chia ra làm các nhóm (13-12+11)+(10-9+8)-(7-6+5) -(4+3+2-1)

hay ta thấy 3 nhóm đầu có tổng  =số trừ trong các nhóm đó tức là 12+9-6+(4+3+2-1)

Hay= 23

ê câu này bằng13

Kết quả bằng 13 đúng không?

11 nha

bằng 13

`(2 1/3 + 3 1/2): (-4 1/6 + 3 1/7) +7,5`

`=(7/3 +7/2) : (-25/6 + 22/7) + 15/2`

`=35/6 : (-43/42) + 15/2`

`=-245/43+15/2`

`=155/86`

\(\text{Thực hiện phép tính}\)

   \(3\text{ × }\left(\frac{1}{7}+\frac{1}{3}-\frac{3}{14}\right) : \frac{11}{4}\)

\(=3\text{ x }\left(\frac{3}{21}+\frac{7}{21}-\frac{3}{14}\right)\text{ : }\frac{11}{4}\)

\(=3\text{ x }\left(\frac{10}{21}-\frac{3}{14}\right)\text{ : }\frac{11}{4}\)

\(=3\text{ x }\left(\frac{140}{294}-\frac{63}{294}\right)\text{ : }\frac{11}{4}\)

\(=3\text{ x }\frac{77}{294}\text{ : }\frac{11}{4}\)

\(=\frac{231}{294}:\frac{11}{4}\)

\(=\frac{231}{294}\text{ x }\frac{4}{11}\)

\(=\frac{924}{3234}=\text{Phân số này có thể rút gọn đó bạn ! Bạn tự rút gọn nha !}\)

   \(\left(\frac{21}{8}+\frac{1}{2}\right)\text{ : }\frac{5}{16}\)

\(=\left(\frac{21}{8}+\frac{4}{8}\right)\text{ : }\frac{5}{16}\)

\(=\frac{25}{8}\text{ : }\frac{5}{16}\)

\(=\frac{25}{8}\text{ x }\frac{16}{5}\)

\(=\frac{400}{40}=10\)

\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{5}\right)\cdot....\cdot\left(1-\frac{1}{2003}\right)\cdot\left(1-\frac{1}{2004}\right)\)

\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot....\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)

\(=\frac{1\cdot2\cdot3\cdot4\cdot....\cdot2002\cdot2003}{2\cdot3\cdot4\cdot5\cdot....\cdot2003\cdot2004}\)

\(=\frac{1}{2004}\)

phân số thứ nhất bằng 0/2 =0

vậy tích đó chắc chắn bằng 0

chuẩn ko mn?

a)\(\frac{5}{9}:\left(\frac{1}{11}-\frac{5}{22}\right)+\frac{5}{9}:\left(\frac{1}{15}-\frac{2}{3}\right)\)

\(=\frac{5}{9}:\left[\left(\frac{1}{11}-\frac{5}{22}\right)+\left(\frac{1}{15}-\frac{2}{3}\right)\right]\)

\(=\frac{5}{9}:\left[-\frac{3}{22}+-\frac{3}{5}\right]\)

\(=\frac{5}{9}:\frac{-81}{110}\)

\(=\frac{-550}{729}\)

a: \(=-\left(10x^2+15x-8x-12\right)=-10x^2-7x+12\)

-10x^2-7x+12

= 12-7x-10x^2

1: ĐKXĐ:  \(x\notin\left\{0;3\right\}\)

\(\dfrac{x+3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}\)

\(=\dfrac{x+3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x\left(x-3\right)}\)

\(=\dfrac{\left(x+3\right)\left(x-3\right)-x^2+9}{x\left(x-3\right)}\)

\(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}\)

=0

2: ĐKXĐ: \(x\notin\left\{0;1\right\}\)

\(\dfrac{3}{x}-\dfrac{5}{x-1}+\dfrac{3x+2}{x^2-x}\)

\(=\dfrac{3}{x}-\dfrac{5}{x-1}+\dfrac{3x+2}{x\left(x-1\right)}\)

\(=\dfrac{3x-3-5x+3x+2}{x\left(x-1\right)}\)

\(=\dfrac{x-1}{x\left(x-1\right)}=\dfrac{1}{x}\)

a) tích có 100 thừa số => n=100 
=> A=0 :))