: ( 9/4 + 16/5 + 25/6 ) - ( 5/4 - 14/5 + 47/8 ) =0. 78125

\(\left(\frac{9}{4}+\frac{16}{5}+\frac{25}{6}\right)-\left(\frac{5}{4}-\frac{14}{5}+\frac{47}{8}\right)\)

\(=\frac{9}{4}+\frac{16}{5}+\frac{25}{6}-\frac{5}{4}+\frac{14}{5}-\frac{47}{8}\)

\(=\left(\frac{9}{4}-\frac{5}{4}\right)+\left(\frac{16}{5}+\frac{14}{5}\right)+\left(\frac{25}{6}-\frac{47}{8}\right)\)

\(=1+6-\frac{41}{24}\)

\(=7-\frac{41}{24}=\frac{127}{24}\)

(\(\frac{9}{4}+\frac{16}{5}+\frac{25}{6}\)) -(\(\frac{5}{4}-\frac{14}{5}+\frac{47}{8}\))

=\(\frac{9}{4}+\frac{16}{5}+\frac{25}{6}-\frac{5}{4}+\frac{14}{5}-\frac{47}{8}\)

=\(\frac{9-5}{4}+\frac{16+14}{5}+\frac{25}{6}-\frac{47}{8}\)

= 1 + 6+\(\frac{200-282}{48}\)

=\(\frac{336-82}{48}\)

=\(\frac{254}{48}\)

=\(\frac{127}{24}\)

Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\right)\)

\(A=1-\frac{1}{2^{100}}\)

\(A=\frac{2^{100}-1}{2^{100}}\)

Tham khảo nhé~

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}}\)

\(\Rightarrow\)\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(\Rightarrow\)\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

\(\Rightarrow\)\(A=1-\frac{1}{2^{100}}\)

Em ơi, bài này yêu cầu gì thế ?

So sánh hay tính ? Phải có đề thì mới giải đc em nhé.

x=1 , y=0 nhé

mk đi

\(y=\frac{x-1}{2x+3}\)

\(\Rightarrow2xy+3y=xy-y\)

\(\Rightarrow2xy+3y-xy+y=0\)

\(\Rightarrow xy+4y=0\)

\(\Rightarrow\left(x+4\right)y=0\)

\(\Rightarrow\hept{\begin{cases}x=-4\\y=0\end{cases}}\)

UCLN la cai quai gi

chi hieu BCNN thoi

ta có:

\(\frac{6n-7}{4n-1}=1.\frac{6n-7}{4n-1}=\frac{3}{3}.\frac{6n-7}{4n-1}=\frac{3\left(6n-7\right)}{3\left(4n-1\right)}\)\(=\frac{12n-14}{12n-3}=\frac{12n-3}{12n-3}-\frac{11}{12n-3}\)

\(=1-\frac{11}{12n-3}=>12n-3\)thuộc tập hợp ước của 11

=>12n-3=1=>n=\(\frac{1}{3}\) (loại) vì ko thuộc N

12n-1=11=>n=1

Vậy n=1

Nhớ tk nha=)))

\(\frac{1}{2.7}+\frac{1}{7.12}+\frac{1}{12.17}+....+\frac{1}{2012.2017}\)

\(=\frac{1}{5}\left(\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+....+\frac{5}{2012.2017}\right)\)

\(=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+....+\frac{1}{2012}-\frac{1}{2017}\right)\)

\(=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{2017}\right)\)

\(=\frac{1}{5}.\frac{2015}{4034}=\frac{403}{4034}\)

ĐẶT A=DÃY SỐ TRÊN=>5A=5/2.7+........+5/2012.2017

=>A=1/2-1/7........-1/2012-1/2017 RÚT GỌN TA ĐƯỢC A=1/2-1/2017

Đặt \(M=\frac{1}{2.7}+\frac{1}{7.12}+\frac{1}{12.17}+...+\frac{1}{2012.2017}\)

\(\Rightarrow\) \(5M=5.(\frac{1}{2.7}+\frac{1}{7.12}+\frac{1}{12.17}+...+\frac{1}{2012.2017})\)

           \(=\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+...+\frac{5}{2012.2017}\)

           \(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{2012}-\frac{1}{2017}\)

           \(=\frac{1}{2}-\frac{1}{2017}\)

\(\Rightarrow\) \(M=(\frac{1}{2}-\frac{1}{2017}):5\)

         \(=\frac{1}{10}-\frac{1}{10085} \)

Vậy \(M=\frac{1}{10}-\frac{1}{10085}\)

\(\frac{-19}{17}:\frac{-3}{8}=\frac{152}{51}\)

\(-\frac{19}{17}:-\frac{3}{8}=\frac{19}{17}:\frac{3}{8}=\frac{19}{17}\times\frac{8}{3}=\frac{152}{51}\)

\(\frac{-19}{17}:\frac{-3}{8}=\frac{152}{51}\)

\(#LTH\)

Do p là số nguyên tố nên ta có các trường hợp:

+ Với \(p=3\)thì \(\hept{\begin{cases}p+8=3+8=11\\p+10=3+10=13\end{cases}}\) là các số nguyên tố (chọn)

\(\orbr{\begin{cases}p=3k+1\\p=3k+2\end{cases}}\) \(\left(k\in N\right)\)

+Với \(p=3k+1\)thì \(p+8=3k+1+8\)

\(=3k+9=3\left(k+3\right)⋮3\)\(\Rightarrow p+8\text{ }\)là hợp số (loại)

+Với \(p=3k+2\)thì \(p+10=3k+2+10\)

\(=3k+12=3\left(k+4\right)⋮3\)\(\Rightarrow p+10\text{ }\)là hợp số (loại)

Vậy \(p=3\)thỏa mãn đề

Sao lại có 3k + 1 và 3k + 2 ạ