Question

a) \((\dfrac{1}{3}-2x)^2+\dfrac{5}{4}=\dfrac{21}{16}\)

b) \(\dfrac{4-x}{3}=\dfrac{5}{2}\)

c) \(7\dfrac{1}{3}-\left|x-1\right|:2=\dfrac{5}{2}\)

Answer 1

a, \((\dfrac{1}{3}-2x)^2+\dfrac{5}{4}=\dfrac{21}{16}\)

\((\dfrac{1}{3})^2\) - (2x)² = \(\dfrac{1}{16}\)

=> \(\dfrac{1}{9}\)- (2x)²=\(\dfrac{1}{16}\)

=> (2x)²=\(\dfrac{7}{144}\)

=> 2².x²=\(\dfrac{7}{144}\)

=> 4.x² =\(\dfrac{7}{144}\)

=> x²= \(\dfrac{7}{576}\)

=>x= +\(\sqrt{\dfrac{7}{576}}\) hoặc - \(\sqrt{\dfrac{7}{576}}\)

b,\(\dfrac{4-x}{3}=\dfrac{5}{2}\)

=> (4-x).2 = 5.3

=>8-x.2 = 15

=> x.2 = 8-15

=>x.2 = -7

=> x= -\(\dfrac{7}{2}\)

c. 7\(\dfrac{1}{3}\)- | x-1| : 2= \(\dfrac{5}{2}\)

=>\(\dfrac{22}{3}\)-|x-1| .\(\dfrac{1}{2}\) =\(\dfrac{5}{2}\)

=> |x-1|.\(\dfrac{1}{2}\)=\(\dfrac{29}{6}\)

=> |x-1| =\(\dfrac{29}{3}\)

+) x-1 = \(\dfrac{29}{3}\)=> x=\(\dfrac{32}{3}\)

+) x-1 = -\(\dfrac{29}{3}\)=> x=-\(\dfrac{26}{3}\)

Vậy x= \(\dfrac{32}{3}\)hoặc x=-\(\dfrac{26}{3}\)