ko qui đồng nha

Lời giải:

a, Đặt \(A=\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)

\(\Rightarrow A=\dfrac{-1}{4.5}+\dfrac{-1}{5.6}+\dfrac{-1}{6.7}+...+\dfrac{-1}{9.10}\)

\(\Rightarrow A=\dfrac{-1}{4}+\dfrac{1}{5}-\dfrac{1}{5}+...-\dfrac{1}{9}+\dfrac{1}{10}\)

\(\Rightarrow A=\dfrac{-1}{4}+\dfrac{1}{10}\)

\(\Rightarrow A=\dfrac{-3}{20}\)

\(\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{9.10}\\ =\dfrac{-1}{4.5}+\dfrac{-1}{5.6}+\dfrac{-1}{6.7}+\dfrac{-1}{7.8}+\dfrac{-1}{8.9}+\dfrac{-1}{9.10}\)

\(=\dfrac{-1}{4}-\dfrac{-1}{5}+\dfrac{-1}{5}-\dfrac{-1}{6}+\dfrac{-1}{6}-\dfrac{-1}{7}+\dfrac{-1}{7}-\dfrac{-1}{8}+\dfrac{-1}{8}-\dfrac{-1}{9}+\dfrac{-1}{9}-\dfrac{-1}{10}\)

\(=\dfrac{-1}{4}-\dfrac{-1}{10}\\=\dfrac{-1}{4}+\dfrac{1}{10}\\=\dfrac{-5}{20}+\dfrac{2}{20}\\=\dfrac{-3}{20}\)

Lời giải:

a)

\(A=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+....+\frac{-1}{90}\)

\(=\frac{-1}{4.5}+\frac{-1}{5.6}+\frac{-1}{6.7}+...+\frac{-1}{9.10}\)

\(=\frac{4-5}{4.5}+\frac{5-6}{5.6}+\frac{6-7}{6.7}+....+\frac{9-10}{9.10}\)

\(=\frac{1}{5}-\frac{1}{4}+\frac{1}{6}-\frac{1}{5}+\frac{1}{7}-\frac{1}{6}+...+\frac{1}{10}-\frac{1}{9}\)

\(=\frac{1}{10}-\frac{1}{4}=-\frac{3}{20}\)

b)

\(2B=5+\frac{8}{11}+\frac{3}{11}+\frac{1}{15}+\frac{13}{15.2}\)

\(=5+\frac{11-3}{11}+\frac{3}{11}+\frac{1}{15}+\frac{15-2}{15.2}\)

\(=5+1-\frac{3}{11}+\frac{3}{11}+\frac{1}{15}+\frac{1}{2}-\frac{1}{15}\)

\(=5+1+\frac{1}{2}=\frac{13}{2}\Rightarrow B=\frac{13}{4}\)

b: \(C=\left(\dfrac{12}{199}+\dfrac{23}{200}-\dfrac{34}{201}\right)\cdot\dfrac{3-2-1}{6}=0\)

2. Tính:

a, \(\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)

=\(\left(\dfrac{-1}{20}+\dfrac{-1}{72}\right)+\left(\dfrac{-1}{30}+\dfrac{-1}{90}\right)+\left(\dfrac{-1}{42}+\dfrac{-1}{56}\right)\)

=\(\left(\dfrac{-18}{360}+\dfrac{-5}{360}\right)+\left(\dfrac{-3}{90}+\dfrac{-1}{90}\right)+\left(\dfrac{-4}{168}+\dfrac{-3}{168}\right)\)

=\(\dfrac{-23}{360}+\dfrac{-4}{90}+\dfrac{-7}{168}\)

=\(\dfrac{-23}{360}+\dfrac{-16}{360}+\dfrac{-15}{360}\)=\(\dfrac{-54}{360}=\dfrac{-3}{20}\)

b, \(\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\)

=\(\dfrac{5}{2}+\dfrac{4}{1}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{3}{2}+\dfrac{1}{2}.\dfrac{1}{15}+\dfrac{1}{15}.\dfrac{13}{4}\)

=\(\dfrac{5}{2}+\dfrac{1}{11}.\left(\dfrac{4}{1}+\dfrac{3}{2}\right)+\dfrac{1}{15}.\left(\dfrac{1}{2}+\dfrac{13}{4}\right)\)

=\(\dfrac{5}{2}+\dfrac{1}{11}.\dfrac{11}{2}+\dfrac{1}{15}.\dfrac{15}{4}\)

=\(\dfrac{5}{2}+\dfrac{1}{2}+\dfrac{1}{4}\)

=\(\dfrac{10}{4}+\dfrac{2}{4}+\dfrac{1}{4}\)

=\(\dfrac{13}{4}\)

3. Tìm x

a, \(\dfrac{x-5}{8}=\dfrac{18}{x-5}\)

\(\left(x-5\right).\left(x-5\right)=8.18\)

\(\left(x-5\right)^2=144\)

\(x-5=\sqrt{144}\)

\(x-5=12\)

\(x=12+5\)

\(x=17\)

b,\(\left(x-2\right)^{10}=\left(2-x\right)^8\)

\(x^{10}-2^{10}=x^8-2^8\)

\(x^{10}+x^8=2^{10}+2^8\)

\(\Rightarrow x=2\)

2b, Đặt giá trị biểu thức trên là B.

Ta có: \(B=\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\)

\(B=\dfrac{7}{7}\left(\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\right)\)

\(B=7.\left(\dfrac{5}{2.7}+\dfrac{4}{7.11}+\dfrac{3}{11.14}+\dfrac{1}{14.15}+\dfrac{13}{15.28}\right)\)

\(B=7.\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{28}\right)\)

\(B=7.\left(\dfrac{1}{2}-\dfrac{1}{28}\right)\)

\(B=7.\left(\dfrac{14}{28}-\dfrac{1}{28}\right)=7.\dfrac{13}{28}=\dfrac{13}{4}\)

    5/2.1+4/1.11+3/11.2+1/2.15+13/15.4

-> 5/2+4/11+3/22+1/30+13/60

-> 1650/660+240/660+90/660+22/660+143/660

-> 2145/660

-> 13/4

5/2.1+4/1.11+3/11.2+1/12.15+13/15.4

=7.(5/2.7+4/7.11+3/11.14+1/14.15+13/15.28)

=7.(1/2-1/7+1/7-1/11+1/11-1/14+1/14-1/15+1/15-1/28)

=7.(1/2-1/28)=13/14=hỗn số 3 1 phần 4

\(n=7\left(\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\right)\)

\(tacocongthuc:\frac{1}{n}-\frac{1}{n+k}=\frac{k}{n\left(n+k\right)}\)

\(\Rightarrow n=7\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-.....-\frac{1}{28}\right)=7\left(\frac{1}{2}-\frac{1}{28}\right)=\frac{7.13}{28}=\frac{13}{4}\)

#)Giải :

( k chép lại đề )

\(\frac{n}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.4}\)

\(\frac{n}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{15}-\frac{1}{28}\)

\(\frac{n}{7}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)

\(\Rightarrow n=\frac{13}{28}.7=\frac{13}{4}\)

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