\(26\times7-17\times9+13\times26-17\times11\)

\(=26\times\left(7+13\right)-17\left(9+11\right)\)

\(=26\times20-17\times20\)

\(=20\times\left(26-17\right)\)

\(=20\times11\)

\(=220\)

a)(125x37x32):4 = 37000

b)374:(17x11)=2 ai k cho minh minh se k lai cho

a) = 37000

b) = 2

Hình như lúc nãi bạn cũng ra bài này Lục Thanh Nguyên ạ

cho ra phép tính rõ ràng

a) = 37000

b) = 2

mik sắp hết âm rồi, chỉ còn 69 nữa thôi, các bạn giúp mik nha, khi nào hết âm mik sẽ ra câu hỏi và giúp các bạn 

mơn các bạn trc và cx mơn các bạn mà đã giúp mik trc kia ~~~~~~~~~~

a = 37000

b = 2

a)(125x37x32):4 = 37000

b)374:(17x11)=2

Lời giải:

$A=-1+2+(-3)+4+(-5)+6+....+(-4025)+4026$

$=[(-1)+2]+[(-3)+4]+[(-5)+6]+...+[(-4025)+4026]$

$=\underbrace{1+1+1+.....+1}_{2013}=1.2013=2013$

Đặt \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{4026}\)

\(=1-\frac{1}{x+1}=\frac{2011}{4026}\)

\(\Rightarrow\)x+1=4026

x=4026-1

x=4025

Vậy x=4025.

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)

=> \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)

=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{4026}\)

=> \(1-\frac{1}{x+1}=\frac{2011}{4026}\)

=> \(\frac{1}{x+1}=\frac{2015}{4026}\Rightarrow x+1=\frac{4026}{2015}\Rightarrow x=\frac{2011}{2015}\)

Tớ làm bị sai đấy nhé. Bạn có thể tham khảo bài của Xyz (nếu đúng).

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)

=> \(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)

=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{4026}\)

=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)

=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}=\frac{1}{2013}\)

=> x + 1 = 2013 => x = 2012

Trả lời:

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{x.\left(x+1\right)}=\frac{2011}{4026}\)

\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{x.\left(x+1\right)}=\frac{2011}{4026}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{4026}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2013}\)

\(\Leftrightarrow x+1=2013\)

\(\Leftrightarrow x=2012\)

Vậy \(x=2012\)

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)

\(\Rightarrow\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{4026}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)

\(\Rightarrow\frac{x+1}{2\left(x+1\right)}-\frac{2}{2\left(x+1\right)}=\frac{2011}{4026}\)

\(\Rightarrow\frac{x-1}{2\left(x+1\right)}=\frac{2011}{4026}\)

\(\Rightarrow\frac{4026\left(x-1\right)}{4026\cdot2\left(x+1\right)}=\frac{2011\cdot2\left(x+1\right)}{4026\cdot2\left(x+1\right)}\)

\(\Rightarrow4026x-4026=4022x+4022\)

\(\Rightarrow4026x-4022x=4022+4026\)

\(\Rightarrow4x=8048\)

\(\Rightarrow x=2012\)