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) 23  .  2+45:3=43+45  .  13=43+415=1615\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) (34)5:(34)4+215  .  32285  .  910=34+215  .  322(23)5  .  (32)10{\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}}}}

=34+215  .  322215  .  320=34+32=34+9=334 = \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) 330,5x=26,75{3^3} - 0,5x = 26,75

270,5x=26,7527 - 0,5x = 26,75

0,5x=2726,750,5x = 27 - 26,75                                      

0,5x=0,250,5x = 0,25

x=0,5x = 0,5

Vậy x=0,5x = 0,5.

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

x132199= 19\left| {x - \frac{1}{3}} \right| \cdot 2 - \frac{{19}}{9} =  - \frac{1}{9}

x132= 19+199\left| {x - \frac{1}{3}} \right| \cdot 2 =  - \frac{1}{9} + \frac{{19}}{9}

x132=2\left| {x - \frac{1}{3}} \right| \cdot 2 = 2

x13=2:2\left| {x - \frac{1}{3}} \right| = 2:2

x13=1\left| {x - \frac{1}{3}} \right| = 1

TH1: x13=1x - \frac{1}{3} = 1

x=1+13x = 1 + \frac{1}{3}

x=43x = \frac{4}{3}

TH2: x13= 1x - \frac{1}{3} =  - 1

x= 1+13x =  - 1 + \frac{1}{3}

x=23x = \frac{{ - 2}}{3}

Vậy x{43;  23}x \in \left\{ {\frac{4}{3};\,\,\frac{{ - 2}}{3}} \right\}.

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