Bài 2: 

a: \(x^2-4x+3=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

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

a, (3x-5)^2 - (x-1)^2 = 0

(3x-5-x+1)(3x-5+x-1) =0

(2x-4)(4x-6)=0

Do đó: 2x-4=0 hoặc 4x-6=0

Th1: 2x-4=0 => 2x=4

=> x=2

Th2: 4x-6=0 => 4x=6

=> x = 4/6 =2/3

Vậy x = 2 ; 2/3

Chọn B

\(\left(2x^2-3x-1\right)^2-3\left(2x^2-3x-5\right)-16=0\)

\(\Leftrightarrow\left(2x^2-3x-1\right)^2-3\left(2x^2-3x-1\right)-4=0\)

\(\Leftrightarrow\left(2x^2-3x-1-4\right)\left(2x^2-3x+1+1\right)=0\)

\(\Leftrightarrow\left(2x^2-3x-5\right)\left(2x^2-3x\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+1\right)\cdot x\cdot\left(2x-3\right)=0\)

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

=>(9x^2+24x-6x-16)(x^2+2x+1)=-16

=>(9x^2+18x-16)(x^2+2x+1)=-16

=>(9x^2+18x+9-25)(x^2+2x+1)=-16

=>[9(x+1)^2-25](x+1)^2=-16

=>9(x+1)^4-25(x+1)^2+16=0

Đặt (x+1)^2=a

=>9a^2-25a+16=0

=>a=1 hoặc a=16/9

=>(x+1)^2=1 hoặc (x+1)^2=16/9

=>\(x\in\left\{0;-2;\dfrac{1}{3};-\dfrac{7}{3}\right\}\)

CẢM ƠN NHÌU NHA

Đặt \(t=2x^2-3x-1\)

\(\Rightarrow t^2-3\left(t-4\right)-16=0\)

\(\Rightarrow t^2-3t+12-16=0\)

\(\Rightarrow t^2-3t-4=0\)

\(\Rightarrow\left\{{}\begin{matrix}t_1=-1\\t_2=4\end{matrix}\right.\)

\(TH_1:t=-1\)

\(\Leftrightarrow2x^2-3x-1=-1\)

\(\Leftrightarrow2x^2-3x=0\)

\(\Leftrightarrow x\left(2x-3\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

\(TH_2:t=4\)

\(\Leftrightarrow2x^2-3x-1=4\)

\(\Leftrightarrow2x^2-3x-5=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1=-1\\x_2=\dfrac{5}{2}\end{matrix}\right.\)

(2x² - 3x - 1)² - 3(2x² - 3x - 5) - 16 = 0

<=> 4x⁴ - 12x³ - x² + 15x = 0

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

<=> x = 0 hoặc x + 1 = 0 hoặc 2x - 3 = 0 hoặc 2x - 5 = 0

<=> x = 0 hoặc x = -1 hoặc x = 3/2 hoặc x = 5/2

\(\left(3x-2\right)\left(3x+8\right)\left(x+1\right)^2+16=0\)

\(\Leftrightarrow\left(9x^2+18x-16\right)\left(x^2+2x+1\right)+16=0\)

\(\Leftrightarrow\left[9\left(x^2+2x+1\right)-25\right]\left(x^2+2x+1\right)+16=0\)

Đặt \(x^2+2x+1=a\ge0\)

\(\left(9a-25\right)a+16=0\)

\(\Leftrightarrow9a^2-25a+16=0\)

\(\Rightarrow\left[{}\begin{matrix}a=1\\a=\frac{16}{9}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2+2x+1=1\\x^2+2x+1=\frac{16}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\left(x+2\right)=0\\\left(x+1\right)^2=\left(\frac{4}{3}\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x+1=\frac{4}{3}\\x+1=-\frac{4}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=\frac{1}{3}\\x=-\frac{7}{3}\end{matrix}\right.\)