Question

1. Tìm x \(\in\) Q, biết

a) (x-\(\dfrac{1}{3}\))\(^2\) = 0

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

c) (2x-1)\(^3\) = 8

Giúp mk vs

Answer 1

a, \(\left(x-\dfrac{1}{3}\right)^2=0\)

=> \(x-\dfrac{1}{3}=0\)

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

Vậy \(x=\dfrac{1}{3}\)

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

=>\(\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{8}\right)^2\)

=> \(x+\dfrac{1}{2}=\dfrac{1}{8}\)

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

c, (2x - 1)^3 = 8

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

=> 2x - 1 = 2

=> 2x = 3

=> x = 3/2

Answer 2

a) (x - \(\dfrac{1}{3}\))²=0

=> x- \(\dfrac{1}{3}\)=0

x=\(\dfrac{1}{3}\)

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

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

^{TH1: x+ \(\dfrac{1}{2}\)}=\(\dfrac{1}{4}\)

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

TH2 : x + \(\dfrac{1}{2}\)= \(\dfrac{-1}{4}\)

x = \(\dfrac{-3}{4}\)

Vậy x = \(\dfrac{-1}{4}\); \(\dfrac{-3}{4}\)

c) (2x-1)³ =8

=> 2x - 1 = 2

2x = 3

x = \(\dfrac{3}{2}\)