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\(\left(1+\dfrac{7}{9}\right).\left(1+\dfrac{7}{20}\right).\left(1+\dfrac{7}{33}.\right)\left(1+\dfrac{7}{48}\right)...\left(1+\dfrac{7}{180}\right)\)
\(=\dfrac{16}{9}.\dfrac{27}{20}.\dfrac{40}{33}.\dfrac{55}{48}...\dfrac{7}{180}\)
\(=\dfrac{2.8}{1.9}.\dfrac{3.9}{2.10}.\dfrac{4.10}{3.11}.\dfrac{5.11}{4.12}...\dfrac{11.17}{10.18}\)
\(=\dfrac{\left(2.3.4.5...11\right).\left(8.9.10.11...17\right)}{\left(1.2.3.4...10\right).\left(9.10.11.12...18\right)}\)
\(=\dfrac{11.8}{1.18}=\dfrac{88}{18}=\dfrac{44}{9}\)
ta có ;
\(\left(1+\dfrac{7}{9}\right)\cdot\left(1+\dfrac{7}{20}\right).\left(1+\dfrac{7}{33}\right)...\left(1+\dfrac{1}{180}\right)\)
=\(\dfrac{16}{9}.\dfrac{27}{20}.\dfrac{40}{33}....\dfrac{187}{180}\)
=\(\dfrac{8.2}{9.1}.\dfrac{9.3}{10.2}.\dfrac{10.4}{3.11}.\dfrac{11.5}{4.12}....\dfrac{17.11}{18.10}\)
=\(\dfrac{8.9.10.11.12.13.14.15.16.17.2.3.4.5.6.7.8.9.10.11}{9.10.11.12.13.14.15.16.17.18.1.2.3.4.5.6.7.8.9.10}\)
=\(\dfrac{8.11}{18}=\dfrac{88}{18}=\dfrac{44}{9}\)
a)
ta có:
\(\left\{{}\begin{matrix}\dfrac{b-a}{b-a}=1..\forall a\ne b\\\dfrac{b-a}{a.b}=\dfrac{1}{a}-\dfrac{1}{b}..\forall a,b\ne0\end{matrix}\right.\)(*)
\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+..+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(\left\{{}\begin{matrix}a=3n-1\\b=3n+2\end{matrix}\right.\)\(\Rightarrow b-a=3..\forall n\)
Thay (*) vào dãy A
\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-....+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)=\dfrac{1}{3}\left(\dfrac{3n+2-2}{2.\left(3n+2\right)}\right)=\dfrac{n}{6n+4}=VP\rightarrow dpcm\)
B) tương tự
e) \(\dfrac{-3}{5}.\dfrac{2}{7}+\dfrac{-3}{5}.\dfrac{5}{7}+2\dfrac{3}{5}\)
= \(\dfrac{-3}{5}.\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\dfrac{13}{5}\)
= \(\dfrac{-3}{5}.1+\dfrac{13}{5}\)
= \(\dfrac{-3}{5}+\dfrac{13}{5}\)
= 2
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{x(x+3)}=\dfrac{6}{19}\)
\(\Rightarrow\)\(\dfrac{1}{3}.(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{x(x+3)})=\dfrac{6}{19}\)
\(\Rightarrow\)\(\dfrac{1}{3}.(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{x}-\dfrac{1}{x+3})=\dfrac{6}{19}\)
\(\Rightarrow\)\(\dfrac{1}{3}.(\dfrac{1}{1}-\dfrac{1}{x+3})=\dfrac{6}{19}\)
\(\Rightarrow\) \(\dfrac{1}{1}-\dfrac{1}{x+3}=\dfrac{6}{19}:\dfrac{1}{3}\)
\(\Rightarrow\) \(\dfrac{1}{1}-\dfrac{1}{x+3}=\dfrac{18}{19}\)
\(\Rightarrow\) \(\dfrac{1}{x+3}=\dfrac{1}{1}-\dfrac{18}{19}\)
\(\Rightarrow\) \(\dfrac{1}{x+3}=\dfrac{1}{19}\)
\(\Rightarrow\) \(x+3=19\)
\(x=19-3\)
\(x=16\)
Vậy \(x=16\)
Ta chi can tach ra , xong ta luoc bot,neu co so bi thua ra thi ta tinh tong.....
