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

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

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

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

         Vậy đa thức vô nghiệm

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

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

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

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

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

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

Vậy: Phương trình vô nghiệm

(2x−1)³−3(x+2)(x−3)=(3+2x)³−3x(x+1)

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

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

<=> \(x=\frac{-21\pm5\sqrt{33}}{48}\)

Thay : \(x=3\) vào phương trình :

\(12-2\cdot\left(1-3\right)^2=4\cdot\left(3-m\right)-\left(3-3\right)\cdot\left(2\cdot3+5\right)\)

\(\Leftrightarrow12-8=12-4m\)

\(\Leftrightarrow4m=8\)

\(\Leftrightarrow m=2\)

thay x=3 vào pt ta được 

\(12-2\left(2-3\right)^2=4\left(3-m\right)-\left(3-3\right)\left(2x+5\right)\)

\(12-2\left(4-12+9\right)=12-4m\)

\(12-8+24-18-12=-4m\)

\(-2=-4m=>m=\dfrac{1}{2}\)

vậy để pt có nghiệm x=3 thì m=\(\dfrac{1}{2}\)

từ nãy mk ghi đề bàu bị sai nhé thông cảm

sửa lại thay x=3 vào pt ta được

\(12-2\left(1-3\right)^2=4\left(3-m\right)-\left(3-3\right)\left(2x+5\right)\)

\(12-8=12-4m\)

\(-8=-4m=>m=2\)

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.

\(\Leftrightarrow4x^2+4x+1-3\left(x^2-4x+4\right)+2\left(x^2+x-2\right)=4-2+2x\)

\(\Leftrightarrow4x^2+4x+1-3x^2+12x-12+2x^2+2x-4=2x+2\)

\(\Leftrightarrow3x^2+18x-15-2x-2=0\)

\(\Leftrightarrow3x^2+16x-17=0\)

\(\text{Δ}=16^2-4\cdot3\cdot\left(-17\right)=460>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-16-2\sqrt{115}}{6}=\dfrac{-8-\sqrt{115}}{3}\\x_2=\dfrac{-8+\sqrt{115}}{3}\end{matrix}\right.\)

\(\frac{\left(2-3x\right)^2}{3}-\frac{\left(1+2x\right)^2}{2}=\frac{3}{4}-2\left(x-1\right)\left(x+2\right)+x\left(1+x\right)\)

\(\frac{2^2-12x-3x^2}{3}-\frac{1^2+4x+2x^2}{2}=\frac{3}{4}-\left(x^2+x-2\right)+3x\)

\(\frac{2.\left(4-12x-3x^2\right)}{6}-\frac{3.\left(1+4x+2x^2\right)}{6}=\frac{11}{4}-x^2+2x\)

\(\frac{8-24x-6x^2}{6}-\frac{3+12x+2x^2}{6}=\frac{11}{4}-x^2+2x\)

\(\frac{8-24x-6x^2-3-12x-2x^2}{6}=\frac{11}{4}-x^2+2x\)

\(\frac{5-36x-8x^2}{6}=\frac{11}{4}-x^2+2x\)

Chỗ đây thì mk chịu