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

.

a: ĐKXĐ: x>=1

Đặt \(a=\sqrt{5x-1};b=\sqrt{x-1}\)

\(a^2-b^2=\left(5x-1\right)-\left(x-1\right)=4x\)

\(\frac{a^2+b^2}{2}=\frac{5x-1+x-1}{2}=\frac{6x-2}{2}=3x-1\)

\(ab=\sqrt{\left(5x-1\right)\left(x-1\right)}=\sqrt{5x^2-6x+1}\)

Thay vào phương trình ban đầu, ta được: \(\left(a+b\right)\left(\frac{a^2+b^2}{2}-ab\right)=\left(a-b\right)\left(a+b\right)\)

=>\(\frac{\left(a+b\right)\left(a-b\right)^2}{2}=\left(a-b\right)\left(a+b\right)\)

=>\(\frac{\left(a-b\right)^2}{2}=\left(a-b\right)\)

=>\(\left(a-b\right)^2=2\left(a-b\right)\)

=>(a-b)(a-b-2)=0

TH1: a-b=0

=>a=b

=>5x-1=x-1

=>x=0(loại)

TH2: a-b-2=0

=>\(\sqrt{5x-1}-\sqrt{x-1}-2=0\)

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

=>\(5x-1=x-1+4+2\cdot2\cdot\sqrt{x-1}\)

=>\(4\sqrt{x-1}+x+3=5x-1\)

=>\(4\sqrt{x-1}=5x-1-x-3=4x-4\)

=>\(\sqrt{x-1}=x-1\)

=>\(\sqrt{x-1}\left(\sqrt{x-1}-1\right)=0\)

=>\(\left[\begin{array}{l}x-1=0\\ x-1=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\left(nhận\right)\\ x=2\left(nhận\right)\end{array}\right.\)

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...

ta có:

pt trên \(< =>x^2+6x+1=\left(2x+1\right)\sqrt{x^2+2x+3}\)

\(< =>\left[\left(x^2+6x\right)+1\right]^2=\left(2x+1\right)^2.\left(x^2+2x+3\right)\)

\(< =>x^4+12x^3+36x^2+2.\left(x^2+6x\right)+1=\left(4x^2+4x+1\right)\left(x^2+2x+3\right)\)

\(< =>x^4+12x^3+38x^2+12x+1=\)

\(4x^4+8x^3+12x^2+4x^3+8x^2+12x+x^2+2x+3\)

\(=4x^4+12x^3+21x^2+14x+3\)

\(< =>-3x^4+17x^2-2x-2=0\)

\(< =>-\left(x^2+2x-1\right)\left(3x^2-6x+2\right)=0\)

đến đây dễ rùi bạn tự giải nhé