Đặt A=1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100

4A=(1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100)4

4A=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+98.99.100(101-97)

4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100

4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-3.4.5.6+...+97.98.99.100-97.98.99.100+98.99.100.101

4A=98.99.100.101

=>A=98.99.100.101/4

=> A=24497550

\(c)\)\(\left(-\dfrac{1}{2}\right)^3:1\dfrac{3}{8}-25\%\cdot\left(-6\dfrac{2}{11}\right)\)

\(=-\dfrac{1}{8}\cdot\dfrac{8}{11}-\dfrac{1}{4}\cdot\left(-\dfrac{68}{11}\right)\)

\(=-\dfrac{1}{11}+\dfrac{17}{11}\)

\(=\dfrac{16}{11}\)

#Ayumu

\(=\dfrac{27}{5}\cdot\dfrac{30}{7}+\dfrac{40}{7}\cdot\dfrac{37}{7}=\dfrac{2614}{49}\)

\(c)\)

\(5\dfrac{2}{5}\times4\dfrac{2}{7}+5\dfrac{5}{7}\times5\dfrac{2}{7}\)

\(=\dfrac{27}{5}\times\dfrac{30}{7}+\dfrac{40}{7}\times\dfrac{37}{7}\)

\(=\dfrac{162}{7}+\dfrac{40}{7}\times\dfrac{37}{7}\)

\(=\dfrac{162}{7}+\dfrac{1480}{49}\)

\(=\dfrac{1134}{49}+\dfrac{1480}{49}\)

\(=\dfrac{2614}{49}\)

Sửa đề: +2023^2-2024^2

C=(1-2)(1+2)+(3-4)(3+4)+...+(2023-2024)(2023+2024)

=-(1+2+3+4+...+2023+2024)

=-2024*2025/2=-2049300

\(c,\dfrac{5}{18}\times\dfrac{2}{15}:\dfrac{1}{5}\\ =\dfrac{1}{9}\times\dfrac{1}{3}\times5\\ =\dfrac{5}{27}\\ d,\dfrac{4}{13}+\dfrac{2}{11}:\dfrac{4}{33}\\ =\dfrac{4}{13}+\dfrac{2}{11}\times\dfrac{33}{4}\\ =\dfrac{4}{13}+\dfrac{3}{2}=\dfrac{8}{26}+\dfrac{39}{26}\\ =\dfrac{47}{26}\)

C = 1³ + 2³ + 3³ + 4³ + ..... + 10³

 \(C ​= 1 + 8+ 27+ 64 + 125 + 216 +343 +512 +729+ 1000 = 4025\)

~~~học tốt nha~~~

\(C=1^3+2^3+...+10^3=\frac{2^3\left(1^3+2^3+...+10^3\right)}{2^3}=\frac{2^3+4^3+...+20^3}{8}=\frac{24200}{8}=3025\)

Thanks các bạn nhiều