\(\frac{\sqrt{49}}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{\sqrt{81}}\)

\(\Rightarrow\frac{7}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{9}\)

\(\Rightarrow\frac{21}{18}< \left|x-\frac{12}{18}\right|< \frac{52}{18}\)

còn lại cậu tự tính nha

\(\frac{\sqrt{49}}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{\sqrt{81}}\)

\(\frac{7}{6}< x-\frac{2}{3}< \frac{26}{9}\)

\(\frac{11}{6}< x< \frac{32}{9}\)

Nhầm rồi phần e 

e, \(\frac{4}{9}-\frac{7}{8}.x=\frac{-2}{3}\) 

              \(\frac{7}{8}.x=\frac{4}{9}-\frac{-2}{3}\) 

               \(\frac{7}{8}.x=\frac{-10}{9}\)

                      \(x=\frac{-10}{9}:\frac{7}{8}\)

                      \(x=\frac{-80}{63}\)

=80/63 không có dấu trừ

a) (3 + 6 + 9 + ... + 303) x (245 - 49 x 5)

    (3 + 6 + 9 + ... + 303) x (245 - 245)

 (3 + 6 + 9 + ... + 303) x 0 = 0

\(\left(3+6+9+...+303\right)\times\left(245-49\times5\right)\)
\(=\left(3+6+9+...+303\right)\times\left(245-245\right)\)
\(=\left(3+6+9+...+303\right)\times0\)
\(=0\)

( 3 + 6 + 9 + .... + 303 ) x ( 245 - 49 x 5 )

= ( 3 + 6 + 9 + .... + 303 ) x ( 245 - 245 )

= ( 3 + 6 + 9 + .... + 303 ) x 0 

= 0

x - 30 : 6 = 8 x 9 

x - 5 = 72

    x  = 72 + 5 

   x   = 77

Vậy x = 77

a. x = 2

b. x = 0;1;2;3;4;5;6

c. x = 0

a)x-2=0 hoăc x+1=0

   x=0+2        x=0-1

   x=2            x=-1

a)(x-2) x (x+1)=0

<=>x-2=0 hoặc x+1=0

+)x-2=0=>x=0+2=>x=2

+)x+1=0=>x=0-1=>x=-1

vậy x=2 hay x=-1

két quả: x=35