Biến đổi tử số  và mẫu số,ta có :

Tử số : 20,2 x 51 - 30,3 x 3,4 + 14,58

          = 103,02 - 103,02 + 14.58

          = 0 + 14,58 = 14,58

Mẫu số : 14,58 x 460 + 7,29 x 540 x 2

         = 14,58 x 460 + 14,58 x 540

         = 14,58 x \((460+540)\)

        = 14.58 x 1000

        = 14580

Thay vào ta có : = 14,58 : 14580

                         = 0,001

Vậy : \(\frac{20,2\cdot5,1-30,3\cdot3,4+14,58}{14,58\cdot460+7,29\cdot540\cdot2}=0,001\)

~Chúc bn học giỏi~

=18.88

=14580 **** mk nha

    20,2*5,1-30,3*3,4+14,58

=103,02-103,02+14,58

=14,58

14,58*460+7,92*540*2

=6706,8+4276,8*2

=6706,8+8553,6

=15260,4

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

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

\(=1-\frac{1}{100}=\frac{99}{100}\)

\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{!}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}=\frac{100}{101}\)

\(C=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....+\frac{1}{1024}+\frac{1}{2048}\)

\(\Rightarrow\)\(2C=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{512}+\frac{1}{1024}\)

\(\Rightarrow\)\(2C-C=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\right)\)

\(\Leftrightarrow\)\(C=1-\frac{1}{2048}=\frac{2047}{2048}\)

Câu A bạn quên 1/4.5 kìa , với câu D đâu >>>
 

a)\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+x=\frac{3}{5}\)

\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+x=\frac{3}{5}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+x=\frac{3}{5}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{10}+x=\frac{3}{5}\)

\(\Rightarrow\frac{2}{5}+x=\frac{3}{5}\)

\(\Rightarrow x=\frac{3}{5}-\frac{2}{5}=\frac{1}{5}\)

b)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)

\(\Rightarrow\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{13}-\frac{2}{15}+x=\frac{1}{3}\)

\(\Rightarrow\frac{2}{3}-\frac{2}{15}+x=\frac{1}{3}\)

\(\Rightarrow\frac{8}{15}+x=\frac{1}{3}\)

\(\Rightarrow x=\frac{1}{3}-\frac{8}{15}=-\frac{1}{5}\)

c)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)

\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{9}{10}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{x+1}=\frac{9}{10}\)

\(\Leftrightarrow\frac{x+1-1}{x+1}=\frac{9}{10}\)

\(\Rightarrow\frac{x}{x+1}=\frac{9}{10}\)

\(\Rightarrow x=9\)

b) \(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)

\(\Leftrightarrow\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{15-13}{13.15}+x=\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+x=\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{15}+x=\frac{1}{3}\)

\(\Leftrightarrow x=\frac{1}{15}\)

1.Ta có:1/2=1/1.2

1/6=1/2.3

1/12=1/3.4

........

=>1/2+1/6+1/12+1/20+1/30=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6

=1-1/6

=5/6

2.hinh nhu sai đe

Cái đoạn mở ngoặc mik viết nhầm nha

a: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}=\dfrac{2-10+3}{15}=\dfrac{-5}{15}=\dfrac{-1}{3}\)

b: \(=\left(6+\dfrac{1}{8}-\dfrac{1}{2}\right)\cdot4=\dfrac{48+1-4}{8}\cdot4=\dfrac{45}{2}\)

c: \(=\dfrac{1}{4}\cdot4-2\cdot\dfrac{1}{4}=1-\dfrac{1}{2}=\dfrac{1}{2}\)

d: \(F=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{2008\cdot2010}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)

\(=2\cdot\dfrac{1004}{2010}=\dfrac{1004}{1005}\)