Tính :
a) (-7).8
b) (-3).(-10).(-5)
c) (-10)2.4
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a,(0,8)5:(0,4)6 = (\(\dfrac{0,8}{0,4}\))5 : 0,4 = 25:0,4 = 80
b, (-25)7: 323 - \(\dfrac{6103515625}{3276}\) = - 186264,5149
c, \(\dfrac{4^2.4^3}{2^{10}}\) = \(\dfrac{4^5}{2^{10}}\) = \(\dfrac{2^{10}}{2^{10}}\) = 1
d, \(\dfrac{9^5.5^7}{45^7}\) = \(\dfrac{9^5.5^7}{9^7.5^7}\) = \(\dfrac{1}{81}\)
\(\dfrac{\left(0.8\right)^5}{\left(0.4\right)^4}\)=\(\dfrac{\left(2.0,4\right)^5}{\left(0.4^4\right)}\)=\(\dfrac{2^5.\left(0.4\right)^5}{\left(0,4\right)^4}\)=\(2^5\).\(\left(0.4\right)^1\)=12,8
b)câu b không biết có sai đề không nhưng đáp án câu b là -186264,5149
c) \(\dfrac{4^2.4^3}{2^{10}}\)=\(\dfrac{4^5}{\left(2^2\right)^5}\)=\(\dfrac{4^5}{4^5}\)=1
d)\(\dfrac{9^5.5^7}{45^7}\)=\(\dfrac{9^5.5^5.5^2}{45^7}\)=\(\dfrac{45^5.5^2}{45^7}\)=\(\dfrac{5^2}{45^2}\)=\(\left(\dfrac{5}{45}\right)^2\)=\(\left(\dfrac{1}{9}\right)^2\)=\(\dfrac{1}{81}\)
a)\(P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{46}-\frac{1}{56}\)
=\(1-\frac{1}{56}=\frac{55}{56}\)
b)\(A.\frac{1}{3}=\frac{1}{3}.\left(\frac{3}{1.2}+\frac{3}{2.3}+....+\frac{3}{99.100}\right)\)
= \(\frac{1}{3}A=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{3}{99.100}\)
=> \(\frac{1}{3}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
=> \(\frac{1}{3}A=1-\frac{1}{100}=\frac{99}{100}\)
=> \(A=\frac{99}{100}.3=\frac{297}{100}\)
c)\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)
=\(1-\frac{1}{103}=\frac{102}{103}\)
d) \(\frac{3}{5}C=\frac{3}{5}.\left(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\right)\)
=\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\)
=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{103}\)
=\(1-\frac{1}{103}=\frac{102}{103}\)
=>\(C=\frac{102}{103}.\frac{5}{3}=\frac{170}{103}\)
e) \(\frac{4}{7}D=\frac{4}{7}.\left(\frac{7}{1.5}+\frac{7}{5.9}+...+\frac{7}{101.105}\right)\)
=\(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{101.105}\)
=\(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\)
=\(1-\frac{1}{105}=\frac{104}{105}\)
=< D=\(\frac{104}{105}.\frac{7}{4}=\frac{26}{15}\)
\(A=1+2+3+...+7+8=\dfrac{\left(8+1\right).\left(\dfrac{8-1}{1}+1\right)}{2}=36\)
\(B=3+4+5+...+10+11=\dfrac{\left(11+3\right).\left(\dfrac{11-3}{1}+1\right)}{2}=63\)
\(C=1+3+5+...+13+15=\dfrac{\left(15+1\right).\left(\dfrac{15-1}{2}+1\right)}{2}=64\)
\(D=2+4+6+...+18+20=\dfrac{\left(20+2\right).\left(\dfrac{20-2}{2}+1\right)}{2}=110\)
\(E=1+4+7+...+22+25=\dfrac{\left(25+1\right).\left(\dfrac{25-1}{3}+1\right)}{2}=117\)
\(G=1+5+9+...+33+37+41=\dfrac{\left(41+1\right).\left(\dfrac{41-1}{4}+1\right)}{2}=231\)
a) 17.13+17.42-17.35
=17.(13+42-35)
=17.20=340
b) [25.(18-42)-10]:4+6
=(25.2-10):4+6
=40:4+6=16
c) 36:32+23.22-32.3
=34+25-33
=81+32-27=86
d) B=3.42-22.3
=3.(16-4)
=3.12=36
e)20220+3.[52.10-(23-13)2]
=1+3.(250-100)
=1+450=451
g) 27.77+24.27-27
=27.(77+24-1)
=27.100=2700
h) 5.23+79:77-12020
=40+72-1
=89-1=88
i) 120:{54[50:2+(32-2.4)]}
=120:[54(25+1)]
=120:1404=10/117
A = 102 - ( 52 . 4 - 43 . 3 ) + 23
A = 100 - ( 25 . 4 - 64 . 3 ) + 8
A = 100 - ( 100 - 192 ) + 8
A = 100 - ( - 92 ) + 8
A = 192 + 8
A = 200
10^2 - ( 5^2 . 4 - 4^3 . 3 ) + 2^3
= 100 - ( 25 . 4 - 64 . 3 ) + 8
= 100 - ( 100 - 192 ) + 8
= 100 - ( -92 ) + 8
= 192 + 8
= 200
= 251
a: 3/4=54/72
-1/9=-8/72
-5/8=-45/72
b: -1/7=-8/56
1/8=7/56
3/4=42/56
c: 1/3=10/30
-1/5=-6/30
-1/10=-3/30
a)\(\dfrac{3}{4}=\dfrac{3.18}{4.18}=\dfrac{54}{72}\)
\(\dfrac{1}{-9}=\dfrac{-1}{9}=\dfrac{-1.8}{9.8}=\dfrac{-8}{72}\)
\(\dfrac{-5}{8}=\dfrac{-5.9}{8.9}=\dfrac{-45}{72}\)
b)\(\dfrac{1}{-7}=\dfrac{-1}{7}=\dfrac{-1.8}{7.8}=\dfrac{-8}{56}\)
\(\dfrac{-1}{-8}=\dfrac{1}{8}=\dfrac{1.7}{8.7}=\dfrac{7}{56}\)
\(\dfrac{3}{4}=\dfrac{3.14}{4.14}=\dfrac{42}{56}\)
c)\(\dfrac{1}{3}=\dfrac{1.10}{3.10}=\dfrac{10}{30}\)
\(\dfrac{-1}{5}=\dfrac{-1.6}{5.6}=\dfrac{-6}{30}\)
\(\dfrac{1}{-10}=\dfrac{-1}{10}=\dfrac{-1.3}{10.3}=\dfrac{-3}{30}\)
A=10^2-(5^2.4-4^3.3)+2^3
A=100-(25.4-64.3)+2^3
A=100-(100-192)+8
A=100-100+192+8
A=0+192+8
A=200
a) (-7).8 = -56
b) (-3) . (-10) . (-5) = 30 . (-5) = -150
c) (-10)2 . 4 = 100 . 4 = 400
a) -56
b) -150
c) 400