Do vế trái dương nên pt chỉ có nghiệm khi \(x\ge\dfrac{3}{4}\), kết hợp điều kiện \(2x^4-3x^2+1\ge0\Rightarrow x\ge1\)

Khi đó:

\(4x-3=\sqrt{2x^4-3x^2+1}+\sqrt{2x^4-x^2}\ge\sqrt{2x^4-3x^2+1+2x^4-x^2}\)

\(\Rightarrow4x-3\ge\sqrt{4x^4-4x^2+1}\)

\(\Rightarrow4x-3\ge\left|2x^2-1\right|=2x^2-1\)

\(\Rightarrow2x^2-4x+2\le0\)

\(\Rightarrow2\left(x-1\right)^2\le0\)

\(\Rightarrow x=1\)

\(ĐK:x\ge\frac{1}{2}\)

Bình phương 2 vế ta dc:

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

\(\Leftrightarrow3x^2+4x+1-x^2-2x-2x+1=2\sqrt{\left(x^2+2x\right)\left(2x-1\right)}\)

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

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

\(\Rightarrow x^4+2x^2+1=2x^3+3x^2-2x\)

\(\Leftrightarrow x^4+2x^2+1-2x^3-3x^2+2x=0\)

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

\(\Delta=\left(-1\right)^2-4.\left(-1\right)=5>0\)

\(\Rightarrow x_1=\frac{1+\sqrt{5}}{2}\left(TM\right);x_2=\frac{1-\sqrt{5}}{2}\left(loai\right)\)

Vậy...

Lời giải:
ĐKXĐ: $x\geq 1$
PT $\Leftrightarrow (x^2-2x)+(\sqrt{4x+1}-3)+(\sqrt{x-1}-1)=0$

$\Leftrightarrow x(x-2)+\frac{4(x-2)}{\sqrt{4x+1}+3}+\frac{x-2}{\sqrt{x-1}+1}=0$
$\Leftrightarrow (x-2)\left[x+\frac{4}{\sqrt{4x+1}+3}+\frac{1}{\sqrt{x-1}+1}\right]=0$

Dễ thấy với mọi $x\geq 1$ thì biểu thức trong ngoặc vuông luôn dương.

$\Rightarrow x-2=0$

$\Leftrightarrow x=2$ (tm)

Đặt \(\sqrt{x^2+1}=t>0\)

\(\Rightarrow\left(4x-1\right)t=2t^2-2x\)

\(\Leftrightarrow2t^2-\left(4x-1\right)t-2x=0\)

\(\Delta=\left(4x-1\right)^2+16x=\left(4x+1\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{4x-1-\left(4x+1\right)}{4}=-\dfrac{1}{2}\left(loại\right)\\t=\dfrac{4x-1+4x+1}{4}=2x\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2+1}=2x\) (\(x\ge0\))

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

\(\Rightarrow x=\dfrac{\sqrt{3}}{3}\)

\(\sqrt{4x^2-4x+1}=\sqrt{\left(2x-1\right)}=\left|2x-1\right|=-\left(2x-1\right)\Rightarrow2x-1\le0\Leftrightarrow x\le\frac{1}{2}\)\(\sqrt{4x^2-1}-2\sqrt{2x+1}=0\Leftrightarrow\sqrt{2x+1}\left(\sqrt{2x-1}-2\right)=0\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\2x-1=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{5}{2}\end{matrix}\right.\)

ĐKXĐ : \(\left\{{}\begin{matrix}4x^2+2y+2\ge0\\3x+y\ge0\end{matrix}\right.\)

Ta có : \(\left(\sqrt{4x^2+3}-2x\right)\left(\sqrt{y^2-2y+4}-y+1\right)=3\)

\(\Leftrightarrow\dfrac{3}{\sqrt{4x^2+3}+2x}.\dfrac{3}{\sqrt{y^2-2y+4}+y-1}=3\)

