\(\frac{2}{3.5}+\frac{3}{5.8}+\frac{11}{8.19}+\frac{13}{19.32}+\frac{25}{32.57}+\frac{30}{57.87}\)

\(=\frac{5-3}{3.5}+\frac{8-5}{3}+\frac{19-8}{8.19}+\frac{32-29}{19.32}+\frac{57-32}{32.57}+\frac{87-57}{57.87}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{32}+\frac{1}{32}-\frac{1}{57}+\frac{1}{57}-\frac{1}{87}\)

\(=\frac{1}{3}-\frac{1}{87}=\frac{28}{87}\)

(x - 2)(x + 1) < 0

<=> x - 2 và x + 1 trái dấu

Ta thấy x - 2 < x + 1 nên \(\left\{\begin{matrix}x-2< 0\\x+1>0\end{matrix}\right.\) <=> -1 < x < 2

=> x = {0;1}

\(a^2+a⋮a-3\)

\(\Rightarrow a^2-3a+3a+a⋮a-3\)

\(\Rightarrow a\left(a-3\right)+4a⋮a-3\)

\(\Rightarrow4a⋮a-3\)

\(\Rightarrow4a-12+12⋮a-3\)

\(\Rightarrow4\left(a-3\right)+12⋮a-3\)

\(\Rightarrow12⋮a-3\)

\(\Rightarrow a-3\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)

Vậy a = {4;2;5;1;6;0;7;-1;9;-3;15;-9}

\(\frac{2n+15}{n+1}=\frac{2n+2+13}{n+1}=\frac{2\left(n+1\right)+13}{n+1}=\frac{2\left(n+1\right)}{n+1}+\frac{13}{n+1}=2+\frac{13}{n+1}\)

Để \(\frac{2n+15}{n+1}\in Z\) <=> \(n+1\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)

Vậy để \(\frac{2n+15}{n+1}\in Z\) thì n = {0;-2;12;-14}

nguyenquocngoc:Phải là phim Nữ hoàng băng giá chứ,đâu phải là Công chúa giá băng

\(\frac{x}{9}=\frac{7}{y}\)

=> xy = 7.9

=> xy = 63

=> x;y thuộc Ư(63) = {\(\pm1;\pm3;\pm7;\pm9;\pm21;\pm63\)}

Ta có bảng:

Vậy các cặp (x;y) là (1;63) ; (-1;-63) ; (3;21) ; (-3;-21) ; (7;9) ; (-7;-9) ; (9;7) ; (-9;-7) ; (21;3) ; (-21;-3) ; (63;1) ; (-63;-1)

\(A=1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)

\(3^2A=9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

\(9A-A=\left(9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)\)

\(8A=9-\frac{1}{3^{100}}\)

=> n = 100

\(\frac{3^8.11+3^{10}.5}{3^9.2^4}=\frac{3^8.11+3^8.3^2.5}{3^9.16}=\frac{3^8.11+3^8.45}{3^8.3.16}=\frac{3^8\left(11+45\right)}{3^8.48}=\frac{3^8.56}{3^8.48}=\frac{56}{48}=\frac{7}{6}\)

Gọi UCLN(n + 2,3n + 5) là d

Ta có: \(\left\{\begin{matrix}n+2⋮d\\3n+5⋮d\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}3\left(n+2\right)⋮d\\3n+5⋮d\end{matrix}\right.\Rightarrow\left\{\begin{matrix}3n+6⋮d\\3n+5⋮d\end{matrix}\right.\)

=> 3n + 6 - (3n + 5) \(⋮\) d

=> 3n + 6 - 3n - 5 \(⋮\) d

=> 1 \(⋮\) d => d = 1

=> UCLN(n + 2,3n + 5) = 1

Vậy \(\frac{n+2}{3n+5}\) là phân số tối giản

Ta có: \(A=\frac{2n-1}{n+3}=\frac{2n+6-7}{n+3}=\frac{2\left(n+3\right)-7}{n+3}=\frac{2\left(n+3\right)}{n+3}-\frac{7}{n+3}=2-\frac{7}{n+3}\)

Để A có giá trị nguyên <=> \(n+3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

Vậy để A có giá trị nguyên thì n = {-2;-4;4;10}