35^10 * 5^10
------------------
49^5 * 125
giúp mình với
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\(\frac{1.3.5+2.6.10+4.12.20+7.21.35}{1.5.7+2.10.14+4.20.28+7.35.49}\)
\(=\frac{1.3.5\left(1+2+4+7\right)}{1.5.7\left(1+2+7+7\right)}=\frac{1.3.5}{1.5.7}=\frac{15}{35}=\frac{3}{7}\)
\(\frac{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+7\cdot21\cdot35}{1\cdot5\cdot7+2\cdot10\cdot14+4\cdot20\cdot28+7\cdot35\cdot49}\)
\(=\)\(\frac{1\cdot3\cdot5\cdot\left(1+2+4+7\right)}{1\cdot5\cdot7\cdot\left(1+2+7+7\right)}\)
\(=\frac{1\cdot3\cdot5}{1\cdot5\cdot7}\)\(=\frac{15}{35}=\frac{3}{7}\)
bn vội quá viết nhầm lun kìa
hj hj chúc bn làm bài tốt nha
\(\frac{35^{10}.5^{10}}{49^5.125}=\frac{5^{10}.7^{10}.5^{10}}{\left(7^2\right)^5.5^3}=\frac{5^{20}.7^{10}}{7^{10}.5^3}=\frac{5^{17}.1}{1.1}=5^{17}\)
a . 24/50 < 51/100 b . 2003/2013 < 2004/2014 c . 2121/3131 = 21212121/31313131
1/2×(2×x-3)+105/2=-137/2
1/2×(2×x-3)=-137/2-105/2
1/2×(2×x-3)=-242/2
2×x-3=-242/2:1/2
2×x-3=-242
2.x=(-242)+3
2.x=239
x=239:2
x=239/2
A,-12×25-25×75-25×13
=[(-12)-75-13].25
=(-100).25
-2500
B,-50/7×49/10-35/2×-10/7+ -25/3× -9/5
=(-35)-(-25)+(-13)
=-23
C,-354+265-156+125
=[(-354)-156]+(265+125)
=(-510)+277
=-233
D,-74+40-50+16-35
=(-74)+40+(-50)+16+(-35)
=[(-74)+(-35)]+[40+(-50)+16]
=(-109)+26
=-83
E,-2/3×4/5+-4/5×4/3-0,125
=-2/3.4/5+4/5.(-4/3)-0,125
=4/5.[-2/3+(-4/3)]-0,125
=4/5.(-2)-0,125
=-8/5-0,125
=(-1,6)+(-0,125)
=-1,725
Lời giải chi tiết:
8 + 1 = 9 | 7 – 5 = 2 | 5 + 2 = 37 | 10 – 6 = 4 |
28 + 1 = 29 | 57 – 5 = 52 | 7 – 35 = 2 | 52 + 0 = 52 |
45 + 2 = 47 | 49 – 6 = 43 | 7 – 2 = 35 | 99 – 8 = 91 |
8+1=9
7-5=2
5+2=7
10-6=4
28+1=29
57-5=52
7-35=-28
52+0=52
OKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKK
1/5+4/10+9/15+25/25+36/30+49/35+64/40+81/45
=1/5 + (4/10:2) + (9/15:3)+25/25 + (36/30:6)+ (49/35:7) + (64/40:8) + (81/45:9)
=1/5 + 1/5 + 1/5 + 1 + 1/5 + 1/5 + 1/5 + 1/5
=1/5 x 7 + 1
=12/5
\(35^{10}x5^{10}=\left(35x5\right)^{10}=175^{10}\)
\(49^5x125=7^3x5^3x7^7=35^3x7^7\)
\(\frac{35^{10}.5^{10}}{49^5.125}=\frac{5^{10}.7^{10}.5^{10}}{\left(7^2\right)^5.5^3}=\frac{5^{20}.7^{10}}{7^{10}.5^3}=5^{17}\)
chúc học tốt :)