(4x - 3)² - (2x + 1)² = 0

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

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

\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\)

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

Vậy ...

3x - 12 - 5x(x - 4) = 0

\(\Leftrightarrow\) 3x - 12 - 5x² + 20x = 0

\(\Leftrightarrow\) -5x² + 23x - 12 = 0

\(\Leftrightarrow\) 5x² - 23x + 12 = 0

\(\Leftrightarrow\) 5x² - 20x - 3x + 12 = 0

\(\Leftrightarrow\) 5x(x - 4) - 3(x - 4) = 0

\(\Leftrightarrow\) (x - 4)(5x - 3) = 0

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

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy ...

(8x + 2)(x² + 5)(x² - 4) = 0

\(\Leftrightarrow\) (8x + 2)(x² + 5)(x - 2)(x + 2) = 0

Vì x² \(\ge\) 0 \(\forall\) x nên x² + 5 > 0 \(\forall\) x

\(\Rightarrow\) (8x + 2)(x - 2)(x + 2) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}8x+2=0\\x-2=0\\x+2=0\end{matrix}\right.\)

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

Vậy ...

Chúc bn học tốt!

a) Ta có: \(\left(4x-3\right)^2-\left(2x+1\right)^2=0\)

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

\(\Leftrightarrow\left(2x-4\right)\left(6x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{2;\dfrac{1}{3}\right\}\)

b) Ta có: \(3x-12-5x\left(x-4\right)=0\)

\(\Leftrightarrow3\left(x-4\right)-5x\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(3-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{4;\dfrac{3}{5}\right\}\)

c) Ta có: \(\left(8x+2\right)\left(x^2+5\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow2\left(4x+1\right)\left(x^2+5\right)\left(x-2\right)\left(x+2\right)=0\)

mà \(2>0\)

và \(x^2+5>0\forall x\)

nên \(\left(4x+1\right)\left(x-2\right)\left(x+2\right)=0\)

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

Vậy: \(S=\left\{-\dfrac{1}{4};2;-2\right\}\)

a) Ta có: 4x-20=0

\(\Leftrightarrow4x=20\)

hay x=5

Vậy: S={5}

b) Ta có: \(2x+x+12=0\)

\(\Leftrightarrow3x+12=0\)

\(\Leftrightarrow3x=-12\)

hay x=-4

Vậy: S={-4}

c) Ta có: x-5=3-x

\(\Leftrightarrow x-5-3+x=0\)

\(\Leftrightarrow2x-8=0\)

\(\Leftrightarrow2x=8\)

hay x=4

Vậy: S={4}

d) Ta có: 7-3x=9-x

\(\Leftrightarrow7-3x-9+x=0\)

\(\Leftrightarrow-2x-2=0\)

\(\Leftrightarrow-2x=2\)

hay x=-1

Vậy: S={-1}

a, 4x-20=0                                         b,   2x+x+12=0

⇔4x      = 20                                        ⇔3x  + 12=0

⇔  x      = 5                                             ⇔3x         = -12

Vậy tập nghiệm S = { 5}                           ⇔ x        = -4

                                                             Vậy tập nghiệm S={ -4}

2: 

a: =>-2x=10

=>x=-5

b: =>(x-3)(2x+5)=0

=>x=3 hoặc x=-5/2

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

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

\(\Leftrightarrow-5\left(4x+3\right)=0\)

\(\Leftrightarrow4x+3=0\)

\(\Leftrightarrow4x=-3\)

\(\Leftrightarrow x=\frac{-3}{4}\)

Vậy tập nghiệm của pt là \(S=\left\{\frac{-3}{4}\right\}\)

h) \(\left(2x-3\right)\left(3x+1\right)-x\left(6x+10\right)=30\)

\(\Leftrightarrow3x\left(2x-3\right)+\left(2x-3\right)-6x^2-10x=30\)

\(\Leftrightarrow6x^2-9x+2x-3-6x^2-10x=30\)

\(\Leftrightarrow-9x+2x-3-10x=30\)

\(\Leftrightarrow-17x-3=30\)

\(\Leftrightarrow-17x=33\)

\(\Leftrightarrow x=\frac{-33}{17}\)

Vậy tập nghiệm của pt là \(S=\left\{\frac{-33}{17}\right\}\)

k) \(\left(2x-1\right)\left(8x+5\right)-\left(4x+3\right)^2=20\)

\(\Leftrightarrow8x\left(2x-1\right)+5\left(2x-1\right)-\left(16x^2+24x+9\right)=20\)

\(\Leftrightarrow16x^2-8x+10x-5-16x^2-24x-9=20\)

\(\Leftrightarrow-8x+10x-5-24x-9=20\)

\(\Leftrightarrow-22x-14=20\)

\(\Leftrightarrow-11x-7=10\)

\(\Leftrightarrow x=\frac{-11}{17}\)

Vậy tập nghiệm của pt là \(S=\left\{\frac{-11}{17}\right\}\)

x + 12 =(-5)-x

x + x = -5 - 12

2x = -17

  \(x=-\frac{17}{2}\)  

x +  5 = 10 - x

x + x = 10 - 5

2x = 5

  \(x=\frac{5}{2}\)

12-x=x+1

12 - 1 = x+x

11=2x

x=\(\frac{11}{2}\)

14+4x=3x-20

4x-3x=-20-14

       x=-34

 \(2x+6=0\\ \Leftrightarrow2x=-6\\ \Leftrightarrow x=-3\)

\(4x+20=0\\ \Leftrightarrow4x=-20\\ \Leftrightarrow x=-5\)

\(2x-3=0\\ \Leftrightarrow2x=3\\ \Leftrightarrow x=\dfrac{3}{2}\)

\(15-7x=9-3x\\ \Leftrightarrow15-7x-9+3x=0\\ \Leftrightarrow6-4x=0\\ \Leftrightarrow-4x=-6\\ \Leftrightarrow x=\dfrac{3}{2}\)

\(3x-1=x+3\\ \Leftrightarrow3x-1-x-3=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)

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