Đk: \(x\ge\dfrac{1}{2}\)

Bpt\(\Leftrightarrow\left(x^2+2x\sqrt{2x-1}+2x-1\right)-\left[4\left(2x-1\right)+4\sqrt{2x-1}+1\right]\ge0\)

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

\(\Leftrightarrow\left(x-\sqrt{2x-1}-1\right)\left(x+3\sqrt{2x-1}+1\right)\ge0\) (1)

Vì \(x\ge\dfrac{1}{2}\Rightarrow x+3\sqrt{2x-1}+1>0\)

Từ (1) \(\Rightarrow x-\sqrt{2x-1}-1\ge0\)

\(\Leftrightarrow\sqrt{2x-1}\le x-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1\ge0\\x-1\ge0\\2x-1\le\left(1-x\right)^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x\in R\backslash\left(2-\sqrt{2};2+\sqrt{2}\right)\end{matrix}\right.\)\(\Rightarrow x\ge2+\sqrt{2}\)

Vậy...

a, ĐKXĐ : \(\left[{}\begin{matrix}x\le-3\\x\ge0\end{matrix}\right.\)

TH1 : \(x\le-3\) ( LĐ )

TH2 : \(x\ge0\)

BPT \(\Leftrightarrow x^2+2x+x^2+3x+2\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge4x^2\)

\(\Leftrightarrow\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge x^2-\dfrac{5}{2}x\)

\(\Leftrightarrow2\sqrt{\left(x+2\right)\left(x+3\right)}\ge2x-5\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\x\ge-2\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\4x^2+20x+24\ge4x^2-20x+25\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}0\le x< \dfrac{5}{2}\\x\ge\dfrac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow x\ge0\)

Vậy \(S=R/\left(-3;0\right)\)

đề câu 2 có sai gì ko v 

a: \(\sqrt{x^2+6x+9}=\sqrt{11+6\sqrt{2}}\)

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

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

=>\(\left[{}\begin{matrix}x+3=3+\sqrt{2}\\x+3=-3-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-6-\sqrt{2}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}2x-y=4\\x+2y=-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x-2y=8\\x+2y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-2y+x+2y=8-3\\2x-y=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5x=5\\y=2x-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\cdot1-4=-2\end{matrix}\right.\)

Do \(x^6-x^3+x^2-x+1=\left(x^3-\dfrac{1}{2}\right)^2+\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}>0\) ; \(\forall x\) nên BPT tương đương:

\(\sqrt{13}-\sqrt{2x^2-2x+5}-\sqrt{2x^2-4x+4}\ge0\)

\(\Leftrightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}\le\sqrt{26}\) (1)

Ta có:

\(VT=\sqrt{\left(2x-1\right)^2+3^2}+\sqrt{\left(2-2x\right)^2+2^2}\ge\sqrt{\left(2x-1+2-2x\right)^2+\left(3+2\right)^2}=\sqrt{26}\) (2)

\(\Rightarrow\left(1\right);\left(2\right)\Rightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}=\sqrt{26}\)

Dấu "=" xảy ra khi và chỉ khi \(2\left(2x-1\right)=3\left(2-2x\right)\Leftrightarrow x=\dfrac{4}{5}\)

Vậy BPT có nghiệm duy nhất \(x=\dfrac{4}{5}\)

Lời giải:

a.

PT $\Leftrightarrow |2x+1|=|x-1|$

$\Leftrightarrow 2x+1=x-1$ hoặc $2x+1=-(x-1)$

$\Leftrightarrow x+2=0$ hoặc $3x=0$

$\Leftrightarrow x=-2$ hoặc $x=0$ (tm)

b.

PT $\Leftrightarrow 9x^2-6x+1=x^2-4x+4$

$\Leftrightarrow 8x^2-2x-3=0$

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

$\Leftrightarrow 4x-3=0$ hoặc $2x+1=0$

$\Leftrightarrow x=\frac{3}{4}$ hoặc $x=\frac{-1}{2}$ (tm)

a: =>|2x+1|=|x-1|

=>2x+1=x-1 hoặc 2x+1=-x+1

=>x=-2 hoặc x=0

b: =>|3x-1|=|x-2|

=>3x-1=x-2 hoặc 3x-1=-x+2

=>2x=-1 hoặc 4x=3

=>x=-1/2 hoặc x=3/4