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

\(\frac{1}{x-2}+3=\frac{3-x}{x-2}\) (ĐKXĐ: x≠2)

⇔ \(\frac{1+3\left(x-2\right)}{x-2}=\frac{3-x}{x-2}\)

⇔ \(1+3x-6=3-x\)

⇔ 4x=8

⇔ x=2 ( không thỏa nãn ĐKXĐ)

Vậy phương trình vô nghiệm

Đặt bt trong ngoặc đầu tiên = t

pt trở thành

\(t\left(t-2\right)-3=0\Leftrightarrow t^2-2t-3=0\) \(\Leftrightarrow\left[{}\begin{matrix}t=3\\t=-1\end{matrix}\right.\)

với t=3, ta có:

\(x^2+2x-1=3\Leftrightarrow x^2+2x-4=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1+\sqrt{5}\\x=-1-\sqrt{5}\end{matrix}\right.\)

t= -1 tương tự

\(pt\Leftrightarrow\left(16x^2+24x+9\right)\left(2x^2+3x+1\right)=810\)

\(\Leftrightarrow32x^4+48x^3+16x^2+48x^3+72x^2+24x+18x^2+27x+9-810=0\)

\(\Leftrightarrow32x^4+96x^3+106x^2+51x-801=0\)

\(\Leftrightarrow32x^4+96x^3+106x^2+318x-267x-801=0\)

\(\Leftrightarrow\left(x+3\right)\left(32x^3+106x-267\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x-3\right)\left(16x^2+24x+189\right)=0\)

Vì \(16x^2+24x+89=\left(4x+3\right)^2+80\ge80\) nên \(\orbr{\begin{cases}x+3=0\\2x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=\frac{3}{2}\end{cases}}\)

Ta có: \(\left(4x+3\right)^2\left(x+1\right)\left(2x+1\right)=810\)

\(\Leftrightarrow\left(16x^2+24x+9\right)\left(2x^2+3x+1\right)=810\)

Đặt \(a=2x^2+3x+1\)

\(\Rightarrow\left(8a+1\right)a=810\)

\(\Leftrightarrow8a^2+a-810=0\)

\(\Leftrightarrow\left(a-10\right)\left(8a+81\right)=0\)

\(\Rightarrow\left(2x^2+3x-9\right)\left(16x^2+24x+189\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x-3\right)\left(16x^2+24x+189\right)=0\)

Lại có: \(16x^2+24x+189=\left(4x+3\right)^2+80>0\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\2x-3=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-3\\x=\frac{3}{2}\end{cases}}\)

Ta có \(4\left(x-1\right)^2-\left(x+1\right)^2=x+13\Leftrightarrow4\left(x^3-3x^2+3x-1\right)-\left(x^2+2x+1\right)=x+13\)

\(\Leftrightarrow4x^3-12x^2+12x-4-x^2-2x-1-x-13=0\)

\(\Leftrightarrow4x^3-13x^2+9x-18=0\)\(\Leftrightarrow\left(4x^3-12x^2\right)-\left(x^2-3x\right)+\left(6x-18\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(4x^2-x+6\right)=0\Leftrightarrow\orbr{\begin{cases}x-3=0\\4x^2-x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\4x^2-x+6=0\left(1\right)\end{cases}}}\)

Ta thấy (1) vô nghiệm vì \(\Delta=1-24=-23< 0\)

Vậy phương trình có nghiệm x=3

Ta có pt

\(\Leftrightarrow\sqrt[3]{x-2}-1+\sqrt{x+1}-2=0\)

\(\Leftrightarrow\dfrac{x-3}{\sqrt[3]{\left(x-2\right)^2}+\sqrt[3]{x-2}+1}+\dfrac{x-3}{\sqrt{x+1}+2}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\dfrac{1}{\sqrt[3]{\left(x-2\right)^2}+\sqrt[3]{x-2}+1}+\dfrac{1}{\sqrt{x+1}+2}\right)=0\)

<=> x=3

X-\(\frac{3}{2}\)+X-\(\frac{5}{6}\)=\(-\frac{1}{3}\)

➜2X=\(-\frac{1}{3}\)+\(\frac{3}{2}+\frac{5}{6}\)

➜ 2X=2

➜X = 1

Vậy....................

a)\(\dfrac{7x-1}{2}+2x=\dfrac{16-x}{3}\)

\(\dfrac{\left(7x-1\right).3}{2.3}+\dfrac{2x.6}{6}=\dfrac{\left(16-x\right)2}{3.2}\)

khử mẫu 

=> (7x-1).3+12x=(16-x).2

=>21x-3+12x=-2x+32

=>21x-3+12x+2x-32=0

=>35x-35=0

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

ĐKXĐ: x khác +-2

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

khử mẫu

(x+1).(x+2)+(x-1)(x-2)=2x²+4

=>x²+x+2+x+2+x²-2x-x+2=2x²+4

=>x²+x+2+x+2+x²-2x-x+2-2x²-4=0

=>(x²+x²-2x²)+(x+x-2x-x)+(2+2+2-4)=0

=>-x+2=0

=>-x=-2

=>x=2(loại)

vậy pt vô nghiệm