Question

tìm x biét

a, \(\left(x+2\right)^2-9=0\)

b, \(\left(x+2\right)^2-x^2+4=0\)

c, \(\left(x-3\right)^2=\left(2-3x\right)^2\)

d, \(x^3-6x^2+12x-8=0\)

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

Answer 1

a, (a, (x + 2)² - 9 = 0

⇒ (x + 2)² = 0 + 9 = 9

⇒ (x + 2)² = \(\left(\pm3\right)^2\)

⇒ x + 2 = \(\pm3\)

\(\Rightarrow\left\{{}\begin{matrix}x+2=3\\x+2=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3-2\\x=-3-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

Vậy x ∈ {1; -5}

b, \(\left(x+2\right)^2-x^2+4=0\)

⇒ x² + 4x + 4 - x² + 4 =0

⇒ 4x + 8 = 0

⇒ 4 (x + 2) = 0

⇒ x + 2 = 0

⇒ x = 0 - 2

⇒ x = -2

Vậy x = -2

c, (x - 3)² = (2 - 3x)²

⇒ (x - 3)² - (2 - 3x)² = 0

⇒ x² - 6x + 9 - 4 + 12x - 9x² = 0

⇒ 6x - 8x² + 5 = 0

⇒2 \(\left(3x-4x^2+\dfrac{5}{2}\right)\)= 0

⇒ 3x - 4x² + \(\dfrac{5}{2}\) = 0

⇒ - (4x²- 3x + \(\dfrac{9}{16}+\dfrac{31}{16}\)) = 0

⇒ - (4x² - 3x + \(\dfrac{9}{16}\)) - \(\dfrac{31}{16}\) = 0

⇒ - (2x - \(\dfrac{3}{4}\))² = \(\dfrac{31}{16}\) (vô lí)

Vậy x ∈ ∅

Answer 2

Answer 3

e: \(\Leftrightarrow\left(x+1+x-2\right)\left(x^2+2x+1-x^2+x+2+x^2-4x+4\right)=0\)

=>2x-1=0

=>x=1/2

a: \(\Leftrightarrow\left(x+2+3\right)\left(x+2-3\right)=0\)

=>(x+5)(x-1)=0

=>x=1 hoặc x=-5

b: \(\Leftrightarrow x^2+4x+4-x^2+4=0\)

=>4x+8=0

=>x=-2

c: \(\Leftrightarrow\left(3x-2-x+3\right)\left(3x-2+x-3\right)=0\)

=>(2x+1)(4x-5)=0

=>x=5/4 hoặc x=-1/2