\(\frac{2}{3.5}+\frac{3}{5.8}+\frac{11}{8.19}+\frac{13}{19.32}+\frac{25}{32.57}+\frac{30}{57.87}\)\(\frac{2}{3.5}+\frac{3}{5.8}+\frac{11}{8.19}+\frac{13}{19.32}+\frac{25}{32.57}+\frac{30}{57.87}\)
\(\frac{2}{3.5}+\frac{3}{5.8}+\frac{11}{8.19}+\frac{13}{8.19}+\frac{25}{32.57}+\frac{30}{57.87}\)
( Dấu . là dấu nhân)
Hình như sai đề rồi bạn ơi ! Ở phân số 13/8.19 thì tớ nghĩ phải là 13/19.32 chứ
Tính A=2/3.5 + 3/5.8 + 11/8.19 + 13/19.32 + 25/32.57 + 30/57.87
\(A=\frac{2}{3.5}+\frac{3}{5.8}+\frac{11}{8.19}+\frac{13}{19.32}+\frac{25}{32.57}+\frac{30}{57.87}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{32}+\frac{1}{32}-\frac{1}{57}+\frac{1}{57}-\frac{1}{87}\)
\(=\frac{1}{3}-\frac{1}{87}=\frac{29}{87}-\frac{1}{87}=\frac{28}{87}\)
Tính nhanh:
C= 2/3.5 + 3/5.8 + 11/8.19 + 13/19.32 + 25/32.57 + 30/57.87
\(C=\dfrac{2}{3\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{11}{8\cdot19}+\dfrac{13}{19\cdot32}+\dfrac{25}{32\cdot57}+\dfrac{30}{57\cdot87}\)\(C=\left(\dfrac{5-3}{3\cdot5}\right)+\left(\dfrac{8-5}{5\cdot8}\right)+\left(\dfrac{19-8}{8\cdot19}\right)+\left(\dfrac{32-19}{19\cdot32}\right)+\left(\dfrac{57-32}{32\cdot57}\right)+\left(\dfrac{87-57}{57\cdot87}\right)\)\(C=\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+\left(\dfrac{1}{5}-\dfrac{1}{8}\right)+\left(\dfrac{1}{8}-\dfrac{1}{19}\right)+\left(\dfrac{1}{19}-\dfrac{1}{32}\right)+\left(\dfrac{1}{32}-\dfrac{1}{57}\right)+\left(\dfrac{1}{57}+\dfrac{1}{87}\right)\)\(C=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{32}+\dfrac{1}{32}-\dfrac{1}{57}+\dfrac{1}{57}-\dfrac{1}{87}\)\(C=\dfrac{1}{3}-\dfrac{1}{87}=\dfrac{28}{87}\)
tính :
S = 2/3.5 + 3/5.8 + 11/8.19 + 13/19.32 + 25/32.57 + 30/57-87
làm nhanh giúp ạ
\(S=\frac{2}{3.5}+\frac{3}{5.8}+\frac{11}{8.19}+\frac{13}{19.32}+\frac{25}{32.57}+\frac{30}{57.85}\)
\(S=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{32}+\frac{1}{32}-\frac{1}{57}+\frac{1}{57}-\frac{1}{87}\)
\(S=\frac{1}{3}-\frac{1}{87}\)
\(S=\frac{28}{87}\)
TÍNH NHANH:
A=\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\)
B=\(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+...+\dfrac{1}{95}\)
C=\(\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+...+\dfrac{1}{9800}\)
*D=\(\dfrac{2}{3.5}+\dfrac{3}{5.8}+\dfrac{11}{8.19}+\dfrac{13}{19.32}+\dfrac{25}{32.57}+\dfrac{30}{57.87}\)
\(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\)
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=1-\dfrac{1}{100}=\dfrac{99}{100}\)
\(B=\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+..+\dfrac{1}{195}\) ( là 195 ms đúng ! )
\(B=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{13\cdot15}\)
\(B=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\)
\(B=\dfrac{1}{2}\left(1-\dfrac{1}{15}\right)=\dfrac{1}{2}\cdot\dfrac{14}{15}=\dfrac{7}{15}\)
\(C=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{98\cdot100}\)
Rồi làm tương tự cân b nha!
\(D=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{32}+\dfrac{1}{32}-\dfrac{1}{57}\)
\(+\dfrac{1}{57}-\dfrac{1}{87}\)
\(D=\dfrac{1}{3}-\dfrac{1}{87}=\dfrac{28}{87}\)
Tính :
\(\frac{3}{5.8}+\frac{11}{8.19}+\frac{12}{19.31}+\frac{70}{31.101}+\frac{99}{101.200}\)
\(\frac{3}{5.8}+\frac{11}{8.19}+\frac{12}{19.31}+\frac{70}{31.101}+\frac{99}{101.200}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\)
\(=\frac{1}{5}-\frac{1}{200}\)
\(=\frac{39}{200}\)
1/5-1/8+1/8-1/19+1/19-1/31+1/31-1/101+1/200=1/5-1/200=195/1000=39/200
\(\frac{3}{5.8}+\frac{11}{8.19}+\frac{12}{19.31}+\frac{70}{31.101}+\frac{99}{101.200}\)=?
cho \(A=\frac{3}{5.8}+\frac{11}{8.19}+\frac{12}{19.31}+\frac{70}{31.101}+\frac{99}{101.200}\) so sánh A với 1
\(\Rightarrow A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\)
\(\Rightarrow A=\frac{1}{5}-\frac{1}{200}\)
\(\Rightarrow A=\frac{39}{200}\)
vì \(\frac{39}{200}< 1\) nên A < 1
\(A=\frac{3}{5.8}+\frac{11}{8.19}+\frac{12}{19.31}+\frac{70}{31.101}+\frac{99}{101.200}\)
Áp dụng công thức \(\frac{b-a}{a.b}=\frac{1}{a}-\frac{1}{b}\) với a < b và a khác b khác 0, ta có:
\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+...+\frac{1}{101}-\frac{1}{200}\\ =\frac{1}{5}-\frac{1}{200}\\ =\frac{40-1}{200}\\ =\frac{39}{200}\\ \frac{39}{200}< 1\\\Rightarrow A< 1\left(đpcm\right)\)
Chúc bạn học tốt!
Cho \(A=\frac{3}{5.8}+\frac{11}{8.19}+\frac{12}{19.31}+\frac{70}{31.101}+\frac{99}{101.200}\)so sánh a và 1
\(\Rightarrow A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\)
\(\Rightarrow A=\frac{1}{5}-\frac{1}{200}\)
\(\Rightarrow A=\frac{39}{200}\)
Vì \(\frac{39}{200}<1\Rightarrow A<1\)