Tính:
a) \(\frac{20^5.5^{10}}{100^5}\) b) \(\frac{\left(0.9\right)^5}{\left(0,3\right)^6}\) c) \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
BÀI 2 TÍNH
A = \(\dfrac{20^5.5^{10}}{100^5}\)
B = \(\dfrac{\left(0,9\right)^5}{\left(0,3\right)^6}\)
GIÚP MÌNH VỚI MN ƠIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII
\(A=\dfrac{\left(20.5\right)^5.5^5}{100^5}=\dfrac{100^5.3125}{100^5}=3125\)
\(B=\dfrac{\left(0,3.3\right)^5}{\left(0,3\right)^5.0,3}=\dfrac{\left(0,3\right)^5.3^5}{\left(0,3\right)^5.0,3}=\dfrac{3^5}{0,3}=810\)
\(A=\dfrac{2^{10}\cdot5^5\cdot5^{10}}{2^{10}\cdot5^{10}}=5^5=3125\\ B=\left(\dfrac{0,9}{0,3}\right)^5\cdot\dfrac{1}{0,3}=\dfrac{3^5}{0,3}=\dfrac{243}{0,3}=810\)
Tính giá trị biểu thức"
a) \(\frac{20^5.5^{10}}{100^5}\) b) \(\frac{\left(0,9\right)^5}{\left(0,3\right)^{^6}}\) c) \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(\frac{20^5.5^{10}}{100^5}.\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(\left(\frac{4}{9}+\frac{1}{3}\right)^2+\left(\frac{3}{4}\right)^3:\left(\frac{3}{4}\right)^2:\left(\frac{-2}{3}\right)^3\)
Tính giá trị của các biểu thúc sau:
a) \(\frac{20^5.5^{10}}{100^5}\) c)\(\frac{6^3+3.6^2+3^3}{-13}\)
b) \(\frac{\left(0,9\right)^5}{\left(0,3\right)^6}\) d) \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
a, 205.510/1005
=205.55.55/1005
=1005.55/1005
=55
=3125
b, (0,9)5/(0,3)6
=(0,3.3)5/0,36
=0,55.35/0,36
=35/0,3
=810
c, 63+3.62+33/-13
=(2.3)3+3.(3.2)2+33/-13
=23.33+3.32.22+33/-13
=33.23+33.22+33/-13
=33(23+22+1)/-13
=27.13/-13
=-27
d, 46.95+69.120/84.312-611
=(22)6.(32)5+(2.3)9.3.23.5/(23)4.312-(2.3)11
=212.310+29.39.3.23.5/212.312-211.311
=212.310+212.310.5/211.311.2.3-211.311
=212.310.(1+5)/211.311(6-1)
=212.310.6/211.311.5
=2.6/3.5
=12/15
=4/5
Tính:
a) \(\left( {\frac{3}{4}:1\frac{1}{2}} \right) - \left( {\frac{5}{6}:\frac{1}{3}} \right)\)
b) \(\left[ {\left( {\frac{{ - 1}}{5}} \right):\frac{1}{{10}}} \right] - \frac{5}{7}.\left( {\frac{2}{3} - \frac{1}{5}} \right)\)
c) \(\left( { - 0,4} \right) + 2\frac{2}{5}.{\left[ {\left( {\frac{{ - 2}}{3}} \right) + \frac{1}{2}} \right]^2}\)
d)\(\left\{ {\left[ {{{\left( {\frac{1}{{25}} - 0,6} \right)}^2}:\frac{{49}}{{125}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 1}}{3}} \right) + \frac{1}{2}} \right]\)
a)
\(\begin{array}{l}\left( {\frac{3}{4}:1\frac{1}{2}} \right) - \left( {\frac{5}{6}:\frac{1}{3}} \right)\\ = \left( {\frac{3}{4}:\frac{3}{2}} \right) - \left( {\frac{5}{6}.3} \right)\\ = \left( {\frac{3}{4}.\frac{2}{3}} \right) - \frac{5}{2}\\ = \frac{1}{2} - \frac{5}{2}\\ = \frac{-4}{2}\\= - 2.\end{array}\)
b)
\(\begin{array}{l}\left[ {\left( {\frac{{ - 1}}{5}} \right):\frac{1}{{10}}} \right] - \frac{5}{7}.\left( {\frac{2}{3} - \frac{1}{5}} \right)\\ = \left( {\frac{{ - 1}}{5}} \right).10 - \frac{5}{7}.\left( {\frac{{10}}{{15}} - \frac{3}{{15}}} \right)\\ = - 2 - \frac{5}{7}.\frac{7}{{15}}\\ = - 2 - \frac{1}{3}\\ = \frac{{ - 6}}{3} - \frac{1}{3}\\ = \frac{{ - 7}}{3}\end{array}\)
c)
