\(x=\frac{1}{2}\) => \(B\left(x\right)=B\left(\frac{1}{2}\right)=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{100}\)

\(x\times B\left(x\right)=x+x^2+x^3+x^4+...+x^{100}+x^{101}\)

\(\frac{1}{2}\times B\left(\frac{1}{2}\right)=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{100}+\left(\frac{1}{2}\right)^{101}\)

\(B\left(\frac{1}{2}\right)-\frac{1}{2}\times B\left(\frac{1}{2}\right)=\frac{1}{2}\times B\left(\frac{1}{2}\right)=1-\left(\frac{1}{2}\right)^{101}\)

\(B\left(x\right)=\frac{1}{2}B\left(x\right)\times2=\left(1-\left(\frac{1}{2}\right)^{101}\right)\times2=2-\left(\frac{1}{2}\right)^{100}\)

Thay \(x=\frac{1}{2}\) vào đa thức B(x) ta có :

     \(B\left(\frac{1}{2}\right)=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+.....+\left(\frac{1}{2}\right)^{100}\)

\(\Leftrightarrow2B\left(\frac{1}{2}\right)=2\left(1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+.....+\left(\frac{1}{2}\right)^{100}\right)\)

\(\Leftrightarrow2B\left(\frac{1}{2}\right)=2+1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+......+\left(\frac{1}{2}\right)^{99}\)

Ta có :

 \(2B\left(\frac{1}{2}\right)-B\left(\frac{1}{2}\right)=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{100}}\right)\)

 \(\Leftrightarrow B\left(\frac{1}{2}\right)=2-\frac{1}{2^{100}}\)

Vậy tại \(x=\frac{1}{2}\) thì đa thức \(B\left(x\right)\) có giá trị là \(2-\frac{1}{2^{100}}\)

\(\Leftrightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=0\)

\(\Leftrightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{97}+\frac{1}{98}+\frac{1}{99}\right)=0\)

\(\Leftrightarrow x+100=0\text{ (do }\frac{1}{97}+\frac{1}{98}+\frac{1}{99}\ne0\text{)}\)

\(\Leftrightarrow x=-100\)

bn tự chép đề lại nha

từ đề bài suy ra  \(1+\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+3=1+1+1+0=3\)

suy ra \(\frac{x+1+99}{99}+\frac{x+2+98}{98}+\frac{x+3+97}{97}+3=3\)

suy ra \(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=3-3=0\)

suy ra \(\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\right)=0\)

mà 1/99 +1/98+1/97 lớn hơn 0

từ 2 điều trên suy ra x+100=0 suy ra x=-100

\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}\)=\(-4\)

<=>\(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=0\)=>\(\frac{x+1+99}{99}+\frac{x+2+98}{98}+\frac{x+3+97}{97}+\frac{x+4+96}{96}=0\)

=>\(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)

=>\(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

Vì: \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)

=>\(x+100=0\)

\(x=0-100\)

\(x=-100\)

Vậy \(x=-100\)

Ta có 51/2.52/2...100/2

       = 1.2.3....100/1.2...50.2.2...2 (nhân cả tử và mẫu với 1.2.3...50)

      = 1.2.3...100/(1.2)(2.2)(3.2)...(50.2)

      = 1.2.3...100/2.4.6...100

      = 1.3.5...99 => đpcm nhớ giữ lời hứa đấy

\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)

\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}=\frac{3.4.5...100}{2.3.4...99}=\frac{100}{2}=50\)

100 / 2=50

b:

3/2 x 4/3 x 5/4 x ......... x 8/7 x 9/8

Ta loai bo so giong nhau o TS va MS

Ta duoc 9/2