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\(Q=\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{47.49}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{47}-\frac{1}{49}\)
\(=\frac{1}{3}-\frac{1}{49}\)
\(=\frac{46}{147}\)
Vậy \(Q=\frac{46}{147}\)
Ta có : \(\frac{2}{3}Q=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{47.49}\)
\(\Rightarrow\frac{2}{3}Q=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}\)
\(\Rightarrow\frac{2}{3}Q=\frac{1}{3}-\frac{1}{49}=\frac{49}{147}-\frac{3}{147}=\frac{46}{147}\)
\(\Rightarrow Q=\frac{46}{147}\div\frac{2}{3}=\frac{138}{294}=\frac{23}{49}\)
Vậy ...
a, Ta có \(A=\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{49.51}\)
\(=\frac{3}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{49.51}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)\)
\(=\frac{1}{2}-\frac{3}{102}=\frac{48}{102}=\frac{24}{51}\)
b,Ta có \(\frac{1}{2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}\)
\(=\frac{2-1}{2}+\frac{4-2}{2.4}+\frac{7-4}{4.7}+\frac{11-7}{7.11}+\frac{16-11}{11.16}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\)
\(=\frac{15}{16}\)
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=3(1/1.3+1/3.5+1/5.7+1/7.9)
=3/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9) vi khoang cach tu 1-3;3-5;5-7;7-9 la 2 nen ta nhan tat ca voi 1/2 ma 3.1/2=3/2
=3/2.(1-1/9) rut gon -1/3+1/3;-1/5+1/5;-1/7+1/7=0
=3/2.8/9=4/3
ta có :3/(1.3)+3/(3.5)+3/(5.7)+3/(7.9)
ta đặt 3 làm chung rồi tự làm đc
NHẦM GIẢI LẠI :
\(A=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{3}{2}.\frac{16}{51}=\frac{8}{17}\)
\(\frac{-1}{2}\)\(-\left(\frac{3}{2}+x\right)=-22\)
\(\frac{-1}{2}\)\(-\frac{3}{2}\)\(-x=-22\)
\(-\left(\frac{1}{2}+\frac{3}{2}\right)-x=-22\)
\(-2-x=-22\)
\(x=-2-\left(-22\right)\)
\(x=20\)
không đúng thì thôi bạn đừng k sai nha!
a, Ta có:
\(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{0,6-\frac{3}{9}+\frac{3}{11}}+\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{14}}{-1-\frac{3}{7}+\frac{3}{28}}=\frac{2\left(0,2-\frac{1}{9}+\frac{1}{11}\right)}{3\left(0,2-\frac{1}{9}+\frac{1}{11}\right)}+\frac{2\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{28}\right)}{-3\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{28}\right)}=\frac{2}{3}+\frac{-2}{3}=0\)
k đúng cho mình nha. Thanks!!!
a, bày cho mình cách viết bằng phân số đi , mình trình bày cách làm cho. k đúng cho mình nha.
\(C=\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{3}{47\cdot49}\)
\(\Rightarrow\frac{2}{3}C=\frac{2}{3}\cdot\left(\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{3}{47\cdot49}\right)\)
\(\Rightarrow\frac{2}{3}C=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{47\cdot49}\)
\(\Rightarrow\frac{2}{3}C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{47}-\frac{1}{49}\)
\(\Rightarrow\frac{2}{3}C=\frac{1}{3}-\frac{1}{49}\)
\(\Rightarrow\frac{2}{3}C=\frac{46}{147}\)
\(\Rightarrow C=\frac{46}{147}:\frac{2}{3}\)
\(\Rightarrow C=\frac{23}{49}\)
3/3.5+3/5.7+3/7.9+.....+3/47.49
=1-1/5+1/5-1/7+...+1/47-1/49
=1-1/49
=48/49