\(2^2.6-18:3^2+15.2\)

\(=4.6-18:9+15.2\)

\(=24-2+30\)

\(=52\)

2².6-18:3²+15.2=

=4.6-18:9+30

=24-2+30

=22+30

=52

Chúc bn hc tốt

a: \(5^6:5^3+2^3:2^2\)

\(=5^3+2\)

=125+2

=127

b: \(3333:3+225:15\)

=1111+15

=1126

c: \(15\cdot2^3+4\cdot3^2-5\cdot7\)

\(=15\cdot8+4\cdot9-35\)

=120+36-35

=121

d: \(5\cdot4^2-18:3^2\)

\(=5\cdot16-18:9\)

=80-2=78

e: \(6^2:4\cdot3+2\cdot5^2\)

\(=36:4\cdot3+2\cdot25\)

=27+50

=77

f: \(80-\left(4\cdot5^2-3\cdot2^3\right)\)

=80-(100-24)

=80-76

=4

\(A=\dfrac{2^{20}\cdot3^{10}+2^9\cdot2^{12}\cdot3^9}{2^{18}\cdot5\cdot3\cdot3^8-2^{19}\cdot3^9}\)

\(=\dfrac{2^{20}\cdot3^9\left(3+2\right)}{2^{18}\cdot3^9\left(5-2\right)}=4\cdot\dfrac{5}{3}=\dfrac{20}{3}\)

a, = 3.25 + 15.4 - 16 : 2 = 75 + 60 - 8 = 135 - 8 = 127

a: =75+60-8=127

Ta có: \(7\cdot6^{10}\cdot2^{21}\cdot3^6-2^{19}\cdot6^{15}\)

\(=7\cdot3^{10}\cdot2^{10}\cdot2^{21}\cdot3^6-2^{19}\cdot3^{15}\cdot2^{15}\)

\(=7\cdot2^{31}\cdot3^{16}-2^{34}\cdot3^{15}\)

\(=2^{31}\cdot3^{15}\left(7\cdot3-2^3\right)=2^{31}\cdot3^{15}\cdot13\)

Ta có: \(9\cdot6^{15}\cdot2^{13}-4\cdot3^{15}\cdot2^{26}\)

\(=3^2\cdot3^{15}\cdot2^{15}\cdot2^{13}-2^2\cdot3^{15}\cdot2^{26}\)

\(=3^{17}\cdot2^{28}-3^{15}\cdot2^{28}=3^{15}\cdot2^{28}\left(3^2-1\right)=3^{15}\cdot2^{31}\)

Ta có: \(B=\frac{7\cdot6^{10}\cdot2^{21}\cdot3^6-2^{19}\cdot6^{15}}{9\cdot6^{15}\cdot2^{13}-4\cdot3^{15}\cdot2^{26}}\)

\(=\frac{2^{31}\cdot3^{15}\cdot13}{3^{15}\cdot2^{31}}=13\)

204-84:2=204-42=162

15.2³+4.3²-5.7=15.8+4.9-35=120+36-35=121

5⁶:5³+2³.2²=15625:125+8.4=125+32=157

164.53+47.164=164.(53+47)=164.100=16400

HỌC TỐT.

mình sai ở đâu vậy mà bạn k sai

Các bạn giúp mk với tối nay nộp rùi 

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2:

=>27:3^x=2*25-16-31=50-47=3

=>3^x=27/3=9

=>3^x=3^2

=>x=2

1:

b: \(=16\cdot55+16\cdot45-1\)

=16(55+45)-1

=1600-1

=1599

c: \(=\dfrac{1800}{49-\left[2\cdot\left(36-34\right)^3-5\right]}\)

\(=\dfrac{1800}{49-2\cdot2^3+5}=\dfrac{1800}{49-16+5}=\dfrac{1800}{38}\)=900/19

d: \(=\dfrac{5\cdot3^{11}\cdot2^{11}-3^{11}\cdot2^{11}}{2^{10}\cdot3^{10}\cdot2^2\cdot3+7\cdot2^{12}\cdot3^{12}}\)

\(=\dfrac{3^{11}\cdot2^{11}\left(5-1\right)}{2^{12}\cdot3^{11}\left(1+7\cdot3\right)}=\dfrac{1}{2}\cdot\dfrac{4}{1+21}=\dfrac{4}{22\cdot2}=\dfrac{1}{11}\)