Thực hiện phép tính (tính hợp lí nếu có thể): a) 1/ 8 − 7 /11 + − 5/ 8 + 6/ 11 ; b) √ 25 /81 . ( − 3 )^2 − √ 1 /64 : ( − 1 /2 )^2 .

a) \[\frac{1}{8} - \frac{7}{{11}} + \frac{{ - 5}}{8} + \frac{6}{{11}} = \left( {\frac{1}{8} + \frac{{ - 5}}{8}} \right) - \left( {\frac{7}{{11}} - \frac{6}{{11}}} \right) = \frac{{ - 1}}{2} - \frac{1}{{11}} = \frac{{ - 13}}{{22}}\];

b) \[\sqrt {\frac{{25}}{{81}}} \,\,.\,\,{\left( { - 3} \right)^2} - \sqrt {\frac{1}{{64}}} :{\left( {\frac{{ - 1}}{2}} \right)^2} = \frac{4}{9}\,\,.\,\,9 - \frac{1}{8}:\frac{1}{4}\]

\[ = 4 - \frac{1}{8}\,\,.\,\,4 = 4 - \frac{1}{2} = \frac{7}{2}\].

2 (1,0 điểm).

a)\(\frac{1}{3}x - \frac{5}{2} = \frac{{ - 19}}{9}\)

\(\frac{1}{3}x = \frac{{ - 19}}{9} + \frac{5}{2}\)

\(\frac{1}{3}x = \frac{7}{{18}}\)        

\(x = \frac{7}{{18}}:\frac{1}{3}\)       

\(x = \frac{7}{6}\)           

Vậy \(x = \frac{7}{6}\).

b) \(\left( {x + 1} \right)\left( {\sqrt x - 5} \right) = 0\)

TH1: \(x + 1 = 0\)

\(x = - 1\)

TH2: \(\sqrt x - 5 = 0\)

\(\sqrt x = 5\)

\(x = 25\)

Vậy \(x \in \left\{ { - 1;\,\,25} \right\}\).