Question

Tìm \(x\in Q,\) \(biết:\)

\((x-\dfrac{1}{8})^3=\dfrac{27}{64}\)

Answer 1

\(\left(x-\dfrac{1}{8}\right)^3=\dfrac{27}{64}\\ \Leftrightarrow\left(x-\dfrac{1}{8}\right)^3=\left(\dfrac{3}{4}\right)^3\)

\(\Leftrightarrow x-\dfrac{1}{8}=\dfrac{3}{4}\\ \Leftrightarrow x=\dfrac{3}{4}+\dfrac{1}{8}\\ \Leftrightarrow x=\dfrac{7}{8}\)

Answer 2

\(\left(x-\dfrac{1}{8}\right)^3=\dfrac{27}{64}\)

\(\Rightarrow\left(x-\dfrac{1}{8}\right)^3=\left(\dfrac{3}{4}\right)^3\)

\(\Rightarrow x-\dfrac{1}{8}=\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{3}{4}+\dfrac{1}{8}\)

Vậy \(x=\dfrac{7}{8}\).

Answer 3

(x-1/8)^3 = (3/4)^3

-> x-1/8 = 3/4

->x = 7/8

Vậy x = 7/8

Answer 4

\(\left(x-\dfrac{1}{8}\right)^3=\dfrac{27}{64}\)

\(\Rightarrow\left(x-\dfrac{1}{8}\right)^3=\left(\dfrac{3}{4}\right)^3\)

\(\Rightarrow x-\dfrac{1}{8}=\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{1}{8}+\dfrac{3}{4}=\dfrac{7}{8}\)

Answer 5

\(\left(x-\dfrac{1}{8}\right)^3=\dfrac{27}{64}\)

\(\left(x-\dfrac{1}{8}\right)^3=\left(\dfrac{3}{4}\right)^3\)

=>\(x-\dfrac{1}{8}=\dfrac{3}{4}\)

Đến đây bn tự làm nha!