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a) \(\left(-15\right)\cdot\left(10-12\right)=\left[\left(-15\right)\cdot10-\left(-15\right)\cdot12\right]=-150+180=30\)
b) \(\left(-17\right)\cdot10-\left(-3\right)\cdot10=\left[-17-\left(-3\right)\right]\cdot10=\left(-17+3\right)\cdot10=-14\cdot10=-140\)
![](https://rs.olm.vn/images/avt/0.png?1311)
C=1+2+3-4+5+6+7-8+...+1999-2000+2001
C=1+2+(-1)+5+6+(-1)+...+1997+1998+(-1)+2001
C=1+1+5+5+...+1997+1997+2001
C=2001+(1.2+5.2+...+1997.2)
C=2001+2(1+5+...+1997)
Tong 1+5+...+1997={(1997+1)[(1997-1):4+1]}:2=499500
C=2001+2.499500
C=2001+999000
C=1001001
he he
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1 )
(-10 + 5) -(4-x)=12 -(5-6)
<=> -10 + 5 - 4 + x = 12 - 5 + 6
<=> x = 12 - 5 + 6 + 10 - 5 + 4
<=> x = 22
2 )
14-(5-8+3-x)=|-7+10|
<=> 14 - 5 + 8 - 3 + x = | 3 |
<=> x = | 3 | -14 + 5 - 8 + 3
<=> x = 3 - 14 + 5 - 8 + 3
<=> x = -11
3 )
-19-(3 +x-5)=|4-15+2|
<=> -19 - 3 - x + 5 = | -9 |
<=> -x = | -9 | - 5 + 19 + 3
<=> -x = 9 - 5 + 19 + 3
<=> -x = 26
<=> x = -26
4 )
45-(x+45-3)=|-7+3|
<=> 45 - x - 45 + 3 = | -4 |
<=> -x + 3 = | -4 |
<=> -x = 4 - 3
<=> -x = 1
<=> x =-1
10 )
-(-12 +4-10)-(4-x)=-(-3)
<=> 12 - 4 + 10 - 5 + x = 3
<=> x = 3 -12 + 4 - 10 + 5
<=> x = -10
Mình là Siêu Phẩm Hacker , rất mong được thi đấu một lần vs Edokawa conan
1) (-10 + 5) - (4 - x) = 12 - (5 - 6)
=> -5 - 4 + x = 12 + 1
=> -5 - 4 + x = 13
=> 4 + x = -5 - 13
=> 4 + x = - 18
=> x = -18 - 4
=> x = -22
2) 14 - (5 - 8 + 3 - x) = |-7 + 10|
=> 14 + 5 + 8 - 3 + x = |3|
=> 24 + x = 3
=> x = 3 - 24
=> x = -21
3) -19 - (3 + x - 5) = |4 - 15 + 2|
=> -19 - 3 - x + 5 = |-9|
=> -17 - x = 9
=> x = -17 - 9
=> x = -26
4) 45 - (x + 45 - 3) = |-7 + 3|
=> 45 - x - 45 + 3 = |-4|
=> 3 - x = 4
=> x = 3 - 4
=> x = -1
5) -(-12 + 4 - 10) - (4 - x) = -(-3)
=> -(-29) - (4 - x) = 3
=> 29 - (4 - x) = 3
=> 4 - x = 29 - 3
=> 4 - x = 26
=> x = 4 - 26
=> x = -22
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có
đặt 3/5+3/7-3/11=N=> N=3*(1/5+1/7-1/11)
đặt 4/5+4/7-4/11=P=> P=4*(1/5+1/7-1/11)
=> N/P=3*(1/5+1/7-1/11)/4*(1/5+1/7-1/11)=> M=3/4
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a)
A = 2 + 22 + 23 + 24 + ... + 2200
2A = 22 + 23 + 24 + 25 + ... + 2200
2A - A = A = 2200 - 2
b) chịu
c)
C = 4 + 42 + 43 + 44 +... + 4100
4C = 42 + 43 + 44 + 45 + ... + 4101
4C - C = 3C = 4101 - 4
\(\Rightarrow\) C = \(\frac{4^{101}-4}{3}\)
d)
D = 5 + 52 + 53 + ... + 5100
5D = 52 + 53 + 54 + ... + 5101
5D - D = 4D = 5101 - 5
\(\Rightarrow\)D = \(\frac{5^{101}-5}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
P = 1/3 + -3/4 + 3/5 + -1/36 + 1/15 + -2/9
=60/180 - 135/180 + 108/180 - 5/180 + 12/180 - 40/180
=60-135+108-5+12-40/180
=0/180=0
Nhớ k cho mik nha, Hok tốt !
\(D=-\dfrac{4}{5}+\dfrac{4}{5^2}-\dfrac{4}{5^3}+...+\dfrac{4}{5^{200}}\)
\(\Rightarrow D=4\left(-\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+...+\dfrac{1}{5^{200}}\right)\)
\(5D=4\cdot\left(-1+\dfrac{1}{5}-\dfrac{1}{5^2}+...+\dfrac{1}{5^{199}}\right)\)
\(\Rightarrow5D+D=4\cdot\left(-1+\dfrac{1}{5}-\dfrac{1}{5^2}+...+\dfrac{1}{5^{199}}-\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+...+\dfrac{1}{5^{200}}\right)\)
\(\Rightarrow6D=4\cdot\left(\dfrac{1}{5^{200}}-1\right)\)
\(\Rightarrow D=\dfrac{2}{3}\cdot\left(\dfrac{1}{5^{200}}-1\right)\)
oki