b)  ta có  \(\left(4x-5\right)^3-\left(2x+5\right)\left(16x^2-25\right)=0\)

        \(\left(4x-5\right)^3-\left(2x+5\right)\left(4x+5\right)\left(4x-5\right)=0\)

    \(\left(4x-5\right)\left[\left(4x-5\right)^2-\left(2x+5\right)\left(4x+5\right)\right]=0\)

 \(\left(4x-5\right)\left(16x^2-40x+5^2-8x^2-10x-20x-5^2\right)=0\)

   \(\left(4x-5\right)\left(8x^2-70x\right)=0\)

=> \(\orbr{\begin{cases}4x-5=0\\8x^2-70x=0\end{cases}=>\orbr{\begin{cases}4x=5\\x\left(8x-70\right)=0\end{cases}< =>}\orbr{\begin{cases}x=\frac{5}{4}\\\orbr{\begin{cases}x=0\\8x-70=0=>x=\frac{35}{4}\end{cases}}\end{cases}}}\) \(\orbr{\begin{cases}4x-5=0\\8x^2-70x=0\end{cases}< =>\orbr{\begin{cases}x=\frac{5}{4}\\x\left(8x-70\right)=0\end{cases}\orbr{\begin{cases}x=\frac{5}{4}\\\orbr{\begin{cases}x=0\\8x-70=0=>x=\frac{35}{4}\end{cases}}\end{cases}}}}\)

\(\orbr{\begin{cases}4x-5=0\\8x^2-70x=0\end{cases}=>\orbr{\begin{cases}x=\frac{5}{4}\\x\left(8x-70\right)=0\end{cases}}}\)

=> \(\orbr{\begin{cases}x=0\\x=\frac{35}{4}\end{cases}}\)     Vậy   \(\orbr{\begin{cases}x=\frac{5}{4}\\x=0\end{cases}}\) hoặc   \(x=\frac{35}{4}\)

a) x = 2 7                         b) x = 2.

c) x = 2                          d) x = 1.

a) 

<=> 3x - 3 + x - 2 = 2x - 2 - x + 1

<=> 3x + x - 2x + x = -2 + 1 + 3 + 2

<=>    3x               = 4

<=>   x                  = 4/3

Các câu sau làm tương tự

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

<=>   \(3x-3+x-2=2x-2-x+1\)

<=>  \(4x-5=x-1\)

<=> \(3x=4\)

<=>  \(x=\frac{4}{3}\)

Vậy....

1.a) (4x - 6y)² - (8xy - 5)² = (4x - 6y - 8xy + 5)(4x - 6y + 8xy - 5)

   b) 16x² - 49y² = (4x)² - (7y)² = (4x - 7y)(4x + 7y)

   c) 36x² + 60x + 25 = (6x)² + 2.6x.5 + 5² = (6x + 5)²

   d) (2x - y)(x - y) - (3y - 4x)² + (y - 2x)(2y - 3x) = (y - 2x)(y - x) + (y - 2x)(2y - 3x) - (3y - 4x)²

     = (y - 2x)[(y - x) + (2y - 3x)] - (3y - 4x)² = (y - 2x)(3y - 4x) - (3y - 4x)² = (3y - 4x)[(y - 2x) - (3y - 4x)] = 2(3y - 4x)(x - y)

2.M = (3x - 4)(9x² - 12x + 16) + (6x - 8)² = (3x - 4)[(3x)² - 2.3x.4 + 4²] + [2(3x - 4)]² = (3x - 4)(3x - 4)² + 4(3x - 4)²

       = (3x - 4)²(3x - 4 + 4) = 3x(3x - 4)²

a) =(4x-6y-8xy+3)(4x-6y+8xy-3)

=[4x(1-2y)+3(1-2y)][4x(1+2y)-3(1+2y)]

=(4x+3)(4x-3)(1-2y)(1+2y)

câu a )Phan Thanh Tịnh sai, tui sửa lại 

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

\(10x-5+32-12x=-5\)

\(-2x=5-5+32\)

\(x=\frac{32}{-2}\)

\(x=-16\)

\(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)

\(36x^2-12x-36x^2+27x=30\)

\(15x=30\)

\(x=\frac{30}{15}\)

x = 2

a)(x+2).(x+3)-(x-2).(x+5)=10

  ( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10

 x^2 +3x+2x+6-x^2 -5x+2x+10-10=0

 2x+6=0

2x=-6

x=-3

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

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

\(=\left(x-1\right)^2\)

c: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)

\(=5x^3+14x^2+12x+8\)