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

\(\Leftrightarrow\left|x-1\right|=3-2x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=3-2x\\x-1=2x-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{3}\\x=2\end{matrix}\right.\)

Nếu x lớn hơn hoặc bằng 2, có:

|x - 2|(x - 1)(x + 1)(x + 2) = 4

(x - 2)(x + 2)(x - 1)(x + 1) = 4

(x² - 4)(x² - 1) = 4

x⁴ - 4x² + 4 = 4

(x² - 2)² = 4 => x² - 2 = 2 => x² = 4 => x = 2

Nếu x nhỏ hơn 2, có:

|x - 2|(x - 1)(x + 1)(x + 2) = 4

(2 - x)(2 + x)(x - 1)(x + 1) = 4

(4 - x²)(x² - 1) = 4

5x² - x⁴ - 4 = 4

x² - (x⁴ - 4x² + 4) = 4

x² - 4 - (x² - 2)² = 0

(x ​- 2)(x + 2) - (x² - 2)² = 0

đặt nhé : đặt \(x+\frac{1}{x}=a.....=>x^2+\frac{1}{x^2}=a^2-2...\)

x(x-1)+2(1-x)=0

<=>x(x-1)-2(x-1)=0

<=>(x-1)(x-2)=0

<=>x-1=0 hoặc x-2=0

<=>x=1 hoặc x=2

Vậy S={1;2}

câu a bài 1:(2x+1)(3x-2)=(5x-8)(2x+1)

<=>(2x+1)(3x-2)-(5x-8)(2x+1)=0

<=>(2x+1)(3x-2-5x+8)=0

<=>(2x+1)(6-2x)=0

bước sau tự làm nốt nha !

câu b:gợi ý: tách 4x^2-1thành (2x-1)(2x+1) rồi làm như câu a

Bài 2: 

a: \(\dfrac{1}{2x-3}-\dfrac{3}{x\left(2x-3\right)}=\dfrac{5}{x}\)

\(\Leftrightarrow x-3=5\left(2x-3\right)=10x-15\)

=>-9x=-12

hay x=4/3

b: \(\Leftrightarrow x\left(x+2\right)-x+2=2\)

=>x²+2x-x+2=2

=>x²+x=0

=>x=0(loại) hoặc x=-1(nhận)

c: \(\dfrac{x+1}{x-2}+\dfrac{x-1}{x+2}=\dfrac{2\left(x^2+2\right)}{\left(x-2\right)\left(x+2\right)}\)

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

=>4=4(luôn đúng)

Vậy: S={x|x<>2; x<>-2}

\(\frac{2\left(x-1\right)}{x\left(x+1\right)}=\frac{1}{x}\) \(ĐKXĐ:x\ne0;x\ne-1\)

\(\Rightarrow2\left(x-1\right)=\left(x+1\right)\)

\(\Leftrightarrow2x-2=x+1\)

\(\Leftrightarrow x=3\left(nhan\right)\)

Vậy \(S=\left\{3\right\}\)