Question

Tìm x biết:

a) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

b) \(\left(x^4\right)^2=\dfrac{x^{12}}{x^5}\left(x\ne0\right)\)

c) \(\left(2x+3\right)^2=\dfrac{9}{121}\)

d) \(\left(2x-1\right)^3=\dfrac{-8}{27}\)

Answer 1

c. \(^{ }\left(2x+3\right)^2=\dfrac{9}{121}\)

=> \(\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)

=> 2x +3 = \(\dfrac{3}{11}\) hoặc 2x+3 = \(\dfrac{-3}{11}\)

=> x= \(\dfrac{-15}{11}\) hoặc x = \(\dfrac{-18}{11}\)

Answer 2

d. \(\left(2x-1\right)^3=\dfrac{-8}{27}\)

=> \(\left(2x-1\right)^3=\left(\dfrac{-2}{3}\right)^3\)

=> 2x-1 = \(\dfrac{-2}{3}\)

=> x= \(\dfrac{1}{6}\)

Answer 3

a. \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

=> \(\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\)

=> x + \(\dfrac{1}{2}\) = \(\dfrac{1}{4}\) hoặc \(x+\dfrac{1}{2}=\dfrac{-1}{4}\)

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

Answer 4

a)(x+\(\dfrac{1}{2}\))²=\(\dfrac{1}{16}\)

(x+\(\dfrac{1}{2}\))²=\(\dfrac{1}{4^2}\)

x+\(\dfrac{1}{2}\) =\(\dfrac{1}{4}\)

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