a) \(1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}.1\frac{1}{24}.........1\frac{1}{99}\)

\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.\frac{25}{24}......\frac{100}{99}\)

\(=\frac{\left(2.2\right).\left(3.3\right).\left(4.4\right).\left(5.5\right)....\left(10.10\right)}{\left(1.3\right).\left(2.4\right).\left(3.5\right).\left(4.6\right).....\left(9.11\right)}\)

\(=\frac{\left(2.3.4.5...10\right).\left(23.4.5....10\right)}{\left(1.2.3.4...9\right).\left(3.4.5.6....11\right)}=\frac{10}{1}.\frac{2}{11}=\frac{20}{11}\)

b) \(\frac{99}{98}-\frac{98}{97}+\frac{1}{97.98}=\frac{99}{98}-\frac{98}{97}+\frac{1}{97}-\frac{1}{98}=\left(\frac{99}{98}-\frac{1}{98}\right)-\left(\frac{98}{97}-\frac{1}{97}\right)\)

\(=\frac{98}{98}-\frac{97}{97}=1-1=0\)

a)  \(\frac{20}{11}\)

b) 0

\(A=\frac{21.99+21}{35.201-35}=\frac{21\left(99+21\right)}{35\left(201-1\right)}=\frac{21.\left(99+1+20\right)}{35.200}=\frac{21.120}{35.200}=\frac{\left(3.7\right).\left(2.2.3.\right)10}{\left(5.7\right).\left(2.2.5\right).10}=\frac{3.3}{5.5}=\left(\frac{3}{5}\right)^2\)

\(B=\frac{\left(-2\right)^{36}.3^{17}}{3^{18}.2^{35}}=\frac{2^{36}.3^{17}}{2^{35}.3^{18}}=2^{\left(36-35\right)}.3^{17-18}=2.3^{-1}=\frac{2}{3}\)

\(C=\frac{7^{100}-7^{99}}{7^{98}-7^{97}}=\frac{7^{99}\left(7^1-7^0\right)}{7^{97}\left(7^1-7^0\right)}=7^{99-97}.\frac{\left(7-1\right)}{7-1}=7^2.1=49\)

mấy bn giải chi tiết giúp mk nhé !

Phân tích mẫu ta có

99/1 + 98/2 +...+1/99 = (98/2 + 1) + (97/3 + 1) +...+(1/99 + 1) +99/1 - 99

( cộng 1 vào mỗi phân số trừ 99/1   do đó phải trừ đi 99 để vẵn được đẳng thức đó)

= 100/2 +100/3 +...+100/99 = 100. (1/2 +1/3 +...+1/99)

Do đó B = [100. (1/2 +1/3 +...+1/99)]/(1/2 +1/3 +..1/99) =100

Phân tích mẫu ta có

99/1 + 98/2 +...+1/99 = (98/2 + 1) + (97/3 + 1) +...+(1/99 + 1) +99/1 - 99

( cộng 1 vào mỗi phân số trừ 99/1   do đó phải trừ đi 99 để vẵn được đẳng thức đó)

= 100/2 +100/3 +...+100/99 = 100. (1/2 +1/3 +...+1/99)

Do đó B = [100. (1/2 +1/3 +...+1/99)]/(1/2 +1/3 +..1/99) =100

100 nhé

\(x-\frac{37}{45}=\frac{4}{5.9}+\frac{4}{9.13}+.....+\frac{4}{41.45}\)

\(\Rightarrow x-\frac{37}{45}=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\)

\(\Rightarrow x-\frac{37}{45}=\frac{1}{5}-\frac{1}{45}\)

\(\Rightarrow x-\frac{37}{45}=\frac{8}{45}\)

\(\Rightarrow x=\frac{37}{45}+\frac{8}{45}\)

\(\Rightarrow x=1\)

Đặt \(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)

\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)

\(=\frac{100}{99}+\frac{100}{3\times97}+\frac{100}{5\times95}+...+\frac{100}{49\times51}\)

\(=100\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)

Đặt \(C=\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{97\times3}+\frac{1}{99\times1}\)

\(=2\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)

\(A=\frac{B}{6}=\frac{100}{2}=50\)

Vậy \(A=50\)

Đặt \(A=\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{98}{2}+\frac{99}{1}\)

\(A=\left(\frac{1}{99}+1\right)+\left(\frac{2}{98}+1\right)+\left(\frac{3}{97}+1\right)+...+\left(\frac{98}{2}+1\right)+1\) ( 99/1 = 99, tất cả 98 ( không tính 99/1) hạng tử trong A đều cộng với 1 , dư ra 1 chỗ cuối)

\(A=\frac{100}{99}+\frac{100}{98}+\frac{100}{97}+...+\frac{100}{2}+\frac{100}{100}\) ( 100/100=1)

