ĐK: `{(2x^2+8x+6>=0),(x^2-1>=0),(2x+2>=0):} <=> {(x=-1),(x>=1):}`

`\sqrt(2x^2+8x+6)+\sqrt(x^2-1)=2x+2`

`<=>(2x^2+8x+6)+(x^2-1)+2\sqrt((2x^2+8x+6)(x^2-1))=(2x+2)^2`

`<=>2(x+3)(x+1)+(x-1)(x+2)+2\sqrt((x+1)^2 (x+3)(x-1))=4(x+1)^2`

`<=> (x+1)[2(x+3)+(x-1)+2\sqrt((x+3)(x-1))-4(x+1)]=0`

`<=> [(x=-1\ (TM)),([2(x+3)+(x-1)+2\sqrt((x+3)(x-1))-4(x+1)]=0\ (1)):}`

(1) `<=> x-1=2\sqrt((x+3)(x-1))`

`<=>x^2-2x+1=4(x+3)(x-1)`

`<=>x=1\ `(TM)

Vậy `S={\pm 1}`.

\(ĐK:x\le-3;x\ge-1\)

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

\(\left(1\right)\Leftrightarrow2\left(x+3\right)+\left(x-1\right)+2\sqrt{2\left(x+3\right)\left(x-1\right)}=4\left(x+1\right)\\ \Leftrightarrow2\sqrt{2\left(x+3\right)\left(x-1\right)}=x-1\\ \Leftrightarrow8\left(x+3\right)\left(x-1\right)-\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(7x+25\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-\dfrac{25}{7}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=1\)

Vậy \(S=\left\{-1;1\right\}\)

\(pt\Leftrightarrow\sqrt{2x^2+8x+6}-4+\sqrt{x^2-1}-2x+2=0\)

\(\Leftrightarrow\frac{2\left(x-1\right)\left(x+5\right)}{\sqrt{2x^2+8x+6}+4}+\sqrt{x^2-1}-2\left(x-1\right)=0\)

Giải nốt nhá

\(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)

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

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

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

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

\(\Leftrightarrow2x^2+8x+6+\left(2x+2\right)\sqrt{2\left(x+3\right)\left(x-1\right)}+\left(x+1\right)\left(x-1\right)=4\left(x+1\right)^2\)

\(\Leftrightarrow\left(2x+2\right)\sqrt{2\left(x+3\right)\left(x-1\right)}=4\left(x+1\right)^2-2x^2-8x-6-\left(x+1\right)\left(x-1\right)\)

\(\Leftrightarrow8\left(x+1\right)^3.\left(x+3\right)\left(x-1\right)=\left(x+1\right)^2.\left(x-1\right)^2\)

\(\Leftrightarrow8x^4-8x^3+24x^3-24x^2+16x^3-16x^2+48x^2-48x+8x^2-8x+24x-24\)\(=x^4-2x^3+x^2+2x^3-4x^2+2x+x-2x+1\)

\(\Leftrightarrow8x^4+32x^3+16x^3-32x=x^4-2x^3+x^2+2x^3-4x^2+2x+x^2-2x+1\)

\(\Leftrightarrow8x^4+32x^3+16x^2-32x-24=x^4-2x^2+1\)

\(\Leftrightarrow8x^4+32x^2+16x^2-32x-24-x^4+2x^2-1=0\)

\(\Leftrightarrow7x^4+32x^3+18x^2-32x-25=0\)

\(\Leftrightarrow\left(7x^3+39x^2+57x+25\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(7x^2+25x+7x+25\right)\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[x\left(7x+25\right)+\left(7x+25\right)\right]\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(7x+25\right)\left(x+1\right)\left(x-1\right)=0\)

Nhưng \(7x+25\ne0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=1\end{cases}}\)

Vậy: nghiệm phương trình là x = 1; x = -1

Câu 1:

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

- Với \(x< -1\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP< 0\end{matrix}\right.\) pt vô nghiệm

- Nhận thấy \(x=-1\) là 1 nghiệm

- Nếu \(x>-1\) kết hợp ĐKXĐ các căn thức ta được \(x\ge1\), pt tương đương:

