Ta có:

\(A=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)

\(\Rightarrow A=9.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)

\(\Rightarrow A=9.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)

\(\Rightarrow A=9.\left(\frac{1}{1.3}-\frac{1}{27.29}\right)\)

\(\Rightarrow A=9.\left(\frac{1}{3}-\frac{1}{783}\right)\)

\(\Rightarrow A=9.\frac{1}{3}-9.\frac{1}{783}\)

\(\Rightarrow A=3-\frac{1}{87}\)

Vì \(3-\frac{1}{87}< 3.\)

\(\Rightarrow A< 3\left(đpcm\right).\)

Chúc bạn học tốt!

a)\(A=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)

=\(\frac{9.4}{1.3.5}+\frac{9.4}{3.5.7}+\frac{9.4}{5.7.9}+...+\frac{9.4}{25.27.29}\)

=\(9.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)

=\(9.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)

=\(9.\left(\frac{1}{3}-\frac{1}{27.29}\right)=9.\left(\frac{1}{3}-\frac{1}{783}\right)=9.\left(\frac{261}{783}-\frac{1}{783}\right)=9.\frac{260}{783}\)

=\(\frac{260}{87}\)

b)

ta có: \(3=\frac{261}{87}>\frac{260}{87}\)

vậy A<3

KẾT BẠN VỚI MÌNH NHÉ NGƯỜI ĐẸP

mk cung dinh hoi bai nay

Áp dụng: \(\frac{4}{n\left(n+2\right)\left(n+4\right)}=\frac{n+4-n}{n\left(n+2\right)\left(n+4\right)}=\frac{1}{n\left(n+2\right)}-\frac{1}{\left(n+2\right)\left(n+4\right)}\)

\(\frac{B}{9}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}=\frac{1}{3}-\frac{1}{27.29}

chứng minh B làm sao z

\(\frac{36}{1\cdot3\cdot5}+\frac{36}{3\cdot5\cdot7}+\frac{36}{5\cdot7\cdot9}+...+\frac{36}{25\cdot27\cdot29}\)

\(=9\left[\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+...+\frac{4}{25\cdot27\cdot29}\right]\)

\(=9\left[\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\right]\)

\(=9\left[\frac{1}{3}-\frac{1}{783}\right]=9\cdot\frac{260}{783}=\frac{260}{87}\)

Đặt \(A=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)

\(\Rightarrow\frac{1}{9}A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\)

\(\Rightarrow\frac{1}{9}A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}\)

\(\Rightarrow\frac{1}{9}A=\frac{1}{1.3}-\frac{1}{27.29}\)

\(\Rightarrow\frac{1}{9}A=\frac{261}{783}-\frac{1}{783}\)

\(\Rightarrow\frac{1}{9}A=\frac{260}{783}\)

\(\Rightarrow A=\frac{260}{783}\div\frac{1}{9}\)

\(\Rightarrow A=\frac{2340}{783}=\frac{260}{87}\)

$\frac{4}{n\left(n+2\right)\left(n+4\right)}=\frac{n+4-n}{n\left(n+2\right)\left(n+4\right)}=\frac{1}{n\left(n+2\right)}-\frac{1}{\left(n+2\right)\left(n+4\right)}$4n(n+2)(n+4) =n+4−nn(n+2)(n+4) =1n(n+2) −1(n+2)(n+4) $\frac{B}{9}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}=\frac{1}{3}-\frac{1}{27.29}<\frac{1}{3}$B9 =11.3 −13.5 +13.5 −15.7 +...+125.27 −127.29 =13 −127.29 <13 $\Rightarrow B<3$

a=bao nhiêu bạn

xin lỗi vì thếu đề 

a bài 1 là bằng 3/11 nha

1

a)\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)\)

\(x-\left[10.\left(\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{53.55}\right)\right]=\frac{3}{11}\)

\(x-\left[10.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)\right]=\frac{3}{11}\)

\(x-\left[10.\frac{4}{55}\right]=\frac{3}{11}\)

\(x-\frac{8}{11}=\frac{3}{11}\)

\(x=1\)

=>S=1+1+1+...+1 (19 số 1)

=>S=19

Tick nha vì mình đang cần

mik đang tính đừng làm phiền

=(-1/2+3/2)+(-1/3+4/3)+.................+(-1/20+21/20)

=      1      +     1       +.................+1[S có :[(20-2):2+1]*2=20 (số)

                                                                => có :20:2=10 (cặp)=> Có 10 chữ số 1]

=10

Bạn viết đề thiếu trầm trọng quá !!!

Đáp án: thiếu đề

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