\(A=2.4+4.6+6.8+...+96.98+98.100\)

\(\Rightarrow6A=2.4.6+4.6.8-2.4.6+6.8.10-4.6.8+...+96.98.100-94.96.98+98.100.102-96.98.100\)

\(\Rightarrow6A=\left(2.4.6+4.6.8+6.8.10+...+96.98.100+98.100.102\right)-\left(2.4.6+4.6.8+...+94.96.98+96.98.100\right)\)\(\Rightarrow6A=98.100.102\)

\(\Rightarrow A=\frac{98.100.102}{6}=98.100.17=166600\)

P/s: Ko chắc

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Câu hỏi của Kanzaki Mizuki -toán lớp 6

A = 1×3+3×5+5×7+...+ 97×99+99×101

 6A= 1×3×6+3×5×6+5×7×6+...+97×99×6+99×101×6

6A= 1×3×(5+1)+3×5×(7-1)+5×7×(9-3)+...+97×99×(101-95)+99×101×(103-97)

6A = 1×3×5-1×3+3×5×7-1×3×5+5×7×9-3×5×7+7×9×11-5×7×9+,,,+97×99×101-95×97×99+99×101×103-97×99×101

6A= 1×3+99×101×103

6A= 1029900

A= 171650

171650

a,6B=2.4.6+4.6.(8-2)+...............+98.100.(102-96)

6B=2.4.6+4.6.8-2.4.6+..............+98.100.102-96.98.100

6B=98.100.102

B=98.100.102:6

B=166600

\(2P=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\)

\(2P=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{98}-\frac{1}{100}\)

\(2P=\frac{1}{2}-\frac{1}{100}\)

=> P =\(\frac{49}{100}:2=\frac{49}{100}\cdot\frac{1}{2}=\frac{49}{200}\)

\(\frac{2.4+4.6+6.8+...+98.100}{1.2+2.3+3.4+...+49.50}=\frac{4.\left(1.2+2.3+3.4+...+49.50\right)}{1.2+2.3+3.4+...+49.50}=\frac{4}{1}=4\)

(0,5+0,5)÷0,5_0,5×0,5=