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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,\(\left(x-2\right)^{10}=\left(2-x\right)^8\)
\(x^{10}-2^{10}=x^8-2^8\)
\(x^{10}+x^8=2^{10}+2^8\)
\(\Rightarrow x=2\)
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
\(C=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(\frac{1}{7}C=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{1}{7}C=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(\frac{1}{7}C=\frac{1}{2}-\frac{1}{28}\)
\(\frac{1}{7}C=\frac{13}{28}\)
\(C=\frac{13}{28}:\frac{1}{7}\)
\(C=\frac{13}{4}\)
Vậy \(C=\frac{13}{4}\)
\(\dfrac{5}{2\cdot1}+\dfrac{4}{1\cdot11}+\dfrac{3}{11\cdot2}+\dfrac{13}{15\cdot4}\\ =\dfrac{5}{2}+\dfrac{4}{11}+\dfrac{3}{22}+\dfrac{13}{60}\\ =\dfrac{193}{60}\)