5+3=28
9+1=810
8+6=214
5+4=19
7+3=?
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a: =(-1)+(-1)+...+(-1)=-1011
b: =(-5)+(-5)+...+(-5)=-175
a)=407.52.34
=21164.34
=719576
b)=197.(52+23+59)
=197.134
=26398
\(a,\left(4^{19}+3^{21}\right).\left(5^{20}-3^{15}\right).\left(2^6-8^2\right)\\ =\left(4^{19}+3^{21}\right).\left(5^{20}-3^{15}\right).\left(2^{3.2}-2^{3.2}\right)\\ =\left(4^{19}+3^{21}\right).\left(5^{20}-3^{15}\right).0=0\)
\(b,2^5.197+197.3^2+197.59\\ =197.32+197.9+197.59\\ =197.\left(32+9+59\right)\\ =197.100=19700\)
a: =5*1,4-4*1,5+3*1,3
=7-6+3,9=4,9
b: =1/3*7+3/4*18-2/3*20
=7/3+54/4-40/3
=-11+54/4
=2,5
c: =5/6*17-1/2*16+2/5*15
=85/6-8+6
=85/6-2
=73/6
d: =15*4/5+12*3/4-18*4/9
=12+9-8
=12+1=13
Bài 9,
62x73+36x33=36x73+36x27=36(73+27)=36x100=3600.
197-\([\)6x(5-1)2+20220\(]\):5=197-\([\)6x16+1\(]\):5=197-97:5=197-97/5=888/5.
Bài 10,
21-4x=13
=>4x=21-13=8
=>x=8:4=2.
30:(x-3)+1=45:43=42=16
=>30:(x-3)=16-1=15
=>x-3=30:15=2
=>x=2+3=5.
(x-1)3+5x6=38
=>(x-1)3+30=38
=>(x-1)3=38-30=8=23
=>x-1=2
=>x=3.
3)
3/5 + 3/7-3/11 / 4/5 + 4/7- 4/11
= 3.( 1/5 + 1/7 - 1/11)/4.(1/5+1/7-1/11)
= 3/4
1,
ta có B = 196+197/197+198 = 196/(197+198) + 197/(197+198)
196/197 > 196/197+198
197/198 > 197/197+198
=> A>B
C=(1-2-3-4)+(5-6-7-8)+...+(197-198-199-200)
=-8x50
=-400
Dễ thấy \(4\times2145=\left(2100+45\right)\times4\\ 3964\times6=\left(4+2\right)\times\left(3000+964\right)\\ 10287\times5=\left(3+2\right)\times10287\)
Vậy a = e, b = g, c = d.
a: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}+\dfrac{-\dfrac{1}{4}\cdot\dfrac{-2}{3}-\dfrac{3}{4}:\dfrac{1}{6}}{\dfrac{3}{2}\cdot\left(\dfrac{-2}{3}-\dfrac{3}{4}\cdot\dfrac{-2}{3}\right)}\)
\(=\dfrac{3}{4}+\dfrac{\dfrac{2}{12}-\dfrac{9}{2}}{\dfrac{3}{2}\cdot\dfrac{-1}{6}}=\dfrac{3}{4}+\dfrac{-13}{3}:\dfrac{-3}{12}\)
\(=\dfrac{3}{4}+\dfrac{13}{3}\cdot\dfrac{12}{3}=\dfrac{3}{4}+\dfrac{156}{9}=\dfrac{217}{12}\)
b: \(A=158\left(\dfrac{12\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{4\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{5\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}{6\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}\right)\cdot\dfrac{50550505}{711711711}\)
\(=158\cdot\left(3\cdot\dfrac{6}{5}\right)\cdot\dfrac{50550505}{711711711}\)
\(\simeq40.39\)
7+3=10
ahihi!
7+3=20