Ta có: \(A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{99^2}{98.100}\)

\(A=\frac{\left(2.3.4.5.....99\right).\left(2.3.4.5.....99\right)}{\left(1.2.3.4.....98\right).\left(3.4.5.6.....100\right)}\)

\(A=\frac{99.2}{100}=\frac{99}{50}\)

Học tốt!!!!

A=1²/1.2 . 2²/2.3 . 3²/3.4 . 4²/4.5 . 5²/5.6

⇒1.1/1.2 . 2.2/2.3 . 3.3/3.4 . 4.4/4.5 . 5.5/5.6

⇒1.2.3.4.5/1.2.3.4.5 . 1.2.3.4.5/2.3.4.5.6

⇒1 . 1/6 =1/6.

Vậy A=1/6

B=2²/1.3 . 3²/2.4 . 4²/3.5 . 5²/4.6

⇒2.2/1.3 . 3.3/2.4 . 4.4/3.5 . 5.5/4.6

⇒2.2.3.3.4.4.5.5/1.3.2.4.3.5.4.6 =48.

Vậy B=48.

A = (2.3.4. .... .999/1.2.3. .... .998) . (2.3.4. .... .999/3.4.5. ..... .1000)

    = 999. 2/1000

    = 999/500

Tk mk nha

1)

2/3.5+2/5.7+...+2/11.13+2/13.15+2/1.2+2/2.3+...+2/9.10

=(2/3.5+...2/13.15)+(2/1.2+...+2/9.10)

= (2/3-2/15)+ [2(1-1/10)]

=8/15+9/5

=7/3

2)

11/12+11/12.24+...+11/88.99

=11-1/9

=10/8/9

d) Ta có: \(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{-37}{45}\)

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

\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{-37}{45}\)

\(\Leftrightarrow x=\dfrac{-37}{45}+\dfrac{1}{45}-\dfrac{1}{5}=\dfrac{-36}{45}-\dfrac{1}{5}=\dfrac{-4}{5}-\dfrac{1}{5}=-1\)

Vậy: x=-1

\(A=\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{999^2}{998\cdot1000}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot...\cdot999^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot998\cdot1000}\\ =\dfrac{\left(2\cdot3\cdot4\cdot...\cdot999\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot999\right)}{\left(1\cdot2\cdot3\cdot...\cdot998\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot1000\right)}\\ =\dfrac{2\cdot3\cdot4\cdot...\cdot999}{1\cdot2\cdot3\cdot...\cdot998}\cdot\dfrac{2\cdot3\cdot4\cdot...\cdot999}{3\cdot4\cdot5\cdot...\cdot1000}\\ =999\cdot\dfrac{1}{500}\\ =\dfrac{999}{500}\)

\(S=\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\)

\(=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\)

\(=\dfrac{1}{1}-\dfrac{1}{11}=\dfrac{11}{11}-\dfrac{1}{11}=\dfrac{10}{11}\)

\(B=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}......\frac{10^2}{9.11}=\frac{\left(1.2.3.....10\right)^2}{\left(1.2.3.....9\right).\left(3.4.5....9.10.11\right)}=\frac{\left(1.2.3....10\right)^2}{\left(1.2\right)\left(3.4.5.....9\right)^2\left(10.11\right)}=\frac{\left(1.2.10\right)^2}{\left(1.2\right).\left(10.11\right)}=\frac{1.2.10}{11}=\frac{20}{11}\)