=1/1.3-1/3.5+1/3.5-1/5.7+...+1/99.11-1/11.13

=1/1.3-1/11.13

=1/3-1/143

=140/429

a.\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Mà: \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\Rightarrow x+1=0\Rightarrow x=-1\)

b.

\(\frac{x+4}{1990}+\frac{x+3}{1991}=\frac{x+2}{1992}+\frac{x+1}{1993}\Rightarrow2+\frac{x+4}{1990}+\frac{x+3}{1991}=2+\frac{x+2}{1992}+\frac{x+1}{1993}\)

\(\Rightarrow\left(1+\frac{x+4}{1990}\right)+\left(1+\frac{x+3}{1991}\right)=\left(1+\frac{x+2}{1992}\right)+\left(1+\frac{x+1}{1993}\right)\)

\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}=\frac{x+1994}{1992}+\frac{x+1994}{1993}\)

\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}-\frac{x+1994}{1992}-\frac{x+1994}{1993}=0\)

\(\Rightarrow\left(x+1994\right)\left(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\right)=0\)

\(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\ne0\Rightarrow x+1994=0\Rightarrow x=-1994\)

cái này ko khó bạn áp dụng wuy luật là tính dc

B==1/4.(4/1.3.5+1/3.5.7+...+1/47.49.51)

B=1/1.3-1/3.5+1/3.5-1/5.7+....+1/47.49-1/49.50

B=1/4.(1/3.5-1/49.50)

câu B nhân lên 4 rồi tính

câu C để đó là tính dc

Các bạn tính cặn kẽ v

Đây là bài tính t nha !!!

Ai nhanh mik T>I>C>K

=4/15+4/105+4/315

=>20/36

tích đúng nha bạn

Mình làm bài 2 nhé:

Ta có: \(\frac{1}{2^2}<\frac{1}{2\times3}=\frac{1}{2}-\frac{1}{3}\)

\(\frac{1}{3^2}<\frac{1}{3\times4}=\frac{1}{3}-\frac{1}{4}\)

....

\(\frac{1}{50^2}<\frac{1}{50\times51}=\frac{1}{50}-\frac{1}{51}\)

Tổng các vế ta sẽ có \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}<\frac{1}{2}-\frac{1}{51}=\frac{49}{102}<1\)

 D = 4/1.3.5+4/3.5.7+4/5.7.9+4/7.9.11+4/9.11.13

D.\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)

D\(=\frac{1}{1.3}-\frac{1}{11.13}\)

D\(=\frac{1}{3}-\frac{1}{143}\Rightarrow D=\frac{140}{429}\)

\(A=\frac{10^{1990}+1}{10^{1991}+1}vàB=\frac{10^{1991}+1}{10^{1992}+1}\)

\(B=\frac{10^{1991}+1}{10^{1992}+1}

\(A=\frac{10^{1990}+1}{10^{1991}+1}\Rightarrow10A=\frac{10^{1991}+1+9}{10^{1991}+1}\Rightarrow10A=1+\frac{9}{10^{1991}+1}\)

\(B=\frac{10^{1991}+1}{10^{1992}+1}\Rightarrow10B=\frac{10^{1992}+1+9}{10^{1992}+1}\Rightarrow10B=1+\frac{9}{10^{1992}+1}\)

=> 10A > 10B

=> A>B