Question

[Đaklak Cao Nguyen]

các bạn ơi, giúp mình bài trên với ạ

Answer 1

[Nguyễn Nam Dương]

Ta có: \(A=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+\dfrac{5}{14\cdot19}+...+\dfrac{5}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{49-4}{4\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{1}{5}\cdot\dfrac{45}{196}\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{9}{196}\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{9}{196}\cdot\dfrac{-623}{89}=-\dfrac{9}{28}\)

Answer 2

[ミ★ғox♥️ʀồɴԍ★彡乡]

A=\(\left(\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{44.49}\right)\).\(\frac{1-3-5-7-...-49}{89}\)

Đặt B=\(\left(\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{44.49}\right)\)

5B=\(5\left(\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{44.49}\right)\)

5B=\(\frac{5}{4.9}+\frac{5}{9.14} +...+\frac{5}{44.49}\)

5B=\(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\)\(\frac{1}{49}\)

5B=\(\left(-\frac{1}{9}+\frac{1}{9}\right)+...+\left(-\frac{1}{44}+\frac{1}{44}\right)+\left(\frac{1}{4}-\frac{1}{49}\right)\)\(\frac{1}{49}\))

5B=\(\frac{1}{4}-\frac{1}{49}\)

B=\(\frac{\left(\frac{1}{4}-\frac{1}{49}\right)}{5}\)=\(\frac{9}{196}\)

Đặt C= \(\frac{1-3-5-7-...-49}{89}\)

C=\(\frac{1-\left(3+5+...+49\right)}{89}\)

C=\(1-\left(49+3\right).24:2\)

C=\(-\frac{623}{89}=-7\)

Thay B;C vào A ta được A=\(\frac{9}{196}x\left(-7\right)=-\frac{9}{28}\)

HT