a: \(\dfrac{-3}{8}+\dfrac{12}{25}-\dfrac{5}{8}+\dfrac{2}{-5}+\dfrac{13}{25}\)

\(=\left(-\dfrac{3}{8}-\dfrac{5}{8}\right)+\left(\dfrac{12}{25}+\dfrac{13}{25}\right)-\dfrac{2}{5}\)

\(=-1+1-\dfrac{2}{5}=-\dfrac{2}{5}\)

b: \(\dfrac{3}{7}\cdot\dfrac{20}{11}+\dfrac{9}{11}\cdot\dfrac{-3}{7}-2\dfrac{3}{7}\)

\(=\dfrac{3}{7}\left(\dfrac{20}{11}-\dfrac{9}{11}\right)-2-\dfrac{3}{7}\)

\(=\dfrac{3}{7}-2-\dfrac{3}{7}=-2\)

c: \(\dfrac{3}{2}+\dfrac{5}{4}-\dfrac{14}{16}\)

\(=\dfrac{12}{8}+\dfrac{10}{8}-\dfrac{7}{8}\)

\(=\dfrac{12+10-7}{8}=\dfrac{15}{8}\)

d: Sửa đề: \(\dfrac{2}{5}\cdot\dfrac{1}{3}-\dfrac{1}{15}:\dfrac{1}{5}\)

\(=\dfrac{2}{15}-\dfrac{1}{15}\cdot5\)

\(=\dfrac{2}{15}-\dfrac{1}{3}=\dfrac{2}{15}-\dfrac{5}{15}=-\dfrac{3}{15}=-\dfrac{1}{5}\)

e: \(\dfrac{15}{7}\cdot\dfrac{1}{3}+\dfrac{12}{7}\cdot\dfrac{2}{3}\)

\(=\dfrac{15}{21}+\dfrac{24}{21}\)

\(=\dfrac{39}{21}=\dfrac{13}{7}\)

f: \(32,8+4,2+\left(-4,3\right)+\left(-32,8\right)+4,3\)

\(=\left(32,8-32,8\right)+\left(4,2\right)+\left(-4,3+4,3\right)\)

=0+4,2+0

=4,2

giúp với aa mai tớ nộp r:<

   (- \(\dfrac{2}{5}\))² + \(\dfrac{1}{2}\) x (4,5 - 2) - 25%

= \(\dfrac{4}{25}\) + \(\dfrac{1}{2}\) x 2,5 - 0,25

= 0,16 + 1,25 - 0,25

= 0,16 + (1,25 - 0,25)

= 0,16 + 1

= 1,16 

Muốn điều chế 400g hạt cà phê in ta cần số gam cà fe là 400.100/2.5= 16000 gam

a: Giá mới của chiếc tivi là:

\(8500000\left(1-15\%\right)=8500000\cdot0,85=7225000\left(đồng\right)\)

b: Giá gốc trước khi giảm là:

\(8000000:\left(1-25\%\right)=8000000:\dfrac{3}{4}=\dfrac{32000000}{3}\left(đồng\right)\)

a: Giá mới của cuốn sách là:

\(48000\left(1-15\%\right)=48000\cdot0,85=40800\left(đồng\right)\)

b: giá mới của cuốn sách là:

\(48000\left(1+10\%\right)=52800\left(đồng\right)\)

m>n

\(A=1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+...+8}\)

\(=1+\dfrac{1}{2\cdot\dfrac{3}{2}}+\dfrac{1}{3\cdot\dfrac{4}{2}}+...+\dfrac{1}{8\cdot\dfrac{9}{2}}\)

\(=\dfrac{2}{2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{8\cdot9}\)

\(=2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{8\cdot9}\right)\)

\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

\(=2\left(1-\dfrac{1}{9}\right)=2\cdot\dfrac{8}{9}=\dfrac{16}{9}\)

mn nhanh giùm em ạ

ĐKXĐ: x<>1

Để A là số nguyên thì \(5⋮x-1\)

=>\(x-1\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{2;0;6;-4\right\}\)

\(A=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{47\cdot49}\\ A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{47}-\dfrac{1}{49}\\ A=\dfrac{1}{1}-\dfrac{1}{49}\\ A=\dfrac{48}{49}\)

Vậy \(A=\dfrac{48}{49}\)

\(A=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{47\cdot49}\)

\(\Rightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{47.49}\)

\(\Rightarrow2A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{47}-\dfrac{1}{49}\)

\(\Rightarrow2A=\dfrac{1}{1}-\dfrac{1}{49}\)

\(\Rightarrow2A=\dfrac{49}{49}-\dfrac{1}{49}\)

\(\Rightarrow2A=\dfrac{48}{49}\)

\(\Rightarrow A=\dfrac{48}{49}:2.\)

\(\Rightarrow A=\dfrac{48}{49}\cdot\dfrac{1}{2}\)

\(\Rightarrow A=\dfrac{24}{49}\).