`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`

`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`

a) x = 1; x = - 1 3                 b) x = 2.

c) x = 3; x = -2.                 d) x = -3; x = 0; x = 2.

a) Rút gọn VT = 45x + 8. Từ đó tìm được x = 2 15 .  

b) Rút gọn VT = -25x – 8. Từ đó tìm được x = − 11 25 .

Bài 2:

a: Ta có: \(A=\left(x+1\right)^3+\left(x-1\right)^3\)

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

\(=2x^3+6x\)

b: Ta có: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)

\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)

\(=27x-55\)

a) Rút gọn được VT = 9x + 7. Từ đó tìm được x = 1.

b) Rút gọn được VT = 2x + 8. Từ đó tìm được x = 7 2 .

a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)

d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

a: \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

=>\(x^2+8x+16-\left(x^2-1\right)=16\)

=>\(x^2+8x+16-x^2+1=16\)

=>8x+17=16

=>8x=-1

=>\(x=-\frac18\)

b: \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

=>\(4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

=>\(5x^2+2x+10-5x^2+245=0\)

=>2x+255=0

=>2x=-255

=>\(x=-\frac{255}{2}\)

c: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)

=>\(x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=49\)

=>\(x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)

=>24x+25=49

=>24x=24

=>x=1

a: \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

=>\(x^2+8x+16-\left(x^2-1\right)=16\)

=>\(x^2+8x-x^2+1=0\)

=>8x+1=0

=>8x=-1

=>\(x=-\frac18\)

b: \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\cdot\left(x-7\right)=0\)

=>\(4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

=>\(5x^2+2x+10-5x^2+245=0\)

=>2x+255=0

=>2x=-255

=>\(x=-\frac{255}{2}\)

c: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)

=>\(x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=49\)

=>\(x^3-6x^2+12x-8-x^3+27+6x^2+12x+6\) =49

=>24x+25=49

=>24x=24

=>x=1