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![](https://rs.olm.vn/images/avt/0.png?1311)
Vì \(\left|x+\dfrac{1}{1\cdot2}\right|+\left|x+\dfrac{1}{2\cdot3}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|\ge0\forall x\)
\(\Rightarrow100x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left|x+\dfrac{1}{1\cdot2}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|=x+\dfrac{1}{1\cdot2}+...+x+\dfrac{1}{99\cdot100}\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{1\cdot2}+...+\dfrac{1}{99\cdot100}\right)=100x\)
\(\Rightarrow99x+\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)=100x\)
\(\Rightarrow\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}=x\)
\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=x\)
\(\Rightarrow x=1-\dfrac{1}{100}=\dfrac{99}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)...\left(1+\dfrac{1}{2021.2023}\right)\)
\(=\dfrac{4}{1.3}.\dfrac{9}{2.4}...\dfrac{4088484}{2021.2023}\)
\(=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}...\dfrac{2022.2022}{2021.2023}\)
\(=\dfrac{2.2022}{1.2023}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
do vế trái luôn luôn lớn hơn hoặc =0
=> vế phải cx luôn luôn lớn hơn hoặc =0
=> bỏ giá trị tuyệt đối =100x
có 99x + ........... = 100x
trừ là ra nha bn
ta có:
|x+1/1.2|+|x+1/2.3|+...+|x+1/99.100|=100x
=>|x+1/1.2+x+1/2.3+...+x+1/99.100|=100x
<=>|(x+x+x+...+x)+1/1.2+1/2.3+....1/99.100|=100x
<=>|x.99+1-1/2+1/2-1/3+1/3-1/4+.....+1/99-1/100|=100x
<=>|x.99+1-1/100|=100x
<=>|99x+99/100|=100x
Có 2 trường hợp
TH1
99x+99/100=100x
=>100x-99x=99/100
<=>x=99/100
=>x=99/100
TH2:
99x+99/100=-100x
-100x-99x=99/100
<=>-199x=99/100
<=>x=99/-19900( loại vì |99x+99/100| là số dương nên 100x là số dương mà x là sô âm nên 100x là số âm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=\left(\dfrac{2^2-1}{2^2}\right)\left(\dfrac{1-3^2}{3^2}\right)\left(\dfrac{4^2-1}{4^2}\right)...\left(\dfrac{1-99^2}{100^2}\right)\left(\dfrac{100^2-1}{99^2}\right)=\left(\dfrac{1.3}{2^2}\right)\left(\dfrac{-2.4}{3^2}\right)\left(\dfrac{3.5}{4^2}\right)...\left(\dfrac{-98.100}{99^2}\right)\left(\dfrac{99.101}{100^2}\right)=-\dfrac{101}{200}\)
\(\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)...\left(1+\dfrac{1}{99.101}\right)\)
\(=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}....\dfrac{100^2}{99.101}\)
\(=\dfrac{2.3.4...100}{1.2.3.4...99}.\dfrac{2.3.4...100}{3.4.5....101}\)
\(=\dfrac{100}{1}.\dfrac{2}{101}\)
\(=\dfrac{200}{101}\)