Câu hỏi tương tự nha !!!!!

a, \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{44}{45}\)

=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{44}{45}\)

=> \(1-\frac{1}{x+1}=\frac{44}{45}\)

=> \(\frac{x}{x+1}=\frac{44}{45}\)

=> x = 44

b, Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)

.................

\(\frac{1}{45^2}< \frac{1}{44.45}=\frac{1}{44}-\frac{1}{45}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{44}-\frac{1}{45}=1-\frac{1}{45}< 1\)

Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1\)

a) 1/1.2+1/2.3+1/3.4+...+1/x(x+1)=1-1/2+1/2-1/3+1/3-1/4+....+1/x-1/(x+1)=1-1/(x+1)=x/(x+1)=44/45

=> x=44

b/ 1/2² < 1/1.2; 1/3² < 1/2.3; ....; 1/45² < 1/44.45

=> A < 1/1.2+1/2.3+...+1/44.45=1-1/45=44/45 < 1

=> A < 1

\(5x+\left(1+2+3+4\right)=45\)

\(5x+10=45\)

\(5x=35\)

\(x=7\)

Vậy \(x=7\)

\(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)=45\)

\(\Leftrightarrow5x+\left(1+2+3+4\right)=45\)

\(\Leftrightarrow5x+10=450\Rightarrow5x=440\Rightarrow x=88.\)

\(5x+10=45\)

\(5x=35\)

\(x=7\)

x+1 + x+2 + x+3 + x +4 + x+5 = 45

x + x + x + x + x = 45 -1 -2 -3 -4 -5

5x = 30

x =6

(x+1)+(x+2)+(x+3)+(x+4)+(x+5)=45 

(x + x + x + x + x) + ( 1 + 2 + 3 + 4 + 5)  = 45

\(x\times5\)+ 15 = 45

\(x\times5\)       = 45 - 15

\(x\times5\)        = 30

x                      = 30 : 5

x                      = 6.

Hok tốt !