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a) Ta có: \(6^6:6^3+4^3\cdot4^3\cdot4^2\)
\(=6^3+4^8\)
\(=216+65536=65752\)
b) Ta có: (-137)+(-129)
=-137-129
=-266
c) Ta có: -84-36
=-(84+36)
=-120
d) Ta có: \(11\cdot37+63\cdot11-49\)
\(=11\cdot\left(37+63\right)-49\)
\(=11\cdot100-49\)
\(=1100-49=1051\)
e) Ta có: \(16:2^3+4:3^3-4\cdot3\)
\(=2^4:2^3+\dfrac{4}{27}-12\)
\(=2+\dfrac{4}{27}-12\)
\(=-10+\dfrac{4}{27}=\dfrac{-270}{27}+\dfrac{4}{27}=\dfrac{-266}{27}\)
1)
a) \(...=4^4-225=256-225=31\)
b) \(...=8.9.5+120=360-120=240\)
c) \(...=3^4-3^3=81-27=54\)
d) \(...=7^2-1=49-1=48\)
2) a) \(...=2^6=64\)
b) \(...=3^{15}:3^{10}=3^5=243\)
c) \(...=3^3-3^3=0\)
d) \(...=6^3+4^5=216+1024=1240\)
a)
204-84:12
=204-7
=197
b)
15.23+4.32-5.7
=15.8+4.9-5.7
=120+36-35
=156-35
=121
c)
56:53+23.22
=56-3+23+2
=53+25
=125+32
=157
d)
164.53+47.164
=(164.(53+47)
=164.100
=16400
tìm x:
219-7(x+1)=100
=>7(x+1)=219-100
=>7(x+1)=119
=>x+1=119:7
=>x+1=17
=>x=17-1
=>x=16
(3x-6).3=34
=>(3x-6).3=81
=>3x-6=81:3
=>3x-6=27
=>3x=27+6
=>3x=33
=>x=33:3
=>x=11
a, (\(\dfrac{9}{10}\) - \(\dfrac{15}{16}\)) \(\times\) ( \(\dfrac{5}{12}\) - \(\dfrac{11}{15}\) - \(\dfrac{7}{20}\))
= (\(\dfrac{72}{80}\) - \(\dfrac{75}{80}\)) \(\times\) (\(\)\(\dfrac{25}{60}\) - \(\dfrac{44}{60}\) - \(\dfrac{21}{60}\))
= - \(\dfrac{3}{80}\) \(\times\) (- \(\dfrac{2}{3}\))
= \(\dfrac{1}{40}\)
b, (-1)3 + (- \(\dfrac{2}{3}\))2 : 2\(\dfrac{2}{3}\) + \(\dfrac{5}{6}\)
= -13 + \(\dfrac{4}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{5}{6}\)
= -1 + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{3}{8}\) + \(\dfrac{5}{6}\)
= -1 + \(\dfrac{1}{6}\) + \(\dfrac{5}{6}\)
= -1 + 1
= 0
\(\dfrac{11}{6}+\dfrac{1}{4}=\dfrac{22}{12}+\dfrac{3}{12}=\dfrac{25}{12}\)
\(\dfrac{2}{5}-\dfrac{3}{8}=\dfrac{16}{40}-\dfrac{15}{40}=\dfrac{1}{40}\)
\(\dfrac{3}{10}-\dfrac{4}{15}=\dfrac{9}{30}-\dfrac{8}{30}=\dfrac{1}{30}\)
\(3+\dfrac{2}{5}=\dfrac{15}{5}+\dfrac{2}{5}=\dfrac{17}{5}\)
\(\dfrac{333}{777}+\dfrac{22}{55}=\dfrac{3}{7}+\dfrac{2}{5}=\dfrac{15}{35}+\dfrac{14}{35}=\dfrac{29}{35}\)
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