Câu hỏi:

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Tự luận

1. Thực hiện phép tính (tính nhanh nếu có thể)

a) 23  .  2+45:3\frac{{ - 2}}{3}\,\,.\,\,2 + \frac{4}{5}:3;                                               

b) (34)5:(34)4+215  .  32285  .  910{\left( {\frac{{ - 3}}{4}} \right)^5}:{\left( {\frac{{ - 3}}{4}} \right)^4} + \frac{{{2^{15}}\,\,.\,\,{3^{22}}}}{{{8^5}\,\,.\,\,{9^{10}}}}.

2. Tìm xx, biết:

a) 330,5x=26,75{3^3} - 0,5x = 26,75;                            

b) x132219= (13)2\left| {x - \frac{1}{3}} \right| \cdot 2 - 2\frac{1}{9} =  - {\left( { - \frac{1}{3}} \right)^2} .

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Lời giải

Hướng dẫn giải:

1.

a) \(\frac{{ - 2}}{3}\,\,.\,\,2 + \frac{4}{5}:3 = \frac{{ - 4}}{3} + \frac{4}{5}\,\,.\,\,\frac{1}{3} = \frac{{ - 4}}{3} + \frac{4}{{15}} = \frac{{ - 16}}{{15}}\);                                           

b) \({\left( {\frac{{ - 3}}{4}} \right)^5}:{\left( {\frac{{ - 3}}{4}} \right)^4} + \frac{{{2^{15}}\,\,.\,\,{3^{22}}}}{{{8^5}\,\,.\,\,{9^{10}}}} = \frac{{ - 3}}{4} + \frac{{{2^{15}}\,\,.\,\,{3^{22}}}}{{{{\left( {{2^3}} \right)}^5}\,\,.\,\,{{\left( {{3^2}} \right)}^{10}}}}\)

\( = \frac{{ - 3}}{4} + \frac{{{2^{15}}\,\,.\,\,{3^{22}}}}{{{2^{15}}\,\,.\,\,{3^{20}}}} = \frac{{ - 3}}{4} + {3^2} = \frac{{ - 3}}{4} + 9 = \frac{{33}}{4}\).

2.

a) \({3^3} - 0,5x = 26,75\)

\(27 - 0,5x = 26,75\)

\(0,5x = 27 - 26,75\)                                      

\(0,5x = 0,25\)

\(x = 0,5\)

Vậy \(x = 0,5\).

b) \(\left| {x - \frac{1}{3}} \right| \cdot 2 - 2\frac{1}{9} =  - {\left( { - \frac{1}{3}} \right)^2}\)

\(\left| {x - \frac{1}{3}} \right| \cdot 2 - \frac{{19}}{9} =  - \frac{1}{9}\)

\(\left| {x - \frac{1}{3}} \right| \cdot 2 =  - \frac{1}{9} + \frac{{19}}{9}\)

\[\left| {x - \frac{1}{3}} \right| \cdot 2 = 2\]

\[\left| {x - \frac{1}{3}} \right| = 2:2\]

\[\left| {x - \frac{1}{3}} \right| = 1\]

TH1: \[x - \frac{1}{3} = 1\]

\[x = 1 + \frac{1}{3}\]

\[x = \frac{4}{3}\]

TH2: \[x - \frac{1}{3} =  - 1\]

\[x =  - 1 + \frac{1}{3}\]

\[x = \frac{{ - 2}}{3}\]

Vậy \[x \in \left\{ {\frac{4}{3};\,\,\frac{{ - 2}}{3}} \right\}\].

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