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![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a. $=62-81-12+59-9=(62-12)+(59-9)-81$
$=50+50-81=100-81=19$
b. $=39+13-26-62-39=(39-39)+13-(26+62)$
$=0+13-88=-(88-13)=-75$
c. $=(32-42)+(36-34)+(40-38)=10+2+2=14$
d. $=92-55+8-45=(92+8)-(55+45)=100-100=0$
Bài 1:
a. $=(387-87)-224=300-224=76$
b. $=-(75+35)+379=-110+379=379-110=269$
c. $=(11+15)-(13+17)=25-30=-5$
d. $=(31-21)-(27-24)=10-3=7$
![](https://rs.olm.vn/images/avt/0.png?1311)
Sorry mình nhầm, đổi vị trí cho dãy số 1221211221121221211212211221211221121221122121121221211221121221 cho dấu ... :))
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 47 x 36 + 64 x 47 + 15
A= 47 x ( 64 + 36 ) + 15 = 47 x 100 + 15 = 4700 + 15 = 4715
vậy A= 4715
B= 27+35 + 65 + 73+ 75
B= (27+ 73) + ( 35 + 65) +75
B= 100 +100 +75 = 275
vậy B= 275
C= 37 +37 x 15 +37 x 84
C= 37 x ( 1+15 +84 )= 37 x 100 = 3700
vậy C= 3700
D = 1/20x21 + 1/21x22 + 1/22x23 + 1/23x24
D= 1/20 - 1/21 + 1/21 - 1/22 + 1/22 - 1/23 + 1/23 - 1/24
D= 1/20 -1/24 = 1/120 vậy D= 1/120
E= 1/1x2 + 1/2x3 + ...... + 1/49x50
E= 1/1 - 1/2 + 1/2 - 1/3 +...... + 1/49 - 1/50
E = 1 - 1/50 = 49/50
vậy E= 49/50
CHÚC HOK TOT
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=47.36+64.47+15\)
\(A=47.\left(36+64\right)+15\)
\(A=47.100+15\)
\(A=4700+15\)
\(A=4715\)
\(B=27+35+65+73+75\)
\(B=\left(27+73\right)+\left(35+65\right)+75\)
\(B=100+100+75\)
\(B=275\)
\(C=37+37.15+84.37\)
\(C=37.\left(1+15+84\right)\)
\(C=37.100\)
\(C=3700\)
\(D=\frac{1}{20.21}+\frac{1}{21.22}+\frac{1}{22.23}+\frac{1}{23.24}\)
\(D=\frac{1}{20}-\frac{1}{21}+\frac{1}{21}-\frac{1}{22}+\frac{1}{22}-\frac{1}{23}+\frac{1}{23}-\frac{1}{24}\)
\(D=\frac{1}{20}-\frac{1}{24}\)
\(D=\frac{24}{480}-\frac{20}{480}\)
\(D=\frac{4}{480}=\frac{1}{120}\)
\(E=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(E=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(E=1-\frac{1}{50}\)
\(E=\frac{49}{50}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(42=2\cdot3\cdot7;70=2\cdot5\cdot7\)
=>\(BCNN\left(42;70\right)=2\cdot3\cdot5\cdot7=210\)
=>\(BC\left(42;70\right)=B\left(210\right)=\left\{0;210;420;...\right\}\)
b: \(70=2\cdot5\cdot7;180=3^2\cdot5\cdot2^2\)
=>\(BCNN\left(70;180\right)=2^2\cdot3^2\cdot5\cdot7=1260\)
=>\(BC\left(70;180\right)=\left\{1260;2520;...\right\}\)
c: \(5=5;7=7;8=2^3\)
=>\(BCNN\left(5;7;8\right)=5\cdot7\cdot8=280\)
=>\(BC\left(5;7;8\right)=\left\{280;560;...\right\}\)
d: \(12=2^2\cdot3;18=3^2\cdot2\)
=>\(BCNN\left(12;18\right)=2^2\cdot3^2=36\)
=>\(BC\left(12;18\right)=\left\{36;72;...\right\}\)
e: \(15=3\cdot5;18=3^2\cdot2\)
=>\(BCNN\left(15;18\right)=3^2\cdot2\cdot5=90\)
=>\(BC\left(15;18\right)=\left\{90;180;...\right\}\)
f: \(84=2^2\cdot3\cdot7;108=3^3\cdot2^2\)
=>\(BCNN\left(84;108\right)=2^2\cdot3^3\cdot7=756\)
=>\(BC\left(84;108\right)=\left\{756;1512;...\right\}\)
j: \(33=3\cdot11;44=2^2\cdot11;55=5\cdot11\)
=>\(BCNN\left(33;44;55\right)=3\cdot2^2\cdot5\cdot11=660\)
=>\(BC\left(33;44;55\right)=\left\{660;1320;...\right\}\)
g: \(1=1;12=2^2\cdot3;27=3^3\)
=>\(BCNN\left(1;12;27\right)=1\cdot2^2\cdot3^3=108\)
=>\(BC\left(1;12;27\right)=\left\{108;216;...\right\}\)
n: \(5=5;9=3^2;11=11\)
=>\(BCNN\left(5;9;11\right)=5\cdot3^2\cdot11=495\)
=>\(BC\left(5;9;11\right)=\left\{495;990;...\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đáp án A
Số chia hết cho 2 nhưng không chia hết cho 5 là các số có tận cùng là 2;4;6;8. Nhận thấy 12;44;56 là các số có tận cùng là 2;4;6. Vậy 12;44;56 chia hết cho 2 nhưng không chia hết cho 5