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

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

\(\Leftrightarrow2x\sqrt{y^2-2y+4}+\left(y-1\right).\sqrt{4x^2+3}=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}4x^2.\left(y^2-2y+4\right)=\left(y^2-2y+1\right).\left(4x^2+3\right)\\2x.\left(1-y\right)\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x^2=y^2-2y+1\\2x\left(1-y\right)\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}2x=y-1\\2x=1-y\end{matrix}\right.\\2x\left(1-y\right)\ge0\end{matrix}\right.\)

Với 2x = 1 - y

Khi đó ta có \(\sqrt{4x^2+2y+2}-\sqrt{3x+y}=2x+1\)

\(\Leftrightarrow\sqrt{4x^2-4x+4}-\sqrt{x+1}=2x+1\)      (ĐK : \(x\ge-1\))

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

\(\Leftrightarrow2\left(\sqrt{x^2-x+1}-1\right)=2x+\sqrt{x+1}-1\)

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

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

Phương trình (1) 

<=> \(\dfrac{2\left(x+1\right)}{\sqrt{x^2-x+1}}=2+\dfrac{1}{\sqrt{x+1}+1}+\dfrac{4}{\sqrt{x^2-x+1}}\)

Xét vế trái : \(\dfrac{2\left(x+1\right)}{\sqrt{x^2-x+1}}=\sqrt{\dfrac{4x^2+4x+1}{x^2-x+1}}=\sqrt{\dfrac{5x^2-5x+5-x^2+9x-4}{x^2-x+1}}\)

\(=\sqrt{5-\dfrac{x^2-9x+4}{x^2-x+1}}< \sqrt{5}\) (2) 

Lại có \(2+\dfrac{1}{\sqrt{x+1}+1}+\dfrac{4}{\sqrt{x^2-x+1}}\)

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

\(\ge2+\dfrac{\left(1+1+1+1+1\right)^2}{\sqrt{x+1}+1+4\sqrt{x^2-x+1}}=2+\dfrac{25}{\sqrt{x+1}+1+4\sqrt{x^2-x+1}}\)

Dấu "=" khi \(\dfrac{1}{\sqrt{x+1}+1}=\dfrac{1}{\sqrt{x^2-x+1}}\Leftrightarrow\left[{}\begin{matrix}x\approx3,498374325\\x\approx-0,7385661113\end{matrix}\right.\)

Khi đó \(VP\ge3,6\) (3) 

Từ (3) và (2) => (1) vô nghiệm 

Vậy x = 0 => y = 1

Với 2x = y - 1 kết hợp điều kiện 2x(1 - y) \(\ge0\)

ta được x = 0 ; y = 1 

Vậy (x ; y) = (0;1) 

ĐKXĐ: ...

\(\Leftrightarrow x^2-4x+1+\sqrt{2x+5}+\sqrt{4-2x}=0\)

Do \(\sqrt{2x+5}+\sqrt{4-2x}\ge\sqrt{2x+5+4-2x}=3\)

\(\Rightarrow VT\ge x^2-4x+1+3=x^2-4x+4=\left(x-2\right)^2\ge0\)

Dấu "=" xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}\left[{}\begin{matrix}2x+5=0\\4-2x=0\end{matrix}\right.\\\left(x-2\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow x=2\)

Vậy pt có nghiệm duy nhất \(x=2\)

\(\leftrightarrow\sqrt{4x^2-1}-\sqrt{2x^2-x}-\sqrt{2x+1}+\sqrt{x}=0\)

\(\leftrightarrow\sqrt{\left(2x+1\right)\left(2x-1\right)}-\sqrt{x\left(2x-1\right)}-\sqrt{2x+1}+\sqrt{x}=0\)

\(\leftrightarrow\sqrt{2x+1}\left(\sqrt{2x-1}-1\right)-\sqrt{x}\left(\sqrt{2x-1}-1\right)=0\)

\(\leftrightarrow\left(\sqrt{2x+1}-\sqrt{x}\right)\left(\sqrt{2x-1}-1\right)=0\)

\(\leftrightarrow\sqrt{2x+1}-\sqrt{x}=0hoặc\sqrt{2x-1}-1=0\)