Question

Tìm x :

a. \(15-2\left|x+\dfrac{1}{3}\right|=\dfrac{4}{7}\)

b. \(\dfrac{13}{11}.\dfrac{22}{26}-x^2=\dfrac{7}{16}\)

Answer 1

a, 15 - 2 | x + \(\dfrac{1}{3}\)| = \(\dfrac{4}{7}\)

=>2| x + \(\dfrac{1}{3}\)| = 15 - \(\dfrac{4}{7}\)

=> 2 | x + \(\dfrac{1}{3}\)| =\(\dfrac{101}{7}\)

=> | x + \(\dfrac{1}{3}\)| = \(\dfrac{101}{14}\)

=> x+ \(\dfrac{1}{3}\)= \(\dfrac{101}{14}\) hoặc x +\(\dfrac{1}{3}\)= - \(\dfrac{101}{14}\)

+) x+\(\dfrac{1}{3}\)= \(\dfrac{101}{14}\)=> x = \(\dfrac{302}{3}\)

+) x + \(\dfrac{1}{3}\)= - \(\dfrac{101}{14}\)=> x =-\(\dfrac{317}{42}\)

Vậy ...........

b, \(\dfrac{13}{11}\).\(\dfrac{22}{26}\)-x²=\(\dfrac{7}{16}\)

=>1 -x² = \(\dfrac{7}{16}\)

=> x²= \(\dfrac{9}{16}\)

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

Vậy ..................

Answer 2

\(a,15-2\left|x+\dfrac{1}{3}\right|=\dfrac{4}{7}\)

\(2\left|x+\dfrac{1}{3}\right|=15-\dfrac{4}{7}\)

\(2\left|x+\dfrac{1}{3}\right|=\dfrac{101}{7}\)

\(\left|x+\dfrac{1}{3}\right|=\dfrac{101}{7}:2\)

\(\left|x+\dfrac{1}{3}\right|=\dfrac{101}{14}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{101}{14}\) hoặc \(x+\dfrac{1}{3}=-\dfrac{101}{14}\)

\(x=\dfrac{101}{14}-\dfrac{1}{3}\) \(x=-\dfrac{101}{14}-\dfrac{1}{3}\)

\(x=\dfrac{302}{3}\) \(x=-\dfrac{317}{42}\)