\(4x^2+4x+1+4x+2-2x^2-x\le0\)

\(\Leftrightarrow2x^2+7x+3\le0\Leftrightarrow\left(2x+1\right)\left(x+3\right)\le0\)

TH1 : \(\left\{{}\begin{matrix}2x+1\ge0\\x+3\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{2}\\x\le-3\end{matrix}\right.\)<=> -1/2 =< x =< -3 

TH2 : \(\left\{{}\begin{matrix}2x+1\le0\\x+3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{1}{2}\\x\ge-3\end{matrix}\right.\)( vô lí ) 

ĐKXĐ: \(x^2\ge2\)

Đặt \(\sqrt{x^2-2}=a\ge0\)

BPT tương đương: \(\dfrac{1}{\sqrt{a^2+3}}+\dfrac{1}{\sqrt{3a^2+11}}\le\dfrac{2}{a+1}\)

Ta có: \(VT^2\le2\left(\dfrac{1}{a^2+3}+\dfrac{1}{3a^2+11}\right)< 2\left(\dfrac{1}{a^2+3}+\dfrac{1}{3a^2+1}\right)=\dfrac{8\left(a^2+1\right)}{\left(3a^2+1\right)\left(a^2+3\right)}\)

Mặt khác ta có: \(\left(a-1\right)^4\ge0\Leftrightarrow a^4-4a^3+6a^2-4a+1\ge0\)

\(\Leftrightarrow3a^4+10a^2+3\ge2a^4+4a^3+4a^2+4a+2\)

\(\Leftrightarrow\left(3a^2+1\right)\left(a^2+3\right)\ge2\left(a^2+1\right)\left(a+1\right)^2\)

\(\Rightarrow\dfrac{8\left(a^2+1\right)}{\left(3a^2+1\right)\left(a^2+3\right)}\le\dfrac{4}{\left(a+1\right)^2}\)

\(\Rightarrow VT^2< \dfrac{4}{\left(a+1\right)^2}\Rightarrow VT< \dfrac{2}{a+1}\)

\(\Rightarrow\) BPT đã cho đúng với mọi \(a\ge0\) hay nghiệm của BPT là \(x^2\ge2\)

ĐKXĐ: \(x>0\)

\(\Leftrightarrow\sqrt{\dfrac{\left(x^2+x+1\right)\left(x^2-x+1\right)}{x\left(x^2+1\right)}}-\sqrt{\dfrac{x^2+x+1}{x^2+1}}+\dfrac{\left(x-1\right)^2}{x}\ge0\)

\(\Leftrightarrow\sqrt{\dfrac{x^2+x+1}{x^2+1}}\left(\sqrt{\dfrac{x^2-x+1}{x}}-1\right)+\dfrac{\left(x-1\right)^2}{x}\ge0\)

\(\Leftrightarrow\dfrac{\left(x-1\right)^2}{\sqrt{x^2-x+1}+\sqrt{x}}.\sqrt{\dfrac{x^2+x+1}{x^2+1}}+\dfrac{\left(x-1\right)^2}{x}\ge0\) (luôn đúng \(\forall x>0\))

Vậy nghiệm của BPT đã cho là \(x>0\)

ĐKXĐ: x>=0

\(\sqrt{2x^2-2x+2}\)

\(=\sqrt{2\left(x^2-x+1\right)}=\sqrt{2\left(x^2-x+\frac14+\frac34\right)}\)

\(=\sqrt{2\left(x-\frac12\right)^2+\frac32}\ge\sqrt{\frac32}>1\forall x\) thỏa mãn ĐKXĐ

=>\(1-\sqrt{2x^2-2x+2}<0\forall x\) thỏa mãn ĐKXĐ

\(\frac{x - \sqrt{x}}{1 - \sqrt{2x^2 - 2x + 2}}\ge1\)

=>\(x-\sqrt{x}\le1-\sqrt{2x^2 - 2x + 2}\)

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

=>\(\begin{cases}2x^2-2x+2\le(1+\sqrt{x}-x)^2\\ 1+\sqrt{x}-x>0\end{cases}\Rightarrow\begin{cases}1+\sqrt{x}-x>0\\ 2x^2-2x+2\le x^2-x+1+2\sqrt{x}(1-x)\end{cases}\)

=>\(\begin{cases}1+\sqrt{x}-x>0\\ x^2-x+1\le2\sqrt{x}(1-x)\left(1\right)\end{cases}\)

Đặt \(t=\sqrt{x}\) (Điều kiện: t>=0)

=>\(x=t^2\)

(1) sẽ trở thành: \(t^4-t^2+1\le2t(1-t^2)\)

=>\(t^4+2t^3-t^2-2t+1\le0\)

=>\(t^2+2t-1-\frac{2}{t}+\frac{1}{t^2}\le0\)

=>\(\left(t^2+\frac{1}{t^2}\right)+2\left(t-\frac{1}{t}\right)-1\le0\)

=>\(\left(t-\frac{1}{t}\right)^2+2+2\left(t-\frac{1}{t}\right)+1\le0\)

=>\(\left(t-\frac{1}{t}\right)^2+2\left(t-\frac{1}{t}\right)+1\le0\)

=>\(\left(t-\frac{1}{t}+1\right)^2\le0\)

=>\(t-\frac{1}{t}+1=0\)

=>\(\frac{t^2+t-1}{t}\) =0

=>\(t^2+t-1=0\)

mà t>=0

nên \(t=\frac{\sqrt5-1}{2}\)

=>\(x = t^2 = \left(\frac{\sqrt{5} - 1}{2}\right)^2 = \frac{3 - \sqrt{5}}{2}\)

`sqrt{x-2}-2>=sqrt{2x-5}-sqrt{x+1}`

`đk:x>=5/2`

`bpt<=>\sqrt{x-2}+\sqrt{x+1}>=\sqrt{2x-5}+2`

`<=>x-2+x+1+2\sqrt{(x-2)(x+1)}>=2x-5+4+4\sqrt{2x-5}`

`<=>2x-1+2\sqrt{(x-2)(x+1)}>=2x-1+4\sqrt{2x-5}`

`<=>2\sqrt{(x-2)(x+1)}>=4\sqrt{2x-5}`

`<=>sqrt{x^2-x-2}>=2sqrt{2x-5}`

`<=>x^2-x-2>=4(2x-5)`

`<=>x^2-x-2>=8x-20`

`<=>x^2-9x+18>=0`

`<=>(x-3)(x-6)>=0`

`<=>` \(\left[ \begin{array}{l}x \ge 6\\x \le 3\end{array} \right.\) 

Kết hợp đkxđ:

`=>` \(\left[ \begin{array}{l}x \ge 6\\\dfrac52 \le x \le 3\end{array} \right.\) 

1.

ĐKXĐ: \(x=2\)

Xét \(x=2\), bất phương trình vô nghiệm

\(\Rightarrow\) bất phương trình đã cho vô nghiệm

\(\Rightarrow\) Không tồn tại \(a,b\) thỏa mãn

Đề bài lỗi chăng.