a) \(\left(3x-4\right)^3=\left(9x-8\right)\left(3x^2-8\right)\)

\(\Leftrightarrow27x^3+108x^2+144x+64=27x^3-72x-24x^2+64\)

\(\Leftrightarrow27x^3+108x^2+144x+64-27x^3+72x+24x^2-64=0\)

\(\Leftrightarrow132x^2+216x=0\)

\(\Leftrightarrow12x\left(11x+18\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\11x+18=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-\frac{18}{11}\end{array}\right.\)

a: 49x^2-25=0

=>(7x-5)(7x+5)=0

=>7x-5=0 hoặc 7x+5=0

=>x=5/7 hoặc x=-5/7

b: Đề thiếu vế phải rồi bạn

c: (3x-2)^2-9(x+4)(x-4)=2

=>9x^2-12x+4-9(x^2-16)=2

=>9x^2-12x+4-9x^2+144=2

=>-12x+148=2

=>-12x=-146

=>x=146/12=73/6

d: x^3-6x^2+12x-8=0

=>(x-2)^3=0

=>x-2=0

=>x=2

e: x^3-9x^2+27x-27=0

=>(x-3)^3=0

=>x-3=0

=>x=3

a) \(-25+49x^2=0\)

\(\Leftrightarrow49x^2-25=0\)

\(\Leftrightarrow\left(7x\right)^2-5^2=0\)

\(\Leftrightarrow\left(7x-5\right)\left(7x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}7x-5=0\\7x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}7x=5\\7x=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{7}\\x=-\dfrac{5}{7}\end{matrix}\right.\)

b) \(16x^2-25\left(x-2\right)^2=0\)

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

\(\Leftrightarrow\left(4x-5x+10\right)\left(4x+5x-10\right)=0\)

\(\Leftrightarrow\left(10-x\right)\left(9x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}10-x=0\\9x=10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=\dfrac{10}{9}\end{matrix}\right.\)

c) \(\left(3x-2\right)^2-9\left(x+4\right)\left(x+4\right)=2\)

\(\Leftrightarrow9x^2-12x+4-9\left(x^2+8x+16\right)=2\)

\(\Leftrightarrow9x^2-12x+4-9x^2-72x-144=2\)

\(\Leftrightarrow-84x-140=2\)

\(\Leftrightarrow-84x=142\)

\(\Leftrightarrow x=-\dfrac{142}{84}\)

\(\Leftrightarrow x=-\dfrac{71}{42}\)

d) \(x^3-6x^2+12x-8=0\)

\(\Leftrightarrow x^3-3\cdot2\cdot x^2+3\cdot2^2\cdot x-2^3=0\)

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

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

e) \(-27+27x-9x^2+x^3=0\)

\(\Leftrightarrow x^3-9x^2+27x-27=0\)

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

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

Mình sửa lại câu c một chút nha:

c: (3x-2)^2-9(x+4)(x+4)=2

=>(3x-2)^2-9(x+4)^2=2

=>(3x-2)^2-(3x+12)^2=2

=>(3x-2-3x-12)(3x-2+3x+12)=2

=>-14*(6x+10)=2

=>6x+10=-1/7

=>6x=-71/7

=>x=-71/42

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

c) x = 2                          d) x = 1.

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

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

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

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

\(x=-16\)

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

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

\(15x=30\)

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

x = 2

2.

a. 3x(12x - 4) - 9x(4x - 3) = 30

<=> 36x² - 12x - 36x² + 27x = 30

<=> 36x² - 36x² - 12x + 27x = 30

<=> 15x = 30

<=> x = 2

b. x(5 - 2x) + 2x(x - 1) = 15

<=> 5x - 2x² + 2x² - 2x = 15

<=> -2x² + 2x² + 5x - 2x = 15

<=> 3x = 15

<=> x = 5

a) x² ( 5x³ - x - 2323x²y=  6969x³y²- 2323x⁴y+ 2323x²y²

c) x² ( 4x^{3 -} 5xy + 2x ) ( -

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....

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

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