Question

[Toàn Nguyễn]

Tìm x thuộc Q, biết rằng:

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

b)  \(\left(x-2\right)^2=1\)

c)  \(\left(2x-1\right)^3=-8\)

d)  \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)

Answer 1

[o0o I am a studious person o0o]

a) \(\left(x-\frac{1}{2}\right)^2=0\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{2}\end{cases}}\)

b) \(\left(x-2\right)^2=1\)

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

\(\Leftrightarrow\left(x-2-1\right)\left(x-2+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)

c) \(\left(2x-1\right)^3=-8\)

\(\Leftrightarrow2x-1=-2\)

\(\Leftrightarrow2x=-1\)

\(\Leftrightarrow x=\frac{-1}{2}\)

d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)

\(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\)

\(\Rightarrow x=-\frac{1}{4}\)

Answer 2

[kakashi]

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

     (x-1/2)^2=0^2

=>x-1/2=0

   x=1/2

b>(x-2)^2=1

     (x-2)^2=1^2

    =>x-2=1

           x=3

(2x-1)^3=-8

=>(2x-1)^3=-2^3

  =>2x-1=-2

        2x=-1

         x=-0,5

d>(x+1/2)^2=1/16

   (x+1/2)^2=(1/4)^2

   =>x+1/2=1/4

             x=1/4-1/2

             x=-1/4

chuc ban may man trong cuoc song!!!!!