2-1}}=\sqrt{x^2-x^2+1}=1\)

Đặt \(\sqrt{x-\sqrt{x^2-1}}=t\Rightarrow\sqrt{x+\sqrt{x^2-1}}=\dfrac{1}{t}\)

Phương trình trở thành:

\(t+\dfrac{1}{t}=2\Rightarrow t^2-2t+1=0\Rightarrow t=1\)

\(\Rightarrow\sqrt{x-\sqrt{x^2-1}}=1\Leftrightarrow x-\sqrt{x^2-1}=1\)

\(\Leftrightarrow x-1=\sqrt{x^2-1}\)

\(\Rightarrow x^2-2x+1=x^2-1\)

\(\Rightarrow x=1\) (thỏa mãn)

\(\sqrt{x+2\sqrt{x}+1}-\sqrt{x-2\sqrt{x}+1}=2\left(x\ge0\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x}+1\right)^2}-\sqrt{\left(\sqrt{x}-1\right)^2}=2\\ \Leftrightarrow\sqrt{x}+1-\left|\sqrt{x}-1\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1-\left(\sqrt{x}-1\right)=2,\forall\sqrt{x}-1\ge0\\\sqrt{x}+1-\left(1-\sqrt{x}\right)=2,\forall\sqrt{x}-1< 0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}0\sqrt{x}=0,\forall x\ge1\\\sqrt{x}=1,\forall x< 1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\in R,x\ge1\\x=1,x< 1\left(loại\right)\end{matrix}\right.\\ \Leftrightarrow x\in R,x\ge1\)

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

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)

`\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\sqrt{x-1}(x>=1)`

`<=>\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=2\sqrt{x-1}`

`<=>\sqrt{(\sqrt{x-1}+1)^2}+\sqrt{(\sqrt{x-1}-1)^2}=2\sqrt{x-1}`

`<=>|\sqrt{x-1}+1|+|\sqrt{x-1}-1|=2\sqrt{x-1}`

`<=>\sqrt{x-1}+1+|\sqrt{x-1}-1|=2\sqrt{x-1}`

`<=>|\sqrt{x-1}-1|=\sqrt{x-1}-1`

`<=>\sqrt{x-1}-1>=0``

`<=>sqrt{x-1}>=1`

`<=>x-1>=1`

`<=>x>=2`

Vậy `S={x|x>=2}`

\(x\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(x\sqrt{x}\left(x-1\right)^2=0\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

B là đáp án đúng