Tính
A=\(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+\frac{1}{5\cdot95}+...+\frac{1}{97\cdot3}+\frac{1}{99\cdot1}}\)
\(A=\frac{\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+\frac{1}{5\cdot95}+...+\frac{1}{97\cdot3}+\frac{1}{99\cdot1}}\)
Giúp mình giải bài này với
\(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+\frac{1}{5\cdot99}+...+\frac{1}{97\cdot3}+\frac{1}{99\cdot1}}\)
\(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.99}+...+\frac{1}{99.1}}\)
\(=\frac{\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)}{2\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}\)
\(=\frac{\frac{100}{1.99}+\frac{100}{3.97}+...+\frac{100}{49.51}}{2\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}\)
\(=\frac{100\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}{2\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}\)
\(=\frac{100}{2}=50\)
\(\frac{1}{99\cdot97}-\frac{1}{97\cdot95}-\frac{1}{95\cdot93}-....-\frac{1}{5\cdot3}-\frac{1}{3\cdot1}\)
Đặt \(A=\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-...-\frac{1}{5.3}-\frac{1}{3.1}\)
\(A=\frac{1}{99.97}-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{93.95}+\frac{1}{95.97}\right)\)
\(A=\frac{1}{99.97}-\left(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{93}-\frac{1}{95}+\frac{1}{95}-\frac{1}{97}\right)\right)\)
\(A=\frac{1}{99.97}-\left(\frac{1}{2}.\left(1-\frac{1}{97}\right)\right)=\frac{1}{99.97}-\frac{1}{2}.\frac{96}{97}=\frac{1}{99.97}-\frac{48}{97}=-\frac{4751}{9603}\)
Tính\(\frac{1}{99\cdot97}-\frac{1}{97\cdot95}-...-\frac{1}{5\cdot3}-\frac{1}{3\cdot1}\)
bây giờ mìh ban rồi, mìh chỉ có thể chỉ cho bn cách làm thôi
dat bieu thuc la A
2A=2*(...)
2A=2/...-2/...
2A=(1/99-1/97)-(1/97-1/95)-...
2A=1/99-1=-98/99
A=...=-49/99
DUYỆT NHÉ
mìh cũng ko chắc chắn lắm đâu đấy nhé
Tính giá trị của biểu thức \(\frac{1}{99\cdot97}-\frac{1}{97\cdot95}-\frac{1}{95\cdot93}-\frac{1}{5\cdot3}-\frac{1}{3\cdot1}\)
Rút gọn \(A=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+\frac{1}{5\cdot95}+...+\frac{1}{49\cdot51}}\)
ghep cac pso o sbc thanh tung cap de mc giong mau cua cac pso tuong ung ở so chia.bien doi sbc :cong tung pso cach deu 2 dau .
ta duoc\(\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+....+\left(\frac{1}{49}+\frac{1}{51}\right)=\\ \frac{100}{1.99}+\frac{100}{3.97}+....+\frac{100}{49.51}\)bieu thuc nay gap 50 lan so chia.suy ra =50! dung 100% do nha!
\(\frac{1}{99\cdot97}-\frac{1}{97\cdot95}-................-\frac{1}{3\cdot1}\)
Tính D = \(\frac{1}{99\cdot97}-\frac{1}{97\cdot95}-\frac{1}{95\cdot93}-....-\frac{1}{3\cdot1}\)
bn tách 1/ 97 .95 = 1/2 . ( 1/95 -1/97) nha! rồi sử dụng phương pháp khử liên tiếp !
Tính :
\(A=\frac{1\cdot98+2\cdot97+3\cdot96+......+98\cdot1}{1\cdot2+2\cdot3+3\cdot4+......+98\cdot99}\)
\(B=\frac{100-\left(1+\frac{1}{2}+\frac{1}{3}+..........+\frac{1}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+.........+\frac{99}{100}}\)