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\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)
\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)
\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)
![](https://rs.olm.vn/images/avt/0.png?1311)
C = 3 - 1 + 4 - 1 + 5 - 1 + .... + 102 - 1 + 103 - 1
= 2 + 3 + 4 + ... + 101 + 102
Số số hạng là : (102 - 2) : 1 + 1 = 100 (số hạng)
Tổng : (102 + 2) . 100 : 2 = 5200
Vậy C = 5200
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A=\(\frac{1}{3}-\frac{3}{4}-\left(\frac{-3}{5}\right)+\frac{1}{72}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
=\(\frac{1}{3}-\frac{3}{4}+\frac{3}{5}+\frac{1}{72}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
=\(\left(\frac{1}{3}+\frac{3}{5}+\frac{1}{15}\right)-\left(\frac{3}{4}+\frac{2}{9}+\frac{1}{36}\right)+\frac{1}{72}\)
=\(\left(\frac{14}{15}+\frac{1}{15}\right)-\left(\frac{35}{36}+\frac{1}{36}\right)+\frac{1}{72}\)
=1 - 1 + \(\frac{1}{72}\)= 0 + \(\frac{1}{72}\)= \(\frac{1}{72}\)
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-5 + (-37 - 45 + 51) - (-37 + 51)
= -5 - 37 - 45 + 51 + 37 - 51
= - (5 + 45) - (37 - 37) + (51 - 51)
= - 50 - 0 + 0
= - 50
- (15 - 47 + 58) + (15 - 47 ) + 3
= - 15 + 47 - 58 + 15 - 47 + 3
= (-15 + 15) + (47 - 47) - (58 - 3)
= 0 + 0 - 55
= - 55
(53 - 45 - 49) - (53 + 45 - 49)
= 53 - 45 - 49 - 53 - 45 + 49
= (53 - 53) + (-49 + 49) - (45 + 45)
= 0 + 0 - 90
= - 90
13 - 15 + 49 - 13 + 15 - 48
= (13 - 13) + (-15 + 15) + (49 - 48)
= 0 + 0 + 1
= 1
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a/
\(A=1.2+1.2+2.3+2.2+3.4+3.2+...+66.67+66.2=\)
\(=\left(1.2+2.3+3.4+...+66.67\right)+2\left(1+2+3+...+66\right)\)
Đặt
\(B=1+2+3+...+66=\dfrac{66\left(1+66\right)}{2}=2211\)
Đặt
\(C=1.2+2.3+3.4+...+66.67\)
\(3C=1.2.3+2.3.3+3.4.3+...+66.67.3=\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+66.67.\left(68-65\right)=\)
\(=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-65.66.67+66.67.68=\)
\(=66.67.68\Rightarrow C=\dfrac{66.67.68}{3}=22.67.68\)
\(\Rightarrow A=C+2B\) Bạn tự tính nhé
b/
\(B=2\left(1.50+2.49+3.48+...+25.26\right)=\)
Ta có
\(C=1.50+2.49+3.48+...+25.26=\)
\(\left(50-49\right).50+\left(50-48\right).49+\left(50-47\right).48+...+\left(50-25\right).26=\)
\(=50.50-49.50+50.49-48.49+50.48-47.48+50.26-25.26=\)
\(=50.\left(26+27+28+...+50\right)-\left(25.26+26.27+27.28+...+49.50\right)\)
Ta có
\(D=26+27+28+...+50=\dfrac{25.\left(26+50\right)}{2}=950\)
Ta có
\(E=25.26+26.27+27.28+...+49.50\)
\(3E=25.26.3+26.27.3+27.28.3+...+49.50.3=\)
\(=25.26.\left(27-24\right)+26.27.\left(28-25\right)+...+49.50.\left(51-48\right)=\)
\(=-24.25.26+25.26.27-25.26.27+26.27.28-...-48.49.50+49.50.51=\)
\(=49.50.51-24.25.26\)
\(\Rightarrow E=\dfrac{49.50.51-24.25.26}{3}\)
\(\Rightarrow C=50D-E\)
\(B=2C\)
Bạn tự tính nhé
a) C = 1/3 + 1/15 + ... + 1/2115
= 1/(1.3) + 1/(3.5) + ... + 1/(45.47)
= 1/2 . (1 - 1/3 + 1/3 - 1/5 + ... + 1/45 - 1/47}
= 1/2 . (1 - 1/47)
= 1/2 . 46/47
= 23/47
\(C=\dfrac{1}{3}+\dfrac{1}{15}+...+\dfrac{1}{2115}\\ \\ \\ \\ \\ C=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{45.47}\\ \\ \\ \\ \\ \Rightarrow2C=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{45.47}\\ \\ \\ \\ \\ 2C=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{45}-\dfrac{1}{47}\\ \\ \\ \\ \\ 2C=\dfrac{1}{1}-\dfrac{1}{47}=\dfrac{46}{47}\\ \\ \\ \\ \\ \Rightarrow C=\dfrac{46}{47}:2\\ \\ \\ \\ \\ C=\dfrac{46}{47}\cdot\dfrac{1}{2}=\dfrac{23}{47}\)