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A = \(\dfrac{2}{5}:\dfrac{3}{4}+\dfrac{2}{5}.\dfrac{3}{6}+\dfrac{2}{5}\)
A=\(\dfrac{2}{5}.\dfrac{4}{3}+\dfrac{2}{5}.\dfrac{3}{6}+\dfrac{2}{5}\)
A=\(\dfrac{2}{5}.\left(\dfrac{4}{3}+\dfrac{3}{6}+1\right)\) =\(\dfrac{2}{5}.\dfrac{17}{6}\)= \(\dfrac{17}{5}\)
B = \(\dfrac{5}{21}.\dfrac{4}{11}+\dfrac{7}{11}.\dfrac{5}{21}-\dfrac{2}{3}\)
B =\(\dfrac{5}{21}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)-\dfrac{2}{3}\)
B= \(\dfrac{5}{21}.1-\dfrac{2}{3}\) = \(\dfrac{5}{21}-\dfrac{2}{3}=\dfrac{-3}{7}\)
A -2/13+-11/26
=-4/26+-11/26
=-15/26
B 3/30+-1/5
=3/30+-6/30
=-3/30=-1/10
C -3/17+(2/3+3/17)
=( -3/17+3/17)+2/3
=2/3
D -5/7+3/4+-1/5+-2/7+1/4
=( -5/7+ -2/7)+(3/4+1/4)+ -1/5
=(-1)+1+ -1/5
=0+-1/5= -1/5
Bài 3:
a. Ta có: \(2^{30}=2^{3.10}=\left(2^3\right)^{10}=8^{10}\)
\(3^{20}=3^{2\cdot10}=\left(3^2\right)^{10}=9^{10}\)
Vì 8 < 9 nên \(8^{10}< 9^{10}\)
Vậy \(2^{30}< 3^{20}\)
b. Ta có: \(3^{400}=3^{4.100}=\left(3^4\right)^{100}=81^{100}\)
\(4^{300}=4^{3.100}=\left(4^3\right)^{100}=64^{100}\)
Vì 81 > 64 nên \(81^{100}>64^{100}\)
Vậy \(3^{400}>4^{300}\)
a)17.85-15.17-120
=17.(85-17)-120
=17.68-12
=1156-120
=1036
b)100:[30-(6-1)^ 2]
=100 :[30-5^ 2]
= 100:[30-25]
=100:5=20
c)/-10/+4+3.2^3+(-14)
=10+4+3.8 +(-14)
=14+24+(-14)
=[14 +(-14)]+24
=0+24=24
d)20-[30-(5-1)^ 2: 2]
=20-[30-4^2:2]
=20-[30-8]
=20-22=-2
e)80-(4.5^2-3.2^3)
= 80-(4.25-3.8)
=80-(4.25-3.2.4)
=80-4.(25-3.2)
= 80-4.19
= 80-76=4
g)35-{12-[-14+6-2)]}
=35 -{12-14-6+2}
=35-12+14+6-2
=23+14+6-2
=37+6-2
=43-2=41
a. \(3^6:3^2+2^3.2^3\)
\(=3^{6-2}+2^{3+3}\)
\(=3^4+2^6\)
\(=81+64=145\)
b. \(3.5^2-16:2^2\)
\(=3.25-16:4\)
\(=75-4=71\)
c. \(3^3.17+3^3.83\)
\(=3^3.\left(17+83\right)\)
\(=27.100=2700\)
d. \(20-\left[30-\left(5-1\right)^2\right]\)
\(=20-\left[30-4^2\right]\)
\(=20-\left[30-16\right]\)
\(=20-14=6\)
Ta có:A=2+22+23+...+220
=>2A=2(2+22+23+...+220)
2A=4+23+24+...+221
=>2A-A=(4+23+24+...+221)-(2+22+23+...+220)
A=221-2
A= 2.( 2+ 2+ 3+ 4+ 5+ 6+ 7+8 +9 + 10= 11+ 12+ 13+ 14+15+ 16+ 17+ 18+ 19+ 20)
A=2.211
A= 422
B= 5.(1+ 5+ 2+ 3+ 4+ 5+ 6+ 7+ 8+ 9+..........+30)
B= 5. 470
B= 2350