\(A=\dfrac{5}{40}+\dfrac{5}{88}+\dfrac{5}{154}+\dfrac{5}{238}+\dfrac{5}{340}\)

\(=\dfrac{5}{5\cdot8}+\dfrac{5}{8\cdot11}+\dfrac{5}{11\cdot14}+\dfrac{5}{14\cdot17}+\dfrac{5}{17\cdot20}\)

\(=\dfrac{5}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}+\dfrac{3}{17\cdot20}\right)\)

\(=\dfrac{5}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)\)

\(=\dfrac{5}{3}\left(\dfrac{1}{5}-\dfrac{1}{20}\right)=\dfrac{5}{3}\cdot\dfrac{3}{20}=\dfrac{1}{4}\)

\(\dfrac{2525}{4848}=\dfrac{25}{48}>\dfrac{24}{48}=\dfrac{1}{2}\)

\(\dfrac{3131}{6464}=\dfrac{31}{64}< \dfrac{32}{64}=\dfrac{1}{2}\)

Do đó: \(\dfrac{2525}{4848}>\dfrac{3131}{6464}\)

mình chưa hiểu lắm ạ

\(\dfrac{1}{2}=\dfrac{1\cdot6}{2\cdot6}=\dfrac{6}{12}\)

\(\dfrac{1}{3}=\dfrac{1\cdot4}{3\cdot4}=\dfrac{4}{12}\)

\(\dfrac{-1}{12}=\dfrac{-1\cdot1}{12\cdot1}=\dfrac{-1}{12}\)

Mẫu chung = 12

\(\dfrac{1}{2}=\dfrac{1.6}{2.6}=\dfrac{6}{12}\)

\(\dfrac{1}{3}=\dfrac{1.4}{3.4}=\dfrac{4}{12}\)

Ta có : \(\dfrac{6}{12},\dfrac{4}{12},\dfrac{-1}{12}\)

\(\dfrac{x+2}{3}=\dfrac{x-4}{5}\)

=>5(x+2)=3(x-4)

=>5x+10=3x-12

=>5x-3x=-12-10

=>2x=-22

=>\(x=-\dfrac{22}{2}=-11\)

\(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{29}{45}\)

=>\(x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{29}{45}\)

=>\(x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{29}{45}\)

=>\(x+\dfrac{8}{45}=\dfrac{29}{45}\)

=>\(x=\dfrac{29-8}{45}=\dfrac{21}{45}=\dfrac{7}{15}\)

Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{200^2}\)

=>\(A< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{199\cdot200}\)

=>\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{199}-\dfrac{1}{200}\)

=>\(A< 1-\dfrac{1}{200}=\dfrac{199}{200}< 1\)

1: (x-2)(y+3)=11

=>\(\left(x-2\right)\left(y+3\right)=1\cdot11=11\cdot1=\left(-1\right)\cdot\left(-11\right)=\left(-11\right)\cdot\left(-1\right)\)

=>\(\left(x-2;y+3\right)\in\left\{\left(1;11\right);\left(11;1\right);\left(-1;-11\right);\left(-11;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(3;8\right);\left(13;-2\right);\left(1;-14\right);\left(-9;-4\right)\right\}\)

2: \(xy+3x-7y-21=0\)

=>\(x\left(y+3\right)-7\cdot\left(y+3\right)=0\)

=>(y+3)(x-7)=0

=>\(\left\{{}\begin{matrix}x-7=0\\y+3=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=7\\y=-3\end{matrix}\right.\)

1, (y + 3) . (x - 2) = 11
Vì x, y ϵ Z
⇒ (y + 3); (x - 2) ϵ Z
⇒ (y + 3); (x - 2) ϵ Ư(11) = {1; -1; 11; -11}
Lập bảng

Vậy (x; y) ϵ {(13; -2); (-9; -4); (3; 8); (1; -14)}

2, xy + 3x - 7y - 21 = 0
x(y + 3) - (7y + 21) = 0
x(y + 3) - 7(y + 3) = 0
(y + 3)(x - 7) = 0
⇒ y + 3 = 0; x - 7 = 0
y = -3; x = 7

(Bạn có thể tự sửa bài nhá)