\(17.8+51.4=34.4+51.4=4\left(51+34\right)=4.84=336\) \(2.2.3.5.19=\left(2.5\right).\left(3.19\right).2=10.2.57=570.2=1140\) \(54.275+825.15+275=54.275+45.275+275=275\left(54+45+1\right)=100.275=27500\) \(\frac{167.198+98}{198.168-100}=\frac{167.198+98}{198.167+198-100}=\frac{167.198+98}{167.198+98}=1\)

\(\frac{1}{n}-\frac{1}{n+k}=\frac{k}{n\left(n+k\right)}\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{2020}=1-\frac{1}{2020}=\frac{2019}{2020}\)

a) 17 x 8 + 51 x 4

= 17 x 4 x 2 + 17 x 3 x 4

= 17 x 4 x ( 2 + 3 )

= 14 x 4 x 5

= 14 x 20

= 280

b) 2 x 2 x 3 x 5 x 19

= ( 2 x 5 ) x ( 3 x 19 ) x 2

= 10 x 57 x 2

= 570 x 2

= 1140

c) 54 x 275 + 825 x 15 + 275

= 54 x 275 + 275 x 3 x 15 + 275 x 1

= 54 x 275 + 275 x 45 + 275 x 1

= 275 x ( 54 + 45 + 1 )

= 275 x 100

= 27500

d) 100 - 99 + 98 - 97 + 96 - 95 + 94 - 93 + ... + 4 - 3 + 2

= (100 - 99) + (98 - 97) + (96 - 95) + (94 - 93) + ... + (4 - 3) + 2

= (1 + 1 + ... + 1) + 2

( 49 số 1 )

= 49 + 2

= 51

k) 1,5 + 2,5 + 3,5 + 4,5 + 5,5 + 6,5 + 7,5 + 8,5

= ( 1,5 + 8,5 ) + ( 2,5 + 7,5 ) + ( 3,5 + 6,5 ) + ( 4,5 + 5,5 )

= 10 + 10 + 10 + 10

= 40

m) Đề bn xem ở trên

= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

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

\(=\frac{1}{1}-\frac{1}{10}\)

\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)

\(a)\) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)

\(\Leftrightarrow\)\(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=-4+4\)

\(\Leftrightarrow\)\(\frac{x+1+99}{99}+\frac{x+2+98}{98}+\frac{x+3+97}{97}+\frac{x+4+96}{96}=0\)

\(\Leftrightarrow\)\(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)

\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)

Nên \(x+100=0\)

\(\Rightarrow\)\(x=-100\)

Vậy \(x=-100\)

Chúc bạn học tốt ~ 

\(b)\) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{2008}{2009}\)

\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)

\(\Leftrightarrow\)\(1-\frac{1}{x+1}=\frac{2008}{2009}\)

\(\Leftrightarrow\)\(\frac{1}{x+1}=1-\frac{2008}{2009}\)

\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{2009}\)

\(\Leftrightarrow\)\(x+1=2009\)

\(\Leftrightarrow\)\(x=2009-1\)

\(\Leftrightarrow\)\(x=2008\)

Vậy \(x=2008\)

Chúc bạn học tốt ~ 

a) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)

\(\Leftrightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\)\(\left(\frac{x+4}{96}+1\right)=-4+4\)

\(\Leftrightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

Mà  \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)

\(\Rightarrow x+100=0\)

\(\Leftrightarrow x=-100\)

ta gọi \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\)là A

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(\Leftrightarrow1.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\right)\)

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

ta gọi B là biểu thức thứ2

\(B=\frac{2.2}{3}\times\frac{3.3}{2.4}\times\frac{4.4}{3.5}\times...\times\frac{10.10}{9.11}\)

\(\Rightarrow\)2 x \(\frac{10}{11}\)\(=\frac{20}{11}\)

\(\Rightarrow\)\(x+\frac{9}{10}=\frac{20}{11}+\frac{9}{110}\)

\(\Rightarrow x=1\)

mk nghĩ vậy bạn ạ, mk mong nó đúng

a, Câu hỏi của Nguyễn Ánh Ngân - Toán lớp 6 - Học toán với OnlineMath

b, Câu hỏi của Vũ Xuân Hiếu - Toán lớp 6 | Học trực tuyến

c)$x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}=\frac{5}{24}$

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

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

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

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

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

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

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

a) = 3/3 x ( -24/54 +45/54 ) : 7/12

   = 1 x 21/54 x 12/7

   = 18/27 

( hiện tại mik chỉ lm đc thế này thui. thông cảm nk )