Question

Tìm x ( bằng phương pháp dùng hằng đẳng thức )

1 ) \(3x^2-75=0\)

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

3 ) \(x^3+3x^2+3x=0\)

Answer 1

1)3.x^2 - 75 = 0

3.x^2 - 3.25 = 0

3.(x^2-25)=0

x^2-5^2=0

(x-5)(x+5)=0

=> x-5=0 hoặc x+5=0

=> x=5 hoặc x=-5

   

Answer 2

1) \(3x^2-75=0\)

\(\Leftrightarrow3\left(x^2-25\right)=0\)

\(\Leftrightarrow x^2-25=0\)

\(\Leftrightarrow x^2=25\)

\(\Leftrightarrow x=\pm\sqrt{25}=\pm5\)

2) \(x^3+9x^2+27x+27=0\)

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

\(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)

3) \(x^3+3x^2+3x=0\)

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

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

\(\Leftrightarrow x+1=1\Leftrightarrow x=0\)

Answer 3

2, x^3 + 9x^2 +27x + 27 = 0

x^3 + 3.x^2.3 + 3.x.3^2 +3^3 = 0

(x+3)^3=0

x+3=0

x=-3