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

\(\Leftrightarrow2x+6+x-1+2\sqrt{2\left(x+3\right)\left(x-1\right)}=4x+4\)

\(\Leftrightarrow2\sqrt{2x^2+4x-6}=x-1\)

\(\Leftrightarrow4\left(2x^2+4x-6\right)=\left(x-1\right)^2\)

\(\Leftrightarrow7x^2+18x-25=0\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{25}{7}< 0\left(l\right)\end{matrix}\right.\)

Vậy pt có nghiệm \(x=\pm1\)

Câu 2:

ĐKXĐ: \(x\ge1\)

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

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

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

- Nếu \(\sqrt{x-1}-1\ge0\Leftrightarrow x\ge2\) pt trở thành:

\(\sqrt{x-1}+1-\sqrt{x-1}+1=2\Leftrightarrow2=2\) (luôn đúng)

- Nếu \(1\le x< 2\) pt trở thành:

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

Vậy nghiệm của pt là \(x\ge2\)

Câu 3:

Bình phương 2 vế ta được:

\(2x^2+2x+5+2\sqrt{\left(x^2+x+4\right)\left(x^2+x+1\right)}=2x^2+2x+9\)

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

\(\Leftrightarrow\left(x^2+x+4\right)\left(x^2+x+1\right)=4\)

Đặt \(x^2+x+1=a>0\) pt trở thành:

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

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

Câu 5:

ĐKXĐ: \(x\ge1\)

\(\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=1\)

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

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

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

\(\Rightarrow VT\ge VP\Rightarrow\) Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}\sqrt{x-1}-2\ge0\\\sqrt{x-1}-3\le0\end{matrix}\right.\) \(\Rightarrow5\le x\le10\)

Vậy nghiệm của pt là \(5\le x\le10\)

\(x^2+2x\sqrt{x+\dfrac{1}{x}}=8x-1\left(x\ne0\right)\)

Vì \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge\dfrac{1}{8}\)

Vì \(x\ne0\Rightarrow\) chia 2 vế cho x,ta được:

\(x+2\sqrt{x+\dfrac{1}{x}}=8-\dfrac{1}{x}\Rightarrow x+\dfrac{1}{x}+2\sqrt{x+\dfrac{1}{x}}=8\)

Đặt \(\sqrt{x+\dfrac{1}{x}}=a\left(a>0\right)\)

pt trở thành \(a^2+2a-8=0\Rightarrow a^2-2a+4a-8=0\)

\(\Rightarrow a\left(a-2\right)+4\left(a-2\right)=0\Rightarrow\left(a-2\right)\left(a+4\right)=0\)

mà \(a>0\Rightarrow a=2\Rightarrow\sqrt{x+\dfrac{1}{x}}=2\Rightarrow x+\dfrac{1}{x}=4\)

\(\Rightarrow\dfrac{x^2-4x+1}{x}=0\Rightarrow x^2-4x+1=0\)

\(\Delta=\left(-4\right)^2-4=12\Rightarrow\left[{}\begin{matrix}x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{4-\sqrt{12}}{2}=2-\sqrt{3}\\x-\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{4+\sqrt{12}}{2}=2+\sqrt{3}\end{matrix}\right.\)

Vậy pt có tập nghiệm \(S=\left\{2-\sqrt{3};2+\sqrt{3}\right\}\)

Cho mình hỏi sao VT lại lớn hơn 0 vậy ạ?

\(1.\sqrt{16-8x+x^2}=4-x\)

\(\sqrt{\left(4-x\right)^2}=4-x\)

\(4-x-4+x=0\)

= 0 phương trình vô nghiệm.

\(2.\sqrt{4x^2-12x+9}=2x-3\)

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

\(2x-3-2x+3=0\)

= 0 phương trình vô nghiệm.

a: Ta có: \(\sqrt{16-8x+x^2}=4-x\)

\(\Leftrightarrow\left|4-x\right|=4-x\)

hay \(x\le4\)

b: Ta có: \(\sqrt{4x^2-12x+9}=2x-3\)

\(\Leftrightarrow\left|2x-3\right|=2x-3\)

hay \(x\ge\dfrac{3}{2}\)