Question

Answer 1

a, \(\left(x-\dfrac{1}{4}\right)^2=0\)

\(\Leftrightarrow x-\dfrac{1}{4}=0\)

\(\Leftrightarrow x=\dfrac{1}{4}\)

b, \(\left(x+\dfrac{1}{3}\right)^2=4\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\)

Answer 2

a,(x-1/4)^2=0

<=>x-1/4=0

<=>x=1/4

b,\(\left(x+\dfrac{1}{3}\right)^2=4\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=2\\x+\dfrac{1}{3}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\)

tick mik nha

Answer 3

a) \(\left(x-\dfrac{1}{4}\right)^2=0\Rightarrow x-\dfrac{1}{4}=0\Rightarrow x=\dfrac{1}{4}\)

b) \(\left(x+\dfrac{1}{3}\right)^2=4\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=2\\x+\dfrac{1}{3}=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\)
 

Answer 4

a) \(\left(x-\dfrac{1}{4}\right)^2=0\Leftrightarrow x-\dfrac{1}{4}=0\Leftrightarrow x=\dfrac{1}{4}\)

b) \(\left(x+\dfrac{1}{3}\right)^2=4\)\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=2\\x+\dfrac{1}{3}=-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\)

c) \(\left(x-\dfrac{1}{5}\right)^3=-27\Leftrightarrow x-\dfrac{1}{5}=-3\Leftrightarrow x=-\dfrac{14}{5}\)

d) \(\left(2x+1\right)^4=\dfrac{1}{16}\)\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{1}{2}\\2x+1=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

e) \(\left(x-1\right)^2-25=0\Leftrightarrow\left(x-1\right)^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

Answer 5

a: Ta có: \(\left(x-\dfrac{1}{4}\right)^2=0\)

\(\Leftrightarrow x-\dfrac{1}{4}=0\)

hay \(x=\dfrac{1}{4}\)

b: Ta có: \(\left(x+\dfrac{1}{3}\right)^2=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=2\\x+\dfrac{1}{3}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\)

c: Ta có: \(\left(x-\dfrac{1}{5}\right)^3=-27\)

\(\Leftrightarrow x-\dfrac{1}{5}=-3\)

hay \(x=-\dfrac{14}{5}\)