Question

a.(x-1/2)^2=0

b.(x-2)^2=1

c.(2x-1)^3=-8

d.(x+1/2)^2=1/6

Answer 1

a) \(\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)

b) \(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

c) \(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)

\(\Leftrightarrow2x-1=-3\Leftrightarrow x=-1\)

d) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{\sqrt{6}}{6}\\x+\dfrac{1}{2}=-\dfrac{\sqrt{6}}{6}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3+\sqrt{6}}{6}\\x=-\dfrac{3+\sqrt{6}}{6}\end{matrix}\right.\)

Answer 2

b: Ta có: \(\left(x-2\right)^2=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

c: Ta có: \(\left(2x-1\right)^3=-8\)

\(\Leftrightarrow2x-1=-2\)

\(\Leftrightarrow2x=-1\)

hay \(x=-\dfrac{1}{2}\)

Answer 3

a. \(\left(x-\dfrac{1}{2}\right)^2=0\)

<=> \(x-\dfrac{1}{2}=0\)

<=> \(x=\dfrac{1}{2}\)

b. (x - 2)² = 1

<=> (x - 2)² - 1² = 0

<=> (x - 2 - 1)(x - 2 + 1) = 0

<=> (x - 3)(x - 1) = 0

<=> \(\left[{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

c. (2x - 1)³ = -8

<=> (2x - 1)³ + 2³ = 0

<=> (2x - 1 + 2)\(\left[\left(2x-1\right)^2-2\left(2x-1\right)+2^2\right]=0\)

<=> (2x + 1)(4x² - 4x + 1 - 4x + 2 + 4) = 0

<=> (2x + 1)(4x² - 8x + 7) = 0

<=> \(\left[{}\begin{matrix}2x+1=0\\4x^2-8x+7=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\S=\varnothing\end{matrix}\right.\)

d. \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{6}\)

<=> \(\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{\left(\sqrt{6}\right)^2}=0\)

<=> \(\left(x+\dfrac{1}{2}-\dfrac{1}{\sqrt{6}}\right)\left(x+\dfrac{1}{2}+\dfrac{1}{\sqrt{6}}\right)=0\)

<=> \(\left(x+\dfrac{3-\sqrt{6}}{6}\right)\left(x+\dfrac{3+\sqrt{6}}{6}\right)=0\)

<=> \(\left[{}\begin{matrix}x+\dfrac{3-\sqrt{6}}{6}=0\\x+\dfrac{3+\sqrt{6}}{6}=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{-3+\sqrt{6}}{6}\\x=\dfrac{-3-\sqrt{6}}{6}\end{matrix}\right.\)