Tính giá trị biểu thức \(M = \left( {\dfrac{2}{3} - \dfrac{1}{4} + 2} \right) - \left( {2 - \dfrac{5}{2} + \dfrac{1}{4}} \right) - \left( {\dfrac{5}{2} - \dfrac{1}{3}} \right)\).
\(M = \left( {\dfrac{2}{3} - \dfrac{1}{4} + 2} \right) - \left( {2 - \dfrac{5}{2} + \dfrac{1}{4}} \right) - \left( {\dfrac{5}{2} - \dfrac{1}{3}} \right)\)
\( = \dfrac{2}{3} - \dfrac{1}{4} + 2 - 2 + \dfrac{5}{2} - \dfrac{1}{4} - \dfrac{5}{2} + \dfrac{1}{3}\)
\( = \left( {\dfrac{2}{3} + \dfrac{1}{3}} \right) + \left( {2 - 2} \right) + \left( {\dfrac{5}{2} - \dfrac{5}{2}} \right) + \left( { - \dfrac{1}{4} - \dfrac{1}{4}} \right)\)
\( = 1 + 0 + 0 - \dfrac{1}{2}\)
\( = \dfrac{1}{2}\)
Vậy \(M = \dfrac{1}{2}\) .
Cho \({x_1}\) là giá trị thỏa mãn \(\dfrac{3}{7} + \dfrac{1}{7}:x = \dfrac{3}{{14}}\) và \({x_2}\) là giá trị thỏa mãn \(\dfrac{5}{7} + \dfrac{2}{7}:x = 1.\) Khi đó, chọn câu đúng.
Ta có: \(\dfrac{3}{7} + \dfrac{1}{7}:x = \dfrac{3}{{14}}\)
\(\dfrac{1}{7}:x = \dfrac{3}{{14}} - \dfrac{3}{7}\)
\(\dfrac{1}{7}:x = \dfrac{3}{{14}} - \dfrac{6}{{14}}\)
\(\dfrac{1}{7}:x = \dfrac{{ - 3}}{{14}}\)
\(x = \dfrac{1}{7}:\left( {\dfrac{{ - 3}}{{14}}} \right)\)
\(x = \dfrac{1}{7}.\dfrac{{14}}{{\left( { - 3} \right)}}\)
\(x = - \dfrac{2}{3}\)
Vậy \({x_1} = - \dfrac{2}{3}\)
* \(\dfrac{5}{7} + \dfrac{2}{7}:x = 1\)
\(\dfrac{2}{7}:x = 1 - \dfrac{5}{7}\)
\(\dfrac{2}{7}:x = \dfrac{2}{7}\)
\(x = \dfrac{2}{7}:\dfrac{2}{7}\)
\(x = 1\)
Vậy \({x_2} = 1\) .
Mà \( - \dfrac{2}{3} < 0 < 1\) nên \({x_1} < {x_2}\) .
Tìm \(x\) biết \(\dfrac{{11}}{{12}} - \left( {\dfrac{2}{5} + x} \right) = \dfrac{2}{3}\)
Ta có \(\dfrac{{11}}{{12}} - \left( {\dfrac{2}{5} + x} \right) = \dfrac{2}{3}\)
\(\dfrac{2}{5} + x = \dfrac{{11}}{{12}} - \dfrac{2}{3}\)
\(\dfrac{2}{5} + x = \dfrac{{11}}{{12}} - \dfrac{8}{{12}}\)
\(\dfrac{2}{5} + x = \dfrac{3}{{12}}\)
\(x = \dfrac{1}{4} - \dfrac{2}{5}\)
\(x = \dfrac{5}{{20}} - \dfrac{8}{{20}}\)
\(x = \dfrac{{ - 3}}{{20}}\)
Vậy \(x = \dfrac{{ - 3}}{{20}}\).
Tính giá trị biểu thức: $A = \dfrac{{\dfrac{2}{3} - \dfrac{2}{5} + \dfrac{2}{{10}}}}{{\dfrac{8}{3} - \dfrac{8}{5} + \dfrac{8}{{10}}}} + \dfrac{1}{2}.$
$A = \dfrac{{\dfrac{2}{3} - \dfrac{2}{5} + \dfrac{2}{{10}}}}{{\dfrac{8}{3} - \dfrac{8}{5} + \dfrac{8}{{10}}}} + \dfrac{1}{2}$
$A = \dfrac{{\left( {\dfrac{2}{3} - \dfrac{2}{5} + \dfrac{2}{{10}}} \right)}}{{4.\left( {\dfrac{2}{3} - \dfrac{2}{5} + \dfrac{2}{{10}}} \right)}} + \dfrac{1}{2}$
$A = \dfrac{1}{4} + \dfrac{1}{2}$
$A = \dfrac{3}{4}.$
Giá trị của biểu thức $\dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + \dfrac{1}{{4.5}} + ... + \dfrac{1}{{2018.2019}}$ là
$\dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + \dfrac{1}{{4.5}} + ... + \dfrac{1}{{2018.2019}}$
$ = 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{5} + ... - \dfrac{1}{{2018}} + \dfrac{1}{{2018}} - \dfrac{1}{{2019}}$
$ = 1 - \dfrac{1}{{2019}}$
$ = \dfrac{{2018}}{{2019}}$ .
Thực hiện phép tính \(\dfrac{2}{9}.\left[ {\dfrac{{ - 4}}{{45}}:\left( {\dfrac{1}{5} - \dfrac{2}{{15}}} \right) + 1\dfrac{2}{3}} \right] - \left( {\dfrac{{ - 5}}{{27}}} \right)\) ta được kết quả là
Ta có \(\dfrac{2}{9}.\left[ {\dfrac{{ - 4}}{{45}}:\left( {\dfrac{1}{5} - \dfrac{2}{{15}}} \right) + 1\dfrac{2}{3}} \right] - \left( {\dfrac{{ - 5}}{{27}}} \right)\)
\( = \dfrac{2}{9}.\left[ {\dfrac{{ - 4}}{{45}}:\left( {\dfrac{3}{{15}} - \dfrac{2}{{15}}} \right) + \dfrac{5}{3}} \right] - \left( {\dfrac{{ - 5}}{{27}}} \right)\)
\( = \dfrac{2}{9}.\left[ {\dfrac{{ - 4}}{{45}}:\dfrac{1}{{15}} + \dfrac{5}{3}} \right] - \left( {\dfrac{{ - 5}}{{27}}} \right)\)
\( = \dfrac{2}{9}.\left[ {\dfrac{{ - 4}}{{45}}.\dfrac{{15}}{1} + \dfrac{5}{3}} \right] - \left( {\dfrac{{ - 5}}{{27}}} \right)\)
\( = \dfrac{2}{9}.\left[ {\dfrac{{ - 4}}{3} + \dfrac{5}{3}} \right] - \left( {\dfrac{{ - 5}}{{27}}} \right)\)
$ = \dfrac{2}{9}.\dfrac{1}{3} - \left( {\dfrac{{ - 5}}{{27}}} \right)$
\( = \dfrac{2}{{27}} + \dfrac{5}{{27}}\)
\( = \dfrac{7}{{27}}\)
