câu a

\(\dfrac{9}{5}\cdot y-\dfrac{3}{5}\cdot\left(y+5\right)=33\\ \dfrac{9}{5}\cdot y-\dfrac{3}{5}\cdot y-3=33\\ \dfrac{6}{5}y-3=33\\ \dfrac{6}{5}y=36\\ y=30\)

câu b

\(75\%\cdot y+y+0,5\cdot y+y:4=25\\ y\cdot\left(75\%+1+0,5\right)+y:4=25\\ y\cdot2,25+y:4=25\\ 9\cdot y+y=100\\ 10y=100\\ y=10\)

x= ........................................................................................................ké

a: 0,9x7+1,8x45+0,9+1,8

=0,9x7+0,9x90+0,9x3

=0,9x(7+90+3)=0,9x100=90

b: 20,24x5,8-20,24x4,7-12,24-8

=20,24x1,1-20,24

=20,24x0,1=2,024

a)                      

\(0,9\times7+1,8\times45+0,9+1,8\) 

\(=0,9\times\left(7+1\right)+1,8\times\left(45+1\right)\)

\(=0,9\times8+0,9\times2\times46\)

\(=0,9\times8+0,9\times92\)

\(=0,9\times\left(8+92\right)\)

\(=0,9\times100\)

\(=90\)

b) \(20,24\times5,8-20,24\times4,7-12,24-8\) 

\(=20,24\times\left(5,8-4,7\right)-\left(12,24+8\right)\)

\(=20,24\times1,1-20,24\)

\(=20,24\times\left(1,1-1\right)\)

\(=20,24\times0,1\) 

\(=2,024\)

nhờ mọi người giải giúp em với ạ,em đang cần gấp

\(12,8\times\dfrac{1}{2}+12,8\times0,25+12,8\times\dfrac{1}{4}\)

\(=12,8\times0,5+12,8\times0,25+12,8\times0,25\)

\(=12,8\times\left(0,5+0,25+0,25\right)\)

\(=12,8\times1\)

\(=12,8\)

\(12,8\times\dfrac{1}{2}+12,8\times0,25+12,8\times\dfrac{1}{4}\\ =12,8\times\dfrac{1}{2}+12,8\times\dfrac{25}{100}+12,8\times\dfrac{1}{4}\\ =12,8\times\dfrac{1}{2}+12,8\times\dfrac{1}{4}+12,8\times\dfrac{1}{4}\\ =12,8\times\left(\dfrac{2}{4}+\dfrac{1}{4}+\dfrac{1}{4}\right)\\ =12,8\times\dfrac{4}{4}=12,8\times1=12,8\)

Sửa đề : \(\dfrac{2025\times2024-1}{2023\times2025+2024}\)

\(=\)\(\dfrac{2025\times\left(2023+1\right)-1}{2023\times2025+2024}\)

\(=\dfrac{2025\times2023+2025-1}{2023\times2025+2024}\)

\(=\dfrac{2025\times2023+\left(2025-1\right)}{2023\times2025+2024}\)

\(=\dfrac{2025\times2023+2024}{2023\times2025+2024}\)

\(=1\)

@\(\text{格雷斯}\)

Đặt \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{1024}\)

\(2\times A=2\times\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{1024}\right)\)

\(=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}\)

\(2\times A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{1024}\right)\)

\(A=1-\dfrac{1}{1024}=\dfrac{1023}{1024}\)

\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2023}\right)\times\left(1-\dfrac{1}{2024}\right)\)

\(=\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2022}{2023}\times\dfrac{2023}{2024}\)

\(=\dfrac{1\times2\times3\times...\times2022\times2023}{2\times3\times4\times...\times2023\times2024}\)

\(=\dfrac{1}{2024}\)

\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2023}\right)\times\left(1-\dfrac{1}{2024}\right)\)

\(=\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2022}{2023}\times\dfrac{2023}{2024}\)

\(=\dfrac{1\times2\times3\times...\times2022\times2023}{2\times3\times4\times...\times2023\times2024}\)

\(=\dfrac{1}{2024}\)

\(Z=\dfrac{3}{3\times5}+\dfrac{3}{5\times7}+\dfrac{3}{7\times9}+...+\dfrac{3}{49\times51}\)

\(=\dfrac{3}{2}\times\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{49\times51}\right)\)

\(=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)

\(=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{51}\right)\)

\(=\dfrac{3}{2}\times\dfrac{16}{51}=\dfrac{8}{17}\)

\(Z=\dfrac{3}{3x5}+\dfrac{3}{5x7}+\dfrac{3}{7x9}+...+\dfrac{3}{49x51}\\ =\dfrac{3}{2}x\left(\dfrac{2}{3x5}+\dfrac{2}{5x7}+\dfrac{2}{7x9}+...+\dfrac{2}{49x51}\right)\\ =\dfrac{3}{2}x\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\\ =\dfrac{3}{2}x\left(\dfrac{1}{3}-\dfrac{1}{51}\right)\\ =\dfrac{3}{2}x\dfrac{16}{51}=\dfrac{8}{17}\)

\(S=\dfrac{2}{1\times2}+\dfrac{2}{2\times3}+\dfrac{2}{3\times4}+...+\dfrac{2}{99\times100}\)

\(=2\times\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{99\times100}\right)\)

\(=2\times\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=2\times\left(1-\dfrac{1}{100}\right)\)

\(=2\times\dfrac{99}{100}=\dfrac{99}{50}\)

CT: \(\dfrac{a}{n\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a};\left(n\ne0;n\ne-a\right)\)

\(S=\dfrac{2}{1\times2}+\dfrac{2}{2\times3}+\dfrac{2}{3\times4}+...+\dfrac{2}{99\times100}\\ \dfrac{S}{2}=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{99\times100}\\ \dfrac{S}{2}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ \dfrac{S}{2}=1-\dfrac{1}{100}=\dfrac{99}{100}\\ S=\dfrac{99}{100}\times2=\dfrac{99}{50}\)

\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{9\times10}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)

CT: \(\dfrac{a}{n\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a}\) (\(n\ne0;n\ne-a\))

\(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+...+\dfrac{1}{9x10}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\\ =1-\dfrac{1}{10}\\ =\dfrac{9}{10}\)