a) \(\frac{-2}{5}+\frac{3}{10}+\frac{-3}{5}\)

\(=\left[\left(-\frac{2}{5}\right)+\left(-\frac{3}{5}\right)\right]+\frac{3}{10}\)

\(=\left(-1\right)+\frac{3}{10}\)

\(=-\frac{7}{10}.\)

c) \(15\frac{1}{4}:\frac{5}{7}-25\frac{1}{4}:\frac{5}{7}\)

\(=\frac{61}{4}:\frac{5}{7}-\frac{101}{4}:\frac{5}{7}\)

\(=\left(\frac{61}{4}-\frac{101}{4}\right):\frac{5}{7}\)

\(=\left(-10\right):\frac{5}{7}\)

\(=-14.\)

Chúc bạn học tốt!

a)A=1/10+1/15+...+1/120

=2(1/20+1/30+...+1/240)

=2(1/4*5+1/5*6+...+1/15*16)

=2*(1/4-1/5+1/5-1/6+...+1/15-1/16)

=2*[(1/4-1/16)+(1/5-1/5)+...+(1/15-1/15)]

=2*[(4/16-1/16)+0+...+0]

=2*3/16=3/8

b) B=1+1/3+1/6+...+1/1225

=2(1/2+1/6+1/12+...+1/2450)

=2(1/1*2+1/2*3+...+1/49*50)

=2*[1-1/2+1/2-1/3+...+1/49-1/50]

=2*[(1-1/50)+(1/2-1/2)+...+(1/49-1/49)]

=2*[(50/50-1/50)+0+...+0]

=2*49/50=49/25

a,\(\frac{1}{2}A=\frac{1}{2}\left(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\right)\)

\(\frac{1}{2}A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\)

\(\frac{1}{2}A=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\)

\(\frac{1}{2}A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\)

\(\frac{1}{2}A=\frac{1}{4}-\frac{1}{16}\)\(\frac{1}{2}A=\frac{3}{16}\)suy ra \(A=\frac{3}{16}:\frac{1}{2}=\frac{3}{8}\)

B thì cậu có thể làm nhiều cách 

nkky6h njki

A  em tự tính nhé 

b) B = 1+ 3 + 3²+...+3⁹⁹

 3B = 3+ 3²+3³+...+3¹⁰⁰

do đó 3B-B= (3+3²+3³+...+3¹⁰⁰) - ( 1+3+3²+...+3⁹⁹)

                  2B= 3¹⁰⁰-1

                      B= (3¹⁰⁰-1) : 2

c) \(C=1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x.\left(x+1\right)}\)

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

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

    \(C=1+\frac{1}{2}.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}\right)\)

     \(C=1+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)\)

   \(C=1+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{x+1}\right)\)

Phần c thế này thôi vì ko có giá trị x cụ thể .

d) \(D=\frac{9}{8}.\frac{16}{15}.\frac{25}{24}.....\frac{8100}{8099}\)

   \(D=\frac{9.16.25....8100}{8.15.24....8099}\)

\(D=\frac{3.3.4.4.5.5....90.90}{2.4.3.5.4.6.....89.91}\)

\(D=\frac{\left(3.4.5...90\right).\left(3.4.5...90\right)}{\left(2.3.5...89\right).\left(4.5.6...91\right)}\)

\(D=\frac{3.4.5...90}{2.3.4...89}.\frac{3.4.5...90}{4.5.6...91}\)

\(D=\frac{90}{2}.\frac{3}{91}\)

\(D=45.\frac{3}{91}=\frac{135}{91}\)

Ta có: \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)....\left(1-\frac{1}{780}\right)\)

\(=\frac{2}{3}.\frac{5}{6}...\frac{779}{780}\)

\(=\frac{4}{6}.\frac{10}{12}....\frac{1558}{1560}\)

\(=\frac{1.4.2.5....38.41}{2.3.3.4....39.40}=\frac{\left(1.2.3..38\right)\left(4.5...41\right)}{\left(2.3.4...39\right)\left(3...40\right)}=\frac{41}{39.3}=\frac{41}{117}\)

 \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)........\left(1-\frac{1}{780}\right)\)

\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}........\frac{779}{780}\)

\(=\frac{4}{6}.\frac{10}{12}\frac{18}{20}.\frac{28}{30}.........\frac{1558}{1560}\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}...............\frac{38.41}{39.40}\)

\(=\frac{\left(1.2.3.4......38\right)\left(4.5.6.7..........41\right)}{\left(2.3.4.5.........39\right)\left(3.4.5.6.........40\right)}\)

\(=\frac{1.41}{39.3}\)

\(=\frac{41}{117}\)

Vậy \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)........\left(1-\frac{1}{780}\right)=\frac{41}{117}\)

\(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{780}\right)\)

\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}...\frac{779}{780}=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}...\frac{1558}{1560}\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{38.41}{39.40}\)

\(=\frac{\left(1.2.3...38\right)\left(4.5.6...61\right)}{\left(2.3.4...39\right)\left(3.4.5...40\right)}=\frac{1.41}{39.3}=\frac{41}{117}\)

(7x6=5+2+6x7) =