\(\begin{array}{l}\left( { - 0,4} \right) + 2\frac{2}{5}.{\left[ {\left( {\frac{{ - 2}}{3}} \right) + \frac{1}{2}} \right]^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left[ {\left( {\frac{{ - 4}}{6}} \right) + \frac{3}{6}} \right]^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left( {\frac{{ - 1}}{6}} \right)^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.\frac{1}{{36}}\\ = \left( { - \frac{2}{5}} \right) + \frac{1}{{15}}\\ = \left( { - \frac{6}{{15}}} \right) + \frac{1}{{15}}\\ = \frac{{ - 5}}{{15}}\\ = \frac{{ - 1}}{3}\end{array}\)
d)
\(\begin{array}{l}\left\{ {\left[ {{{\left( {\frac{1}{{25}} - 0,6} \right)}^2}:\frac{{49}}{{125}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 1}}{3}} \right) + \frac{1}{2}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{1}{{25}} - \frac{3}{5}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 2}}{6}} \right) + \frac{3}{6}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{{ 1}}{{25}}-\frac{15}{25}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left\{ {\left[ {{{\left( {\frac{{ - 14}}{{25}}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left\{ {\frac{{196}}{{{{25}^2}}}.\frac{{25.5}}{{49}}.\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left( {\frac{{4.49.25.5.5}}{{{{25}^2}.49.6}}} \right) - \frac{1}{6}\\ = \frac{4}{6} - \frac{1}{6}\\ = \frac{3}{6}\\ = \frac{1}{2}\end{array}\)
Tính
a) \(\left(\frac{2}{5}\right)^5:\left(\frac{9}{25}\right)^5\)
\(b.25.5^3.\frac{1}{625}.5^2\)
c. \(\frac{20^5.5^{10}}{100^5}\)
\(\left(\frac{1}{7}\right)^2.\frac{1}{7}.49^2\)
\(a.\left(\frac{2}{5}\right)^5:\left(\frac{9}{25}\right)^5=\left(\frac{2\cdot25}{9\cdot5}\right)^5=\frac{10}{9}^5\)
\(b.25\cdot5^3\cdot\frac{1}{625}\cdot5^2=\frac{5^7}{5^4}=5^3\)
\(c.\frac{20^5\cdot5^{10}}{100^5}=\frac{2^{10}\cdot5^{15}}{2^{10}\cdot5^{10}}=5^5\)
\(d.\frac{1}{7}^2\cdot\frac{1}{7}\cdot49^2=\frac{7^4}{7^3}=7\)
Tính:
a)\(1\frac{1}{2} + \frac{1}{5}.\left[ {\left( { - 2\frac{5}{6} + \frac{1}{3}} \right)} \right];\)
b)\(\frac{1}{3}.\left( {\frac{2}{5} - \frac{1}{2}} \right):{\left( {\frac{1}{6} - \frac{1}{5}} \right)^2}.\)
a)
\(\begin{array}{l}1\frac{1}{2} + \frac{1}{5}.\left[ {\left( { - 2\frac{5}{6} + \frac{1}{3}} \right)} \right]\\ = \frac{3}{2} + \frac{1}{5}.\left[ {\left( { - \frac{{17}}{6} + \frac{2}{6}} \right)} \right]\\ = \frac{3}{2} + \frac{1}{5}.\frac{{ - 15}}{6}\\ = \frac{3}{2} + \frac{{ - 1}}{2}\\ = \frac{2}{2}\\=1\end{array}\)
b)
\(\begin{array}{l}\frac{1}{3}.\left( {\frac{2}{5} - \frac{1}{2}} \right):{\left( {\frac{1}{6} - \frac{1}{5}} \right)^2}\\ = \frac{1}{3}.\left( {\frac{4}{{10}} - \frac{5}{{10}}} \right):{\left( {\frac{5}{{30}} - \frac{6}{{30}}} \right)^2}\\ = \frac{1}{3}.\frac{{ - 1}}{{10}}:{\left( {\frac{{ - 1}}{{30}}} \right)^2}\\ = \frac{{ - 1}}{{30}}:\frac{1}{{{{30}^2}}}\\ = \frac{{ - 1}}{{30}}{.30^2}\\ = - 30\end{array}\)
Bỏ dấu ngoặc rồi tính:
a)\(\left( {\frac{{ - 3}}{7}} \right) + \left( {\frac{5}{6} - \frac{4}{7}} \right);\)
b)\(\frac{3}{5} - \left( {\frac{2}{3} + \frac{1}{5}} \right);\)
c)\(\left[ {\left( {\frac{{ - 1}}{3} + 1} \right) - \left( {\frac{2}{3} - \frac{1}{5}} \right)} \right];\)
d)\(1\frac{1}{3} + \left( {\frac{2}{3} - \frac{3}{4}} \right) - \left( {0,8 + 1\frac{1}{5}} \right)\).
