Đặt x² + 3x + 3 = a ;  x² - x - 1 = b ; -2x² - 2x - 1 = c ; -1 = d

Ta nhận thấy a³ + b³ + c³ + d³ = 0 (1) 

và a + b + c + d = 0

Khi đó ta có (1) <=>  (a + b)³ + (c + d)³ - 3ab(a + b) - 3cd(c + d) = 0

<=> ab(a + b) + cd(c + d) = 0

<=> (a + b)(ab - cd) = 0   

<=> \(\left[{}\begin{matrix}a=-b\\ab=cd\end{matrix}\right.\)

Với a = -b ta được x² + 3x + 3 = -x² + x + 1

<=> x² + x + 1 = 0 

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

=> Phương trình vô nghiệm

Với ab = cd 

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

\(\Leftrightarrow\) \(x^4+2x^3-3x^2-8x-4=0\)

\(\Leftrightarrow\left(x^4+2x^3+x^2\right)-\left(4x^2+8x+4\right)=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\pm2\end{matrix}\right.\)

x = -1

f/ \(3xy\left(x+y\right)-\left(x+y\right)\left(x^2+y^2+2xy\right)+y^3=27\)

\(3x^2y+3xy^2-\left(x+y\right)\left(x+y\right)^2+y^3=27\)

\(3x^2y+3xy^3-\left(x+y\right)^3+y^3=27\)

\(3x^2y+3xy^3-\left(x^3+3x^2y+3xy^2+b^3\right)+y^3=27\)

\(-x^3=27\)

\(x=-3\)

Bài 1:

a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(6x-9+4-2x=-3\)

\(4x=-2\)

\(x=-\frac{1}{2}\)

b/ \(2x\left(x^2-2\right)+x^2\left(1-2x\right)-x^2=-12\)

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

\(-4x=-12\)

\(x=\frac{1}{3}\)

Bài 2:

a/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(6x^2+9x-6x^2-15x+4x+10=8\)

\(-2x=8\)

\(x=-4\)

b/ \(4x\left(x-1\right)-3\left(x^2-5\right)-x^2=\left(x-3\right)-\left(x+4\right)\)

\(4x^2-4x-3x^2+15-x^2=-7\)

\(-4x=-22\)

\(x=\frac{11}{2}\)

c/ \(2\left(3x-1\right)\left(2x+5\right)-6\left(2x-1\right)\left(x+2\right)=-6\)

\(6x-2\left(2x+5\right)-12x+6\left(x+2\right)=-6\)

\(6x-4x-10-12x+6x+12=-6\)

\(-4x=-8\)

\(x=2\)

`@` `\text {Ans}`

`\downarrow`

`a)`

`3x(4x-1) - 2x(6x-3) = 30`

`=> 12x^2 - 3x - 12x^2 + 6x = 30`

`=> 3x = 30`

`=> x = 30 \div 3`

`=> x=10`

Vậy, `x=10`

`b)`

`2x(3-2x) + 2x(2x-1) = 15`

`=> 6x- 4x^2 + 4x^2 - 2x = 15`

`=> 4x = 15`

`=> x = 15/4`

Vậy, `x=15/4`

`c)`

`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`

`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`

`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`

`=> 40x^2 -17x - 1 = 1`

`d)`

`(x+2)(x+2)-(x-3)(x+1)=9`

`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`

`=> 6x + 7 =9`

`=> 6x = 2`

`=> x=2/6 =1/3`

Vậy, `x=1/3`

`e)`

`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`

`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`

`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`

`=> 12x +8 = 0`

`=> 12x = -8`

`=> x= -8/12 = -2/3`

Vậy, `x=-2/3`

`g)`

`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`

`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`

`=> -3x + 4 =14`

`=> -3x = 10`

`=> x= - 10/3`

Vậy, `x=-10/3`

Hello các bạn còn đó ko?

1) 2x.(5x-3x)+2x.(3x-5)-3.(x-7)=3

   10x-6x^2+6x^2-10x-3x+21=3

    -3x                             =-18

suy ra x=6

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

     3x^2 +3x-2x^2-4x =-1-x

     x^2 =-1

suy ra không có giá trị nào của x thỏa mãn đề bài

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

  2x^2 +3x^2-3 =5x^2+5x

  -5x      =3

x=-3/5

giải rồi đấy

nhớ tích đúng nha :)

bạn coi lại đề câu 1 đi

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