Question

Dạo này học zoom nhìu wé , mn giúp mình zới ạ :

Answer 1

B=(1/4.9+1/9.14+...+1/44.49).1-3-5-...-49/89

B=1/5(1/4-1/9+1/9-1/14+...+1/44-1/49).1-(3+5+...+49)/89

B=1/5(1/4-1/49).1-24.52:2/89

B=9/196.-7

B=-9/28

Answer 2

Ta có \(\frac{1-3-5-..-49}{89}=\frac{1-\left(3+5+7+...+49\right)}{89}\)

\(=\frac{1-\left[\left(49-3\right):2+1\right].\left(\frac{49+3}{2}\right)}{89}=\frac{1-624}{89}=-7\)

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

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right)=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)=\frac{1}{5}.\frac{45}{196}=\frac{9}{196}\)

Khi đó \(B=\frac{9}{196}.\left(-7\right)=-\frac{9}{28}\)

Answer 3

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

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

\(B=\left[\left(\frac{1}{4}-\frac{1}{49}\right).\frac{1}{5}\right].\frac{1-3-5-...-49}{89}\)

\(B=\left[\frac{45}{196}.\frac{1}{5}\right].\frac{1-3-5-....-49}{89}\)

\(B=\frac{9}{196}.\frac{1-3-5-..-49}{86}\)

\(B=\frac{9}{196}.\frac{1-3-5-...-49}{2}\)

\(B=-\frac{801}{2408}\)

Answer 4

B=\((\frac{1}{4}\)\(-\)\(\frac{1}{9}\)\(+\)\(\frac{1}{9}\)\(-\)\(\frac{1}{14}\)+...+\(\frac{1}{44}\)-\(\frac{1}{49}\)\()\).(\(\frac{1-3-5-...-49}{89}\)

B=(\(\frac{1}{4}\)-\(\frac{1}{49}\)).\(\frac{-2+2...-2}{89}\)

B=(\(\frac{1}{4}\)-\(\frac{1}{49}\)).\(\frac{0}{89}\)

B=(\(\frac{1}{4}\)-\(\frac{1}{49}\)).0

B=0