Question

Answer 1

a: Ta có: \(\left(x+3\right)^2-9=0\)

\(\Leftrightarrow\left(x+3\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=3\\x+3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

b: Ta có: \(\left(x-6\right)^2=x-6\)

\(\Leftrightarrow\left(x-6\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=7\end{matrix}\right.\)

c: Ta có: \(\left(2x-3\right)^2-\left(2x+3\right)^2=0\)

\(\Leftrightarrow-24x=0\)

hay x=0

Answer 2

a. (x + 3)² - 9 = 0

<=> (x + 3)² - 3² = 0

<=> (x + 3 - 3)(x + 3 + 3) = 0

<=> x(x + 6) = 0

<=> \(\left[{}\begin{matrix}x=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

b. x - 6 = (6 - x)²

<=> x - 6 = 36 - 12x + x²

<=> -x² + 12x + x - 6 - 36 = 0

<=> -x² + 13x - 42 = 0

<=> -(-x² + 13x - 42) = 0

<=> x² - 13x + 42 = 0

<=> x² - 6x - 7x + 42 = 0

<=> x(x - 6) - 7(x - 6) = 0

<=> (x - 7)(x - 6) = 0

<=> \(\left[{}\begin{matrix}x-7=0\\x-6=0\end{matrix}\right.\left[{}\begin{matrix}x=7\\x=6\end{matrix}\right.\)

c. (2x - 3)² - (3 + 2x)² = 0

<=> (2x - 3 - 3 - 2x)(2x - 3 + 3 + 2x) = 0

<=> -6 . 4x = 0

<=> 4x = 0

<=> x = 0

d. x³ - 6x² + 12x - 8 = 0

<=> x³ - 3.2.x² + 3.2².x - 2³ = 0

<=> (x - 2)³ = 0

<=> x - 2 = 0

<=> x = 0 + 2

<=> x = 2

Answer 3

a. \(\left(x+3\right)^2-9=0\)

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

\(\Leftrightarrow x\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

b. \(x-6=\left(6-x\right)^2\)

\(\Leftrightarrow x-6=\left(x-6\right)^2\)

\(\Leftrightarrow\left(x-6\right)\left(1-x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\7-x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=7\end{matrix}\right.\)

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

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

\(\Leftrightarrow-24x=0\) \(\Leftrightarrow x=0\)

d. \(x^3-6x^2+12x-8=0\)

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

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Answer 4

d: ta có: \(x^3-6x^2+12x-8=0\)

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

\(\Leftrightarrow x-2=0\)

hay x=2