Question

2.Tìm x biết:

a)(x-4)(x+8)-(x+3)^2=1

b)(3-x)(3+x)+(x-2)^2=5

c)(x-3)^3-(x-3)^3-18x^2-36x=5

d)(x-3)^3-x^2(x-9)=0

Answer 1

a: \(\Leftrightarrow x^2+4x-32-x^2-6x-9=1\)

=>-2x-41=1

=>-2x=42

hay x=-21

b: \(\Leftrightarrow9-x^2+x^2-4x+4=5\)

=>-4x+13=5

=>-4x=-8

hay x=2

c: \(\Leftrightarrow x^3-9x^2+27x-27-x^3+9x^2-27x+27-18x^2-36x=5\)

=>-36x=5

hay x=-5/36

Answer 2

a) \(\left(x-4\right)\left(x+8\right)-\left(x+3\right)^2=1\)

\(x^2+8x-4x-32-x^2-6x-9=1\)

\(-2x=42\)

\(x=-21\)

Vậy \(x=-21\)

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

\(9-x^2+x^2-4x+4=5\)

\(-4x=-8\)

\(x=2\)

Vậy \(x=2\)

c) Sửa đề: \(\left(x+3\right)^3-\left(x-3\right)^3-18x^2-36x=5\)

\(x^3+9x^2+27x+27-x^3+9x^2-27x+27-18x^2-36x-5=0\)

\(-36x+49=0\)

\(x=\dfrac{49}{36}\)

Vậy \(x=\dfrac{49}{36}\)

d) \(\left(x-3\right)^3-x^2\left(x-9\right)=0\)

\(x^3-9x^2+27x-27-x^3+9x^2=0\)

\(27x-27=0\)

\(x=1\)

Vậy \(x=1\)