Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, (x + 1) + (x + 4) + ... + (x + 28) = 155
x + 1 + x + 4 + ... + x + 28 = 155
(x + x + x + ... + x) + (1 + 4 + ... + 28) = 155
x . 10 + 145 = 155
x . 10 = 155 - 145
x . 10 = 10
x = 10 : 10
x = 1
cái này cồng kềnh nhưng ko khó - >ngại. Bn tính 2/11*13+.... trước bằng cách rút gọn sau đó nhân ra -> dễ
\(\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{19\cdot21}\right)\cdot462-\left[0,04:\left(x+1,05\right)\right]:0,12=19\)\(\Leftrightarrow\dfrac{2}{2}\cdot\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\cdot462-\left[\dfrac{1}{25}:\left(x+\dfrac{21}{20}\right)\right]:\dfrac{3}{25}=19\)\(\Leftrightarrow\left(\dfrac{1}{11}-\dfrac{1}{21}\right)\cdot462-\left[\dfrac{1}{25}:\left(x+\dfrac{21}{20}\right)\right]:\dfrac{3}{25}=19\)
\(\Leftrightarrow20-\left[\dfrac{1}{25}:\left(x+\dfrac{21}{20}\right)\right]:\dfrac{3}{25}=19\)
\(\Leftrightarrow\left[\dfrac{1}{25}:\left(x+\dfrac{21}{20}\right)\right]:\dfrac{3}{25}=1\)
\(\rightarrow\dfrac{1}{25}:\left(x+\dfrac{21}{20}\right)=\dfrac{3}{25}\)
\(\Leftrightarrow x+\dfrac{21}{20}=\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{1}{3}-\dfrac{21}{20}=-\dfrac{43}{60}\)
Vậy \(x=-\dfrac{43}{60}\)
a: \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}\cdot x=\dfrac{16}{5}\)
=>2/5x=8/5
=>x=4
b: \(\Leftrightarrow\left(\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}+...+\dfrac{1}{39}-\dfrac{1}{40}\right)\cdot120+\dfrac{1}{3}x=-4\)
\(\Leftrightarrow x\cdot\dfrac{1}{3}+2=-4\)
=>1/3x=-6
=>x=-18
c: =>2|x-1/3|=0,24-4/5=-0,56<0
A=2.(1/1.3 + 1/3.5 + 1/5.7 +.......+1/99.101)
=2.(1/1 + 1/3 + 1/5 + 1/5 + 1/7 +...+1/99 + 1/101)
=2.(1-1/101)
=2.(101/101-1/101)
=2.100/101
200/101
B=2.(1/1.3+1/3.5+1/3.1+....+1/99.101)
=2.(1/1+1/3+1/3+1/5+1/3+1/7+....+1/99+1/101)
=2.(1/1+1/101)
=2.(101/101+1/101)
=2.102/101
=204/101
\(G=\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{95.98}+\dfrac{2}{98.101}\)
\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{95.98}+\dfrac{3}{98.101}\right)\)
\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{101}\right)\)
\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)
\(\Rightarrow G=\dfrac{2}{3}.\dfrac{96}{505}\)
\(\Rightarrow G=\dfrac{64}{505}\)
1.
a,
\(\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{19\cdot21}\right)\cdot462-\left[2,04:\left(x+1,05\right)\right]:0,12=19\\ \left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\cdot462-\left[2,04:\left(x+1,05\right)\right]:0,12=19\\ \left(\dfrac{1}{11}-\dfrac{1}{21}\right)\cdot462-\left[2,04:\left(x+1,05\right)\right]:0,12=19\\ \dfrac{10}{231}\cdot462-\left[2,04:\left(x+1,05\right)\right]:0,12=19\\ 20-\left[2,04:\left(x+1,05\right)\right]:0,12=19\\ \left[2,04:\left(x+1,05\right)\right]:0,12=1\\ 2,04:\left(x+1,05\right)=0,12\\ x+1,05=17\\ x=15,95\)
b,
\(\dfrac{1}{24\cdot25}+\dfrac{1}{25\cdot26}+...+\dfrac{1}{29\cdot30}+x:\dfrac{1}{3}=-4\\ \dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}+...+\dfrac{1}{29}-\dfrac{1}{30}+x\cdot3=-4\\ \dfrac{1}{24}-\dfrac{1}{30}+x\cdot3=-4\\ \dfrac{1}{120}+x\cdot3=-4\\ 3x=\dfrac{-481}{120}\\ x=\dfrac{-481}{360}\)
2.
a,
\(\dfrac{15}{28}-\dfrac{186}{1116}-\dfrac{121}{462}+\dfrac{189}{198}\\ =\dfrac{15}{28}-\dfrac{1}{6}-\dfrac{11}{42}+\dfrac{21}{22}\\ =\dfrac{495}{924}-\dfrac{154}{924}-\dfrac{242}{924}+\dfrac{882}{924}\\ =\dfrac{495-154-242+882}{924}\\ =\dfrac{981}{924}\\ =\dfrac{327}{308}\)
b,
\(\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{1\cdot3}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{2\cdot4}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{3\cdot5}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{99\cdot101}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{2^2-1}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{3^2-1}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{4^2-1}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{100^2-1}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\)\(=\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{100^2}{99\cdot101}\\ =\dfrac{2\cdot2}{1\cdot3}\cdot\dfrac{3\cdot3}{2\cdot4}\cdot\dfrac{4\cdot4}{3\cdot5}\cdot...\cdot\dfrac{100\cdot100}{99\cdot101}\\ =\dfrac{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot100\cdot100}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot99\cdot101}\\ =\dfrac{\left(2\cdot3\cdot4\cdot...\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot100\right)}{\left(1\cdot2\cdot3\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot101\right)}\\ =\dfrac{100\cdot2}{1\cdot101}\\ =\dfrac{200}{101}\)
mk sửa lại đề :D
2.b phải là 1/99.101