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

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

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

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

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

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

Vậy: Phương trình vô nghiệm

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.

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

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

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

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

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

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

Chỗ đây thì mk chịu

(x-2)³+2.(1+2x)²=(1+x)³-3(x-2)²-(x-1)

<=>x³-6x²+12x-8+2.(1+4x+4x²)=1+3x²+3x+x³-3.(x²-4x+4)-x+1

<=>x³-6x²+12x-8+2+8x+8x²=1+3x²+3x+x³-3x²+12x-12-x+1

<=>x³+2x²+20x-6=x³+14x+2

<=>2x²+6x-8=0

<=>2x²-2x+8x-8=0

<=>2x.(x-1)+8.(x-1)=0

<=>2(x-1)(x+4)=0

<=>x-1=0 hoặc x+4=0

<=>x=1 hoặc x=-4

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

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

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

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

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

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

(x -2)\(^3\) +(3x-2)\(^2\) -5x (x+1) = (1+x)\(^3\) - 2(2x+1)\(^2\)

<=> (x\(^3\) -3.x\(^2\).2+3.x.2\(^2\) -2\(^3\)) + [(3x)\(^2\) - 2.3x.2 +2\(^2\)] - (5x.x+ 5x .1) = (1\(^3\) + 3.1\(^2\).x+ 3.1.x\(^2\) + x\(^3\) )- [2((2x)\(^2\) +2.2x.1+ 1\(^2\))]

<=> (x\(^3\) - 6x\(^2\) + 12x - 8) + (9x\(^2\) -12x+ 4)- (5x\(^2\) + 5x) = (1+3x + 3x\(^2\) + x\(^3\)) - [ 2.(4x\(^2\) + 4x +1]= (1+3x + 3x\(^2\) + x\(^3\)) - ( 8x\(^2\)+ 8x +2)

<=> x\(^3\) - 6x\(^2\) + 12x - 8 + 9x\(^2\) -12x+ 4 - 5x\(^2\) - 5x        = 1+3x + 3x\(^2\) + x\(^3\) -  8x\(^2\) -8x - 2

<=>  x\(^3\) +(- 6x\(^2\) + 9x\(^2\) - 5x\(^2\) ) +(12x- 12x - 5x) + (-8 +4) = (1-2) + ( 3x-8x) +( 3x\(^2\) -  8x\(^2\) ) + x\(^3\)