\(a,x^2-16=0\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x=\orbr{\begin{cases}4\\-4\end{cases}}\)
\(b,x^3+\frac{1}{125}=0\)
\(\Rightarrow x^3=-\frac{1}{125}\)
\(\Rightarrow x=-\frac{1}{5}\)
a. x2 - 16 = 0
x2 = 0 + 16 = 16
=> x = 4 ; -4
b.x3 + 1/125 = 0
x3 = 0 - 1/125 = -1/125
=> x = -1/5
Vậy x ...
x2–16=0
==> x2=16
==> x2=42 hoặc x2=(—4)2
==> x=4 hoặc x=—4
x3+1/125=0
x3+ 1/53=0
x3=0–1/53
x3=—1/53
==> x=—1/5
\(a,x-16=0\)
\(\Rightarrow x^2=0+16=16\)
\(\Rightarrow x=4=-4\)
Vậy ...
\(b,x^3+\frac{1}{125}=0\)
\(\Rightarrow x^3=\frac{-1}{125}\)
\(\Rightarrow x=\frac{-1}{5}\)
Vậy ...
a, \(a^2-16=0\)
\(a^2=0+16\)
\(a^2=16\)
\(a^2=4^2\)
\(a=4\)
Vậy \(a=4\)
b, \(x^3+\frac{1}{125}=0\)
\(x^3=0+\frac{1}{125}\)
\(x^3=\frac{1}{125}\)
\(x^3=\left(\frac{1}{5}\right)^3\)
\(x=\frac{1}{5}\)
a) \(x^2-16=0\)
\(x^2=0+16\)
\(x^2=16\)
\(\Rightarrow x=4^2\)hoặc \(x=\left(-4\right)^2\)
Vậy \(x=4\)hoặc \(x=-4\)
b) \(x^3+\frac{1}{125}=0\)
\(x^3=0+\frac{1}{125}\)
\(x^3=\frac{1}{125}\)
\(x^3=\left(\frac{1}{5}\right)^3\)
\(\Rightarrow x=\frac{1}{5}\)
Vậy \(x=\frac{1}{5}\)