1. (3x - 5)² - (3x + 1)² = 8

=> (3x - 5 - 3x - 1)(3x - 5 + 3x + 1) = 8

=> -6(6x - 4) = 8

=> 6x - 4 = \(\dfrac{-4}{3}\)

\(\Rightarrow x=\dfrac{4}{9}\)

2) 2x(8x - 3) - (4x - 3)² = 27

=> 16x² - 6x - 16x² + 24x - 9 = 27

=> 18x - 9 = 27

=> x = 2

3) (2x - 3)² - (2x + 1)² = 3

=> (2x - 3 - 2x - 1)(2x - 3 + 2x +1) = 3

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

=> 4x - 2 = \(\dfrac{-3}{4}\)

\(\Rightarrow x=\dfrac{5}{16}\)

4) (x + 5)² - x² = 45

=> (x + 5 - x)(x + 5 + x) = 45

=> 5(2x + 5) = 45

=> 2x + 5 = 9

=> x = 2

5) (x - 3)³ - (x - 3)(x² + 3x + 9) + 9(x + 1)² = 18

=> x³ - 9x² + 27x - 27 - x³ + 27 + 9(x² + 2x + 1) = 18

=> -9x² + 27x + 9x² + 18x + 9 = 18

=> 45x + 9 = 18

=> 45x = 9

=> x = \(\dfrac{1}{5}\)

6) x(x - 4)(x + 4) - (x - 5)(x² + 5x + 25) = 13

=> x (x² - 16) - (x³ - 125) = 13

=> x³ - 16x - x³ + 125 = 13

=> -16x = -112

=> x = 7.

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

\(\Leftrightarrow8x^3-12x^2+6x-1-8x^3-36x^2-54x-27-3\left(9x^2+6x+1\right)-2\left(x^2-4x+4\right)+x^2+x-2=0\)

\(\Leftrightarrow-48x^2-48x-28-27x^2-18x-3-2x^2+8x-8+x^2+x-2=0\)

\(\Leftrightarrow-76x^2-57x-41=0\)

\(\Leftrightarrow76x^2+57x+41=0\)

\(\text{Δ}=57^2-4\cdot76\cdot41=-9215< 0\)

Vậy: Phương trình vô nghiệm

phân tích theo hằng đẳng thức rồi rút gọn là ra thôi bạn

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

\(8x^3-12x^2+6x-1-3\left(1-6x+9x^2\right)=27+54x+36x^2+8x^3-2\left(x^2+3x-2x-6\right)\)\(8x^3-12x^2+6x-1-3+18x-27x^2=27+54x+36x^2+8x^3-2x^2-6x+4x+12\)\(8x^3-39x^2+24x-4=8x^3+34x^2+52x+39\)

\(8x^3-39x^2+24x-4-8x^3-34x^2-52x-39=0\)

\(-73x^2-28x-43=0\)

         Vậy đa thức vô nghiệm

(2x−1)³−3(x+2)(x−3)=(3+2x)³−3x(x+1)

<=>\(8x^3-12x^2+6x-1-3x^2+3x+18=9+54x+36x^2+8x^3-3x^2-3x\)

<=>\(48x^2+42x-8=0\)

<=> \(x=\frac{-21\pm5\sqrt{33}}{48}\)

a/ \(x=\dfrac{-5}{12}\)

b/ \(x\approx-1,9526\)

c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)

d/ \(x=\dfrac{-20}{13}\)

a) (x-2)³+6(x+1)²-x³+12=0

⇒ x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

⇒ x³-6x²+12x-8+6x²+12x+6-x³+12=0

⇒ 24x+10=0

⇒ 24x=-10

⇒ x=-5/12

a.

PT \(\Leftrightarrow x^3-6x^2+12x-8+6(x^2+2x+1)-x^3+12=0\)

\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3+12=0\)

\(\Leftrightarrow 24x+10=0\Leftrightarrow x=\frac{-5}{12}\)

b. Bạn xem lại đề, nghiệm khá xấu không phù hợp với mức độ tổng thể của bài.

c.

PT $\Leftrightarrow (4x^2+12x+9)+(x^2-1)=5(x^2+4x+4)+(x^2-4x-5)+9(x^2+6x+9)$
$\Leftrightarrow 10x^2+42x+64=0$

$\Leftrightarrow x^2+(3x+7)^2=-15< 0$ (vô lý) 

Do đó pt vô nghiệm.

d.

PT $\Leftrightarrow (1-6x+9x^2)-(9x^2-17x-2)=(9x^2-16)-9(x^2+6x+9)$

$\Leftrightarrow 11x+3=-54x-97$

$\Leftrightarrow 65x=-100$

$\Leftrightarrow x=\frac{-20}{13}$

\(\Leftrightarrow20\left(x^2-4x+3\right)-24\left(4x^2-4x+1\right)=15\left(9x^2+6x+1\right)+90x\left(x-1\right)\)

\(\Leftrightarrow20x^2-80x+60-96x^2+96x-24=135x^2+90x+15+90x^2-90x\)

\(\Leftrightarrow-301x^2+16x+21=0\)

\(\text{Δ}=16^2-4\cdot\left(-301\right)\cdot21=25540\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là 

\(\left\{{}\begin{matrix}x_1=\dfrac{-16-\sqrt{25540}}{-602}=\dfrac{16+\sqrt{25540}}{602}\\x_2=\dfrac{16-\sqrt{25540}}{602}\end{matrix}\right.\)

b: \(=2x^2-3x+10x-15=2x^2+7x-15\)