\(A=100.\left(\frac{1}{2}+...+\frac{1}{97}+\frac{1}{98}+\frac{1}{99}+\frac{1}{100}\right)\)

Thay A vào E, có:

\(E=\frac{100.\left(\frac{1}{2}+...+\frac{1}{97}+\frac{1}{98}+\frac{1}{99}+\frac{1}{100}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

\(E=100\)

\(\Rightarrow E=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+....+\frac{98}{2}+1+1+...+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)     ( Có 99 số 1)

\(\Rightarrow\frac{\frac{1}{99}+1+\frac{2}{98}+\frac{3}{97}+1+...+\frac{98}{2}+1+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)(Nhóm 98 số 1 với 98 phân số đầu ở trên tử)mik viết thiếu nha sorry *-*

\(\Rightarrow E=\frac{\frac{100}{99}+\frac{100}{98}+\frac{100}{97}+...+\frac{100}{2}+\frac{100}{100}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

\(\Rightarrow E=\frac{\frac{100}{2}+\frac{100}{3}+\frac{100}{4}+...+\frac{100}{99}+\frac{100}{100}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

\(\Rightarrow E=\frac{100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

\(\Rightarrow E=\frac{100.1}{1}=100\)

~Chúc bạn hok tốt~

Trả lời :

\(E=100\)

ủng hộ nha !!!!

a) Đặt B = \(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)

\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)

\(=\frac{100}{1.99}+\frac{100}{3.97}+...+\frac{100}{49.51}\)

\(=100\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}\right)\)

Đặt C = \(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}\)

\(=\left(\frac{1}{1.99}+\frac{1}{99.1}\right)+\left(\frac{1}{3.97}+\frac{1}{97.3}\right)+...+\left(\frac{1}{49.51}+\frac{1}{51.49}\right)\)

\(=2\cdot\frac{1}{1.99}+2\cdot\frac{1}{3.97}+...+2\cdot\frac{1}{49.51}\)

\(=2\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)\)

Thay B và C vào A 

\(\Rightarrow A=\frac{100\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}{2\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}=\frac{100}{2}=50\)

b) Đặt E = \(\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{1}{99}\)

\(=\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{1}{99}+1\right)+1\)

\(=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+\frac{100}{100}\)

\(=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)\)

Thay E vào B

\(\Rightarrow B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)}=\frac{1}{100}\)

a)50

b)1/100

tk ủng hộ nha

a,

\(A=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}}\)

\(A=\frac{\left[1+\frac{1}{99}\right]+\left[\frac{1}{3}+\frac{1}{97}\right]+...+\left[\frac{1}{49}+\frac{1}{51}\right]}{2\left[\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}\right]}\)

\(A=\frac{\frac{100}{1.99}+\frac{100}{3.97}+\frac{100}{5.95}+...+\frac{100}{99.1}}{2\left[\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}\right]}\)

\(A=\frac{100\left[\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}\right]}{2\left[\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}\right]}=\frac{100}{2}=50\)

b, Ta có:

\(\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}=\left[1+\frac{98}{2}\right]+\left[1+\frac{97}{3}\right]+...+\left[1+\frac{1}{99}\right]+1\)

\(=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+\frac{100}{100}=100\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right]\)

Thế vào:

\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right]}=\frac{1}{100}\)

\(B=\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{98}{2}+\frac{99}{1}\)

\(B=\left(1+\frac{1}{99}\right)+\left(1+\frac{2}{98}\right)+...+\left(1+\frac{98}{2}\right)+1\)

\(B=\frac{100}{99}+\frac{100}{98}+...+\frac{100}{2}+\frac{100}{100}\)

\(B=100\left(\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}+\frac{1}{100}\right)\)

Ta có: \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}\right)}=\frac{1}{100}\)

Vậy...

P/s: Hoq chắc

#)Giải :

\(B=\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{98}{2}+\frac{99}{1}\)

\(B=1+\left(\frac{1}{99}+1\right)+\left(\frac{2}{98}+1\right)+\left(\frac{3}{97}+1\right)+...+\left(\frac{98}{2}+1\right)\)

\(B=\frac{100}{100}+\frac{100}{99}+\frac{100}{98}+...+\frac{100}{2}\)

\(B=100\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}\right)\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}\right)}=100\)

Ta có:

\(B=\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{98}{2}+\frac{99}{1}\)

\(B=1+\left(\frac{1}{99}+1\right)+\left(\frac{2}{98}+1\right)\left(\frac{3}{97}+1\right)+...+\left(\frac{98}{2}+1\right)\)

\(B=\frac{100}{100}+\frac{100}{99}+\frac{100}{98}+\frac{100}{97}+...\frac{100}{2}\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\times\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}\right)}\)

\(\Rightarrow\frac{A}{B}=\frac{1}{100}\)