ừ Vy Nguyễn, mik làm nè:
e, \(\dfrac{-2}{3}-\dfrac{1}{3}\left(2x-5\right)=\dfrac{3}{2}.\)
\(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-2}{3}-\dfrac{3}{2}.\)
\(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-4}{6}+\dfrac{-9}{6}.\)
\(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-13}{6}.\)
\(2x-5=\dfrac{-13}{6}:\dfrac{1}{3}.\)
\(2x-5=\dfrac{-13}{6}.3.\)
\(2x-5=\dfrac{-13}{2}.\)
\(2x=\dfrac{-13}{2}+5.\)
\(2x=\dfrac{-13}{2}+\dfrac{10}{2}.\)
\(2x=\dfrac{-3}{2}.\)
\(x=\dfrac{-3}{2}:2.\)
\(x=\dfrac{-3}{2.2}=\dfrac{-3}{4}.\)
g, \(\dfrac{2}{5}x+\dfrac{1}{2}=\dfrac{-3}{4}.\)
\(\dfrac{2}{5}x=\dfrac{-3}{4}-\dfrac{1}{2}.\)
\(\dfrac{2}{5}x=\dfrac{-3}{4}+\dfrac{-2}{4}.\)
\(\dfrac{2}{5}x=\dfrac{-5}{4}.\)
\(x=\dfrac{-5}{4}:\dfrac{2}{5}.\)
\(x=\dfrac{-5}{4}.\dfrac{5}{2}.\)
\(x=\dfrac{-25}{8}.\)
h, \(\left(2x-2\dfrac{4}{5}\right):3\dfrac{1}{8}=1\dfrac{3}{5}.\)
\(\left(2x-2\dfrac{4}{5}\right)=\dfrac{8}{5}.\dfrac{25}{8}.\)
\(\left(2x-2\dfrac{4}{5}\right)=5.\)
\(2x=5+2\dfrac{4}{5}.\)
\(2x=7\dfrac{4}{5}.\)
\(x=7\dfrac{4}{5}:2.\)
\(x=\dfrac{39}{10}.\)
(còn tiếp ở phần sau!!!)
Tiếp:
i, \(3,2x-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):3\dfrac{2}{3}=\dfrac{7}{20}.\)
\(\dfrac{16}{5}x-\left(\dfrac{4}{5}+\dfrac{2}{3}\right)=\dfrac{7}{20}.\dfrac{11}{3}.\)
\(\dfrac{16}{5}x-\left(\dfrac{4}{5}+\dfrac{2}{3}\right)=\dfrac{77}{60}.\)
\(\dfrac{16}{5}x-\left(\dfrac{12}{15}+\dfrac{10}{15}\right)=\dfrac{77}{60}.\)
\(\dfrac{16}{5}x-\dfrac{22}{15}=\dfrac{77}{60}.\)
\(\dfrac{16}{5}x=\dfrac{77}{60}+\dfrac{22}{15}.\)
\(\dfrac{16}{5}x=\dfrac{77}{60}+\dfrac{88}{60}.\)
\(\dfrac{16}{5}x=\dfrac{165}{60}=\dfrac{11}{4}.\)
\(x=\dfrac{11}{4}:\dfrac{16}{5}.\)
\(x=\dfrac{11}{4}.\dfrac{5}{16}=\dfrac{55}{64}.\)
k, \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}.\)
\(\left(\dfrac{3x}{7}+1\right)=\dfrac{-1}{28}.\left(-4\right).\)
\(\left(\dfrac{3x}{7}+1\right)=\dfrac{1}{7}.\)
\(\dfrac{3x}{7}=\dfrac{1}{7}-1.\)
\(\dfrac{3x}{7}=\dfrac{1}{7}-\dfrac{7}{7}.\)
\(\dfrac{3x}{7}=\dfrac{-6}{7}.\)
\(\Rightarrow3x=-6.\)
\(\Rightarrow x=-6:3=-2.\)
~ Chúc bn học tốt!!! ~
Bài mik đúng thì nhớ tik mik nha!!!
(1+1/1.3)...(1+1/2016.2018)
=(1.3+1/1.3)...(2016.2018+1/2016.2018)
=(2.2/1.3)...(2017.2017/2016.2018)
=(2...2017).(2..2017)/(1.2.....2016).(3...2018)
=2017.2/2018
=2017.2/1006.2
=2017/1006
thông cảm nha :))
\(T=\left(\dfrac{1}{2}+1\right).\left(\dfrac{1}{3}+1\right).\left(\dfrac{1}{4}+1\right).......\left(\dfrac{1}{98}+1\right).\left(\dfrac{1}{99}+1\right) \) \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}....\dfrac{99}{98}\cdot\dfrac{100}{99}\)
\(=\dfrac{100}{2}=50\)
\(T=\left|\dfrac{1}{2}+1\right|\left|\dfrac{1}{3}+1\right|\left|\dfrac{1}{4}+1\right|.....\left|\dfrac{1}{98}+1\right|\left|\dfrac{1}{99}+1\right|\)
\(T=\left|\dfrac{3}{2}\right|.\left|\dfrac{4}{3}\right|.\left|\dfrac{5}{4}\right|......\left|\dfrac{99}{98}\right|.\left|\dfrac{100}{99}\right|\)
\(T=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}.....\dfrac{99}{98}.\dfrac{100}{99}\)
\(T=\dfrac{3.4.5.....99.100}{2.3.4.....98.99}=\dfrac{100}{2}=50\)