\(\left(\frac{11}{1.12}+\frac{11}{12.23}+...+\frac{11}{89.100}\right)+x=\frac{5}{3}\)

\(\Leftrightarrow\left(\frac{1}{1}-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+...+\frac{1}{89}-\frac{1}{100}\right)+x=\frac{5}{3}\)

\(\Leftrightarrow\left(1-\frac{1}{100}\right)+x=\frac{5}{3}\Leftrightarrow\frac{99}{100}+x=\frac{5}{3}\Leftrightarrow x=\frac{5}{3}-\frac{99}{100}\)

\(\Leftrightarrow x=\frac{297}{300}=\frac{ }{100}\)

mk nhầm \(x=\frac{203}{300}\)

Pn oi tim x o dau do , day chi co n thoi ma

a) \(\frac{22}{7}\div\left(11-\chi\right)=\frac{7}{5}-\frac{2}{3}\)

\(\frac{22}{7}\div\left(11-\chi\right)=\frac{11}{15}\)

\(\left(11-\chi\right)=\frac{22}{7}\div\frac{11}{15}\)

\(\left(11-\chi\right)=\frac{30}{7}\)

\(\chi=11-\frac{30}{7}\)

\(\chi=\frac{47}{7}\)

b) (x+1)+(x+2)+(x+3)+...+(x+100)=5550

Từ 1 đến 100 có 100 số hạng => Có 100 x

(x + x + x + .... + x) + (1 + 2 + 3 + .. + 100) = 5550

Áp dụng tính chất cộng dãy số cách đều, ta có

(100.x) + 5050 = 5550

100.x = 5550 - 5050

100.x = 500

x = 500 : 100

x = 5 

a, (x-1) . 0,5 = 7,5 : (x-1) 

=> = ( x  - 1 ) 0,5 = \(\frac{x-1}{2}\)

\(=\frac{7,5}{x-1}=\frac{15}{2\left(x-1\right)}\)

=> x = - 1 \(\sqrt{15}\)

x = \(\sqrt{15+1}\)

đề sao sao ý

\(\frac{x-5}{3}=\frac{6}{5}\)

\(x-5=\frac{6}{5}.3\)

\(x-5=\frac{18}{5}\)

\(x=\frac{18}{5}+5\)

\(\Rightarrow x=\frac{43}{5}\)

Vậy ...

\(\frac{1}{3}=\frac{2-x}{4}\)

\(\frac{4}{3}=2-x\)

\(x=2-\frac{4}{3}\)

\(\Rightarrow x=\frac{2}{3}\)

Vậy ...

x - 5 = 6 * 3 / 5 => x - 5 = 3.6 => x = 8.6

2 - x = 4 / 3 => x = 2 / 3

a) (x-1)+(x-2)+(x-3)+...+(-100)=101

(x+x+x+...+x)-(1+2+3+...+100)=101

=> 100x-5050=101

100x=101+5050

100x=5151

x=5151:100

x=5151/100

a, \(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{x\cdot\left(x+1\right)\cdot\left(x+2\right)}=\frac{2018}{2019}\)

\(=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot3}+...+\frac{1}{x\cdot\left(x+1\right)}-\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=\frac{2018}{2019}\)

\(=1-\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=\frac{2018}{2019}\)

\(\Rightarrow\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=1-\frac{2018}{2019}\)

\(\Rightarrow\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=\frac{2019}{2019}-\frac{2018}{2019}=\frac{1}{2019}\)

Đến đây bn tự tính nhé !!

Nhưng \(\frac{1}{1.2.3}\)ko bằng \(\frac{1}{1.2}-\frac{1}{2.3}\)ạ

Bn có thể suy nghĩ lại giúp mik đc ko????