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a) Vì \(\dfrac{x+5}{3}\)= \(\dfrac{x-6}{7}\) nên 7(x+5) = 3(x-6)
=> 7x+ 35 = 3x - 18
7x - 3x = -18 -35
4x = -53
x = -53:4
x = \(\dfrac{-53}{4}\)
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
b: \(\Leftrightarrow x-10\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{53\cdot55}\right)=\dfrac{3}{11}\)
\(\Leftrightarrow x-10\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)
\(\Leftrightarrow x-10\cdot\dfrac{4}{55}=\dfrac{3}{11}\)
=>x=3/11+20/55=3/11+4/11=7/11
c: \(\Leftrightarrow\left(\dfrac{x-1}{99}-1\right)+\left(\dfrac{x-2}{98}-1\right)+\left(\dfrac{x-5}{95}-1\right)=\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{95}\)
\(\Leftrightarrow x-100=1\)
hay x=101
a, sai đề
b, \(\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Rightarrow\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{9}\) ( nhân cả 2 vế với \(\dfrac{1}{2}\) )
\(\Rightarrow\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{9}\)
\(\Rightarrow\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\Rightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\Rightarrow x+1=18\Rightarrow x=17\)
Vậy x = 17
Câu a thiếu đề rồi bạn ơi mik giải câu b đây:
\(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\dfrac{2}{42}+\dfrac{2}{56}+\dfrac{2}{72}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\dfrac{2}{6.7}+\dfrac{2}{7.8}+\dfrac{2}{8.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+....+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)
\(2\left(\dfrac{1}{6}-\dfrac{1}{x+2}\right)=\dfrac{2}{9}\)
\(\dfrac{1}{6}-\dfrac{1}{x+2}=\dfrac{2}{9}:2\)
\(\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\)
\(\dfrac{1}{x+1}=\dfrac{1}{18}\)
\(\Rightarrow x+1=18\Rightarrow x=17\)
Vậy x = 17
Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)
Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}\)
\(=\dfrac{4}{15}\)
d) Ta có: \(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x=\dfrac{-37}{45}+\dfrac{1}{45}-\dfrac{1}{5}=\dfrac{-36}{45}-\dfrac{1}{5}=\dfrac{-4}{5}-\dfrac{1}{5}=-1\)
Vậy: x=-1