Ta có: \(\left(x+\frac{1}{2}\right)^2-\frac{1}{16}=0\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
Mà \(\frac{1}{16}=\left(\frac{1}{4}\right)^2\)
\(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\Rightarrow x=\frac{-1}{4}\)
Vậy ....
\(\left(3x+\frac{1}{2}\right)^2+\frac{25}{16}=0\)
\(\Rightarrow\left(3x+\frac{1}{2}\right)^2=\frac{-25}{16}\)
Vì \(\left(3x+\frac{1}{2}\right)^2\ge0\left(\forall x\in Z\right)\)
Nên x thuộc rỗng (không có giá trị của x)
a) (x + 1/2)^2 - 1/16 = 0
(x+1/2)^2 = 1/16 = (1/4)^2 = (-1/4)^2
TH1: x + 1/2 = 1/4
x = -1/4
TH2: x + 1/2 = -1/4
x = -3/4
KL:...
b) (3x+1/2)^2 + 25/16 = 0
(3x + 1/2)^2 = -25/16
=> không tìm được x
a.\(\left(x+\frac{1}{2}\right)^2-\frac{1}{16}=0\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Leftrightarrow x+\frac{1}{2}=\frac{1}{4}\)
\(\Leftrightarrow x=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}\)
Vậy \(x=-\frac{1}{4}\)
b.\(\left(3x+\frac{1}{2}\right)^2+\frac{25}{16}=0\)
\(\Leftrightarrow\left(3x+\frac{1}{2}\right)^2=-\frac{25}{16}\)
vì \(\left(3x+\frac{1}{2}\right)^2\ge0\) mà \(-\frac{25}{16}< 0\) nên \(x\in\varnothing\)
