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13 tháng 7 2018
Tính:
\(A=\left(-2\right)^3.\left(-5\right)^2-41.\left(-8\right)\)
\(=\left(-8\right).25-41.\left(-8\right)\)
\(=\left(-8\right).\left(25-41\right)\)
\(=\left(-8\right).\left(-16\right)\)
\(=128\)
\(B=2018-2018:\left(4^2-3^2\right)\)
\(=2018-2018:\left(16-9\right)\)
\(=2018-2018:5\)
\(=2018-403,6\)
\(=1614,4\)
\(C=\left(-5\right).\left(-25\right).\left(-50\right)+5.25.50\)
\(=\left(-6250\right)+6250\)
\(=0\)
19 tháng 9 2017
a.=47-[45.16-25.12]:14]
=47-420:14
=47-30
=17
b.=50-[12:2+34]
=50-40
=10
c.=100-60:10
=100-6
=94
d.50-[(50-40);2+3]
=50-(10:2+3)
=50-8
=42
2 tháng 10 2015
bài A và B nè bạn!
A=1+3+32+...+3100
3A=3+32+33+...+3101
=>3A+1=1+3+32+...+3100+3101=A+3101
=>3A-A=3101-1
2A=3101-1
A=(3101-1)/2
B=1+4+42+...+450
4B=4+42+...+451
4B+1=1+4+42+...+450+451=B+451
=>4B-B=451-1
3B=451-1
B=(451-1)/3
HD
28 tháng 7 2017
\(a)47-\left[\left(45.2^4-5^2.12\right):14\right]\)
\(=47-\left[\left(45.16-25.12\right):14\right]\)
\(=47-\left[\left(720-300\right):14\right]\)
\(=47-\left[420:14\right]\)
\(=47-30\)
\(=17\)
\(b)50-\left[\left(20-2^3\right):2+34\right]\)
\(=50-\left[\left(20-8\right)\right]:2+34\)
\(=50-\left[12:2+34\right]\)
\(=50-\left[6+34\right]\)
\(=50-40\)
\(=10\)
\(c)10^2-\left[60:\left(5^6:5^4-3,5\right)\right]\)
\(=10^2-\left[60:\left(5^2-3,5\right)\right]\)
\(=10^2-\left[60:\left(25-3,5\right)\right]\)
\(=10^2-\left[60:21,5\right]\)
\(=100-\dfrac{120}{43}\)
\(=\dfrac{4180}{43}\)
\(d)50-\left[\left(50-2^3.5\right):2+3\right]\)
\(=50-\left[\left(50-8.5\right):2+3\right]\)
\(=50-\left[\left(50-40\right):2+3\right]\)
\(=50-\left[10:2+3\right]\)
\(=50-5+3\)
\(=50-8\)
\(=42\)
\(e)10-\left[\left(8^2-48\right).5+\left(2^3.10+8\right)\right]:28\)
\(=10-\left[\left(64-48\right).5+\left(8.10+8\right)\right]:28\)
\(=10-\left[16.5+\left(80+8\right)\right]:28\)
\(=10-\left[80+88\right]:28\)
\(=10-168:28\)
\(=10-6\)
\(=4\)
\(f)8697-\left[3^7:3^5+2.\left(13-3\right)\right]\)
\(=8697-\left[3^2+2.\left(13-3\right)\right]\)
\(=8697-\left[9+2.10\right]\)
\(=8697-9+20\)
\(=8697-29\)
\(=8668\)
\(g)2011+5\left[300-\left(17-7\right)^2\right]\)
\(=2011+5.\left[300-10^2\right]\)
\(=2011+5.\left[300-100\right]\)
\(=2011+5.200\)
\(=2011+1000\)
\(=3011\)
\(h)695-\left[200+\left(11-1\right)^2\right]\)
\(=695-\left[200+10^2\right]\)
\(=695-200+100\)
\(=695-300\)
\(=395\)
\(i)129-5\left[29-\left(6-1\right)^2\right]\)
\(=129-5\left[29-5^2\right]\)
\(=129-5\left[29-25\right]\)
\(=129-5.4\)
\(=129-20\)
\(=109\)
VS
0
HM
1
23 tháng 5 2017
C1:A = \(\frac{10^{50}+2}{10^{50}-1}=\frac{10^{50}-1+3}{10^{50}-1}=\frac{10^{50}-1}{10^{50}-1}+\frac{3}{10^{50}-1}\)
= \(1+\frac{3}{10^{50}-1}\)
B = \(\frac{10^{50}}{10^{50}-3}=\frac{10^{50}-3+3}{10^{50}-3}=\frac{10^{50}-3}{10^{50}-3}+\frac{3}{10^{50}-3}\)
= \(1+\frac{3}{10^{50}-3}\)
Vì \(\frac{3}{10^{50}-1}< \frac{3}{10^{50}-3}\)=) \(1+\frac{3}{10^{50}-1}< 1+\frac{3}{10^{50}-3}\)=) \(A< B\)
C2: Áp dụng tính chất : Nếu \(\frac{a}{b}>1\)=) \(\frac{a}{b}>\frac{a+m}{b+m}\)
Vì B > 1 =) B > \(\frac{10^{50}+2}{10^{50}-3+2}=\frac{10^{50}+2}{10^{50}-1}=A\)
(=) B > A
TN
13 tháng 7 2018
\(\left(-2\right)^3.\left(-5\right)^2-41.\left(-8\right)\)
\(=\)\(-8.25-41.\left(-8\right)\)
\(=-8\left(25-41\right)\)
\(=-8.\left(-16\right)\)
\(=128\)
\(2018-1018:\left(4^2-3^2\right)\)
\(=2018-1018:7\)
\(=2018-\frac{1018}{7}\)
\(=\frac{13108}{7}\)
\(-5.\left(-25\right).\left(-50\right)+5.25.50\)
\(=-6250+6250\)
\(=0\)
28 tháng 11 2015
a,
2.(-25).(-4).50
=(2.50).[(-4).(-25)]
=100.100
=10 000
b,
(-5)2.(-3)3.23
=25 . (-27).8
=(25.8).(-27)
=200 .(-27)
=-5400
=100-[60:(25-15)]=100-(60:10)=100-6=94
\(=100-\left[60:\left(25-15\right)\right]=94\)