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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\dfrac{1}{4}=\dfrac{10}{40}=\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}\)
Mà \(\dfrac{1}{31}>\dfrac{1}{40}\)
\(\dfrac{1}{32}>\dfrac{1}{40}\)
\(\dfrac{1}{33}>\dfrac{1}{40}\)
\(\dfrac{1}{34}>\dfrac{1}{40}\)
\(\dfrac{1}{35}>\dfrac{1}{40}\)
\(\dfrac{1}{36}>\dfrac{1}{40}\)
\(\dfrac{1}{37}>\dfrac{1}{40}\)
\(\dfrac{1}{38}>\dfrac{1}{40}\)
\(\dfrac{1}{39}>\dfrac{1}{40}\)
\(\Rightarrow\) \(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{39}+\dfrac{1}{40}>\dfrac{10}{40}=\dfrac{1}{4}\)
Vậy \(S>\dfrac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\left(1+3\right)+...+3^8\left(1+3\right)=4\left(1+...+3^8\right)⋮4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\left(1+3+3^2\right)+...+3^7\left(1+3+3^2\right)\)
\(=13\left(1+...+3^7\right)⋮13\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=1+3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9\)
\(S=\left(1+3\right)+\left(3^2+3^3\right)+\left(3^4+3^5\right)+\left(3^6+3^7\right)+\left(3^8+3^9\right)\)
\(S=4+3^2\left(1+3\right)+3^4\left(1+3\right)+3^6\left(1+3\right)+3^8\left(1+3\right)\)
\(S=4+3^2.4+3^4.4+3^6.4+3^8.4\)
\(S=4\left(3^2+3^4+3^6+3^8\right)\)
\(4⋮4\\ \Rightarrow4\left(3^2+3^4+3^6+3^8\right)⋮4\\ \Rightarrow S⋮4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=1.\left(1+3\right)+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^8\left(1+3\right)\)
\(S=4x\left(1+3^2+...+3^8\right)\)
Vì 4 chia hết cho 4 nên S chia hết cho 4
![](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)
a: =39-13-62-39=-75
b: =(-2)+(-2)+(-2)=-6
c: 92-55+8-45=100-100=0
d: M={-20;-19;-18;...;18;19;20}
Tổng là (-20)+(-19)+(-18)+...+18+19+20=0
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: \(387+\left(-224\right)+\left(-87\right)\)
\(=\left(387-87\right)+\left(-224\right)\)
=300-224
=76
b: \(39+\left(13-26\right)-\left(62+39\right)\)
\(=39+13-26-62-39\)
\(=\left(39-39\right)+\left(13-26-62\right)\)
=0-75
=-75
c: \(32-34+36-38+40-42\)
\(=\left(32-34\right)+\left(36-38\right)+\left(40-42\right)\)
\(=\left(-2\right)+\left(-2\right)+\left(-2\right)\)
=-6
d: \(92-\left(55-8\right)+\left(-45\right)\)
\(=92-55+8-45\)
\(=\left(92+8\right)-\left(55+45\right)\)
=100-100
=0
e: -20<=x<=20
mà x nguyên
nên \(x\in\left\{-20;-19;-18;...;18;19;20\right\}\)
=>M={-20;-19;-18;-17;...;18;19;20}
Tổng các phần tử của M là:
\(\left(-20\right)+\left(-19\right)+\left(-18\right)+\left(-17\right)+...+18+19+20\)
\(=\left(-20+20\right)+\left(-19+19\right)+...+\left(-2+2\right)+\left(-1+1\right)+0\)
=0+0+...+0
=0
\(A=1+6+6^2+6^4+...+6^{100}\)
\(\Rightarrow6A=6+6^2+6^4+...+6^{100}+6^{101}\)
\(\Rightarrow6A-A=\left(6+6^2+6^4+....+6^{102}\right)-\left(1+6+6^2+6^4+...+6^{100}\right)\)
\(\Rightarrow5A=6^{101}-1\)
\(\Rightarrow A=\frac{6^{101}-1}{5}\)
\(B=1+3^2+3^4+3^6+3^8+...+3^{100}.\)
\(\Rightarrow3B=3^2+3^4+3^6+...+3^{101}\)
\(\Rightarrow3B-B=\left(3^2+3^4+...+3^{101}\right)-\left(1+3^2+3^4+...+3^{100}\right)\)
\(\Rightarrow2B=3^{101}-1\)
\(\Rightarrow B=\frac{3^{101}-1}{2}\)