1/1.2 + 1/2.3 + 1/3.4 + ... + 1/x(x + 1) = 99/100
1- 1/2 +1/2-1/3+1/3-1/4+...+ 1/x - 1/ x+ 1 = 99/100
1 - 1/ x+1 = 99/ 100
=> (100 - 1)/ x+1 = 99 / 100
=> x+1 = 100 => x=99
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{99}{100}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{100}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{99}{100}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{99}{100}=\frac{1}{100}\)
\(\Rightarrow x+1=100\)
\(\Rightarrow x=99\)
mình sửa đề mới làm đc cái chỗ 1/2 phải là 1/1.2 ( đúng ko . xem lại )
A = 1/1.2 + 1/2.3 +....+ 1/x.(x+1)=99/100
A=1/1 - 1/2 + 1/2 - 1/3 +...+ 1/x - 1/x+1 =99/100
A = 1 - 1/x+1 = 99/100
A=x+1 - 1/x+1 = 99/100
A=x/x+1 = 99/100
=> x=99
\(\frac{1}{2}+\frac{1}{2.3}+............+\frac{1}{x.\left(x+1\right)}=\frac{99}{100}\)
=>\(\frac{1}{1.2}+\frac{1}{2.3}+............+\frac{1}{x.\left(x+1\right)}=\frac{99}{100}\)
=>\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..............+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{100}\)
=>\(\frac{1}{1}-\frac{1}{x+1}=\frac{99}{100}\)
=>\(\frac{x+1}{x+1}-\frac{1}{x+1}=\frac{99}{100}\)
=>\(\frac{x+1-1}{x+1}=\frac{99}{100}\)
=>\(\frac{x}{x+1}=\frac{99}{100}\)
=>x=99
Vậy x=99
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