a)
\(\begin{array}{l}\left( {\frac{{ - 3}}{7}} \right) + \left( {\frac{5}{6} - \frac{4}{7}} \right)\\ = \left( {\frac{{ - 3}}{7}} \right) + \frac{5}{6} - \frac{4}{7}\\ = \left[ {\left( {\frac{{ - 3}}{7}} \right) - \frac{4}{7}} \right] + \frac{5}{6}\\ =\frac{-7}{7}+\frac{5}{6}\\= - 1 + \frac{5}{6}\\ = \frac{{ - 1}}{6}\end{array}\)
b)
\(\begin{array}{l}\frac{3}{5} - \left( {\frac{2}{3} + \frac{1}{5}} \right)\\ = \frac{3}{5} - \frac{2}{3} - \frac{1}{5}\\ = (\frac{3}{5} - \frac{1}{5}) - \frac{2}{3}\\ = \frac{2}{5} - \frac{2}{3}\\ = \frac{6}{{15}} - \frac{{10}}{{15}}\\ = \frac{{ - 4}}{{15}}\end{array}\)
c)
\(\begin{array}{l}\left[ {\left( {\frac{{ - 1}}{3}} \right) + 1} \right] - \left( {\frac{2}{3} - \frac{1}{5}} \right)\\ = \left( {\frac{{ - 1}}{3}} \right) + 1 - \frac{2}{3} + \frac{1}{5}\\ = \left( {\frac{{ - 1}}{3} - \frac{2}{3}} \right) + 1 + \frac{1}{5}\\ = \frac{-3}{3}+1+\frac{1}{5}\\= - 1 + 1 + \frac{1}{5}\\ = \frac{1}{5}\end{array}\)
d)
\(\begin{array}{l}1\frac{1}{3} + \left( {\frac{2}{3} - \frac{3}{4}} \right) - \left( {0,8 + 1\frac{1}{5}} \right)\\ = 1 + \frac{1}{3} + \frac{2}{3} - \frac{3}{4} - \left( {\frac{4}{5} + 1 + \frac{1}{5}} \right)\\=1+\frac{3}{3}-\frac{3}{4}-(\frac{5}{5}+1)\\ = 1 + 1 - \frac{3}{4} - (1+1)\\ = - \frac{3}{4}\end{array}\).
Tính giá trị biểu thức:
a. \(\frac{20^5\times5^{10}}{100^5}\)
b. \(\frac{\left(0,9\right)^5}{\left(0,3\right)^6}\)
c. \(\frac{6^3+3\times6^2+3^3}{-13}\)
d. \(\frac{4^6\times9^5+6^9\times120}{8^4\times3^{12}-6^{11}}\)
e. \(\left(2^{-1}+3^{-1}\right):\left(2^{-1}-3^{-1}\right)+\left(2^{-1}\times2^0\right)\times2^3\)
f. \(\left(\frac{-1}{3}\right)^{-1}-\left(\frac{-3}{5}\right)^6+\left(\frac{1}{2}\right)^2:2\)
a. \(\frac{20^5.5^{10}}{100^5}\)
\(=\frac{20^5.\left(5^2\right)^5}{100^5}\)
\(=\frac{20^5.25^5}{100^5}\)
\(=\frac{500^5}{100^5}\)
\(=\left(\frac{500}{100}\right)^5\)
\(=5^5=3125\)
b. \(\frac{\left(0,9\right)^5}{\left(0,3\right)^6}\)
\(=\frac{\left(0,9\right)^5}{\left(0,3\right)^5.0,3}\)
\(=\left(\frac{0,9}{0,3}\right)^5.\frac{1}{0,3}\)
\(=3^5.\frac{1}{0,3}\)
\(=810\)
c. \(\frac{6^3+3.6^2+3^3}{-13}\)
\(=\frac{\left(3.2\right)^3+3.\left(3.2\right)^2+3^3}{-13}\)
\(=\frac{3^3\left(2^3+2^2+1\right)}{-13}\)
\(=\frac{3^3.13}{-13}\)
\(=\left(-3\right)^3\)
\(=-27\)