(200:20):2=200:(20:2)

(200 - 20) - 2 = 200 - (20 - 2)               : Sai

(200 x 20) x 2 = 200 x (20 x 2)              : Đúng

(200 : 20) : 2 = 200 : (20 : 2)                : Sai

(200 + 20) + 2 = 200 + (20 + 2)              : Đúng

số mủ của 1 dù lớn đến bao hiêu thì cũng bằng 1 suy ra x=1 vậy thay x=1 vào biễu thức là ra

\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}.\)

\(x-\frac{20}{2}.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10.\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10\cdot\frac{4}{55}=\frac{3}{11}\)

\(x-\frac{8}{11}=\frac{3}{11}\)

 x = 1

a) \(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(x-\frac{20}{2}.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10.\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10.\frac{4}{55}=\frac{3}{11}\)

\(x-\frac{8}{11}=\frac{3}{11}\)

x = 1

b) \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\) ( nhân cho cả tử và mẫu của các số hạng trên ( ngoại trừ 2/x.(x+1) ) là 2)

\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

\(2.\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(2.\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\frac{1}{x+1}=\frac{1}{18}\)

=> x + 1 = 18

x = 17

\(a,x-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(\Rightarrow x-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)

\(\Rightarrow x-10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)

\(\Rightarrow x-\frac{8}{11}=\frac{3}{11}\)

\(\Rightarrow x=1\)

\(b,\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{18}=\frac{1}{x+1}\)

\(\Rightarrow x+1=18\Leftrightarrow x=17\)

\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)

\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(x-\left[10.\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)\right]=\frac{3}{11}\)

\(x-\left[10.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\right]=\frac{3}{11}\)

\(x-\left[10.\left(\frac{1}{11}-\frac{1}{55}\right)\right]=\frac{3}{11}\)

\(x-\left[\frac{10}{11}-\frac{10}{55}\right]=\frac{3}{11}\)

\(x-\left[\frac{10}{11}-\frac{2}{11}\right]=\frac{3}{11}\)

\(x-\frac{8}{11}=\frac{3}{11}\)

\(x=\frac{3}{11}+\frac{8}{11}=1\)

1 , ủng hộ mk nha

a) x+15 = 20-4x

=> x=1

b) 17-x=7-6x

=> x=-2

c) -12+x=5x-20

=> x=2

d) 4x-5=15-x

=> x=4

e) 9x-7=20-6x

=>x= \(\frac{9}{5}\)

g)2.(x-10)=3.(x-20)=x-4

=> x thuộc ∅

h) (x^2+2).(x-3) <0

=> x=3,...

Trả lời nhanh giúp mình nhé

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câu a: x=1

Ta có:

\(\left|x-1\right|\ge0;\left|x-2\right|\ge0;\left|x-3\right|\ge0;.....;\left|x-10\right|\ge0\)

\(\Rightarrow\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+....+\left|x-10\right|>0\) vì không xảy ra dấu "="

\(\Rightarrow x-11>0\Rightarrow x>11>0\)

Khi đó bài toán trở thành:

\(x-1+x-2+x-3+.....x-10=x-11\)

\(\Leftrightarrow10x-55=x-11\)

\(\Leftrightarrow9x=44\)

\(\Leftrightarrow x=\frac{44}{9}\)