\(\frac{50}{51}+\frac{49}{51}+\frac{48}{51}+...+\frac{1}{51}\)= ?
Tính: \(B=\frac{100^2+1^2}{100\cdot1}+\frac{99^2+2^2}{99\cdot2}+\frac{98^2+3^2}{98\cdot3}+...+\frac{52^2+49^2}{52\cdot49}+\frac{51^2+50^2}{51\cdot50}\)
Tính bằng cách hợp lý:
Tính B=\(\frac{48+\frac{34}{31}-\frac{34}{33}+\frac{34}{35}}{72+\frac{51}{31}-\frac{51}{33}+\frac{51}{35}}:\frac{-242423}{363636}\)
Tính \(1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4}+4\frac{4}{5}+...+50\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{51}\)
Từ dãy trên ta có:
(\(\frac{3}{2}\)+\(\frac{1}{2}\))+(\(\frac{8}{3}\)+\(\frac{2}{3}\))+......+(\(\frac{2600}{51}\)+\(\frac{1}{51}\)) < vì không có cách nhập hỗn số nên mình đổi ra phân số >
= 2 + 3 + 4 + 5 + 6 + ..........................+ 51
Từ 2 -> 51 có :( 51 - 2 ) : 1 + 1 = 50 số
Chia ra : 50 : 2 = 25 cặp
ta có( 51 + 2 ) x 25 =1325
Vậy tổng trên có kết quả bằng 1325 (tớ chỉ nghĩ thế thôi chứ sai đừng trách nhá.Đùa thôi,đúng đấy )
\(1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4}+...+50\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...\frac{1}{51}\)
Ta có :
\(1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4}+...+50\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{51}\)
= \(\left(1\frac{1}{2}+\frac{1}{2}\right)+\left(2\frac{2}{3}+\frac{1}{3}\right)+\left(3\frac{3}{4}+\frac{1}{4}\right)+...+\left(49\frac{49}{50}+\frac{1}{50}\right)+\left(50\frac{50}{51}+\frac{1}{51}\right)\)
= \(2+3+4+5+...+49+50+51\)
= \(\left(\frac{51-2}{1}+1\right).\frac{51+2}{2}\)
= \(50.26,5\)
= 1325
Tính: \(1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4}+...+50\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{51}\)= ________?
Tính \(1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4}+4\frac{4}{5}+...+50\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{51}\)
\(=\left(1\frac{1}{2}+\frac{1}{2}\right)+\left(2\frac{2}{3}+\frac{1}{3}\right)+...+\left(50\frac{50}{51}+\frac{1}{51}\right)\)
\(=2+3+...+51\)
\(=\frac{\left(2+51\right)50}{2}\)
\(=1325\)
Tính \(1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4}+4\frac{4}{5}+...+50\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{51}\)
\(1\dfrac{1}{2}+2\dfrac{2}{3}+3\dfrac{3}{4}+...+50\dfrac{50}{51}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{51}\)
\(=\left(1\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(2\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(3\dfrac{3}{4}+\dfrac{1}{4}\right)+...+\left(50\dfrac{50}{51}+\dfrac{1}{51}\right)\)
\(=2+3+4+...+51\)
\(=\dfrac{50\left(51+2\right)}{2}\)
=1325
\(\frac{x+52}{21}+\frac{x+51}{22}=\frac{x+50}{23}+\frac{x+49}{24}\)
\(\frac{x+52}{21}+\frac{x+51}{22}=\frac{x+50}{23}+\frac{x+49}{24}\)
Bài làm
\(\frac{x+52}{21}+\frac{x+51}{22}=\frac{x+50}{23}+\frac{x+49}{24}\)
\(\Leftrightarrow\frac{x+52}{21}+\frac{x+51}{22}+2=\frac{x+50}{23}+\frac{x+49}{24}+2\)
\(\Leftrightarrow\left(\frac{x+52}{21}+1\right)+\left(\frac{x+51}{22}+1\right)=\left(\frac{x+50}{23}+1\right)+\left(\frac{x+49}{24}+1\right)\)
\(\Leftrightarrow\left(\frac{x+52}{21}+\frac{21}{21}\right)+\left(\frac{x+51}{22}+\frac{22}{22}\right)=\left(\frac{x+50}{23}+\frac{23}{23}\right)+\left(\frac{x+49}{24}+\frac{24}{24}\right)\)
\(\Leftrightarrow\frac{x+52+21}{21}+\frac{x+51+22}{22}=\frac{x+50+23}{23}+\frac{x+49+24}{24}\)
\(\Leftrightarrow\frac{x+73}{21}+\frac{x+73}{22}=\frac{x+73}{23}+\frac{x+73}{24}\)
\(\Leftrightarrow\frac{x+73}{21}+\frac{x+73}{22}-\frac{x+73}{23}-\frac{x+73}{24}=0\)
\(\Leftrightarrow\left(x+73\right)\left(\frac{1}{21}+\frac{1}{22}-\frac{1}{23}-\frac{1}{24}\right)=0\)
\(\Leftrightarrow x+73=0\)
\(\Leftrightarrow x=-73\)
Vậy x = -73
# Học tốt #
\(\frac{x+52}{21}+\frac{x+51}{22}=\frac{x+50}{23}+\frac{x+49}{24}\)
\(\Leftrightarrow\frac{x+52}{21}+\frac{x+51}{22}-\frac{x+50}{23}-\frac{x+49}{24}=0\)
\(\Leftrightarrow\left(\frac{x+52}{21}+1\right)+\left(\frac{x+51}{22}+1\right)-\left(\frac{x+50}{23}+1\right)-\left(\frac{x+49}{24}+1\right)=0\)
\(\Leftrightarrow\frac{x+73}{21}+\frac{x+73}{22}-\frac{x+73}{23}-\frac{x+73}{24}=0\)
\(\Leftrightarrow\left(x+73\right)\left(\frac{1}{21}+\frac{1}{22}-\frac{1}{23}-\frac{1}{24}\right)=0\)
=> \(x+73=0\left(\frac{1}{21}+\frac{1}{22}-\frac{1}{23}-\frac{1}{24}\ne0\right)\)
<=> x=-73
Thanks các bn nha