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\(\dfrac{x+2017}{x+2018}=\dfrac{2020}{2021}\)
\(\Leftrightarrow1-\dfrac{x+2017}{x+2018}=1-\dfrac{2020}{2021}\)
\(\Leftrightarrow\dfrac{x+2018}{x+2018}-\dfrac{x+2017}{x+2018}=\dfrac{2021}{2021}-\dfrac{2020}{2021}\)
\(\Leftrightarrow\dfrac{\left(x+2018\right)-\left(x+2017\right)}{x+2018}=\dfrac{2021-2020}{2021}\)
\(\Leftrightarrow\dfrac{x+2018-x-2017}{x+2018}=\dfrac{1}{2021}\)
\(\Leftrightarrow\dfrac{\left(2018-2017\right)+\left(x+x\right)}{x+2018}=\dfrac{1}{2021}\)
\(\Leftrightarrow\dfrac{1}{x+2018}=\dfrac{1}{2021}\)
\(\Leftrightarrow x+2018=2021\)
\(\Leftrightarrow x=3\left(tm\right)\)
vậy ....
a: \(\dfrac{x+2}{27}=\dfrac{x}{-9}\)
=>x+2=-3x
=>4x=-2
hay x=-1/2
b: \(\dfrac{-7}{x}=\dfrac{21}{34-x}\)
=>-7(34-x)=21x
=>34-x=-3x
=>2x=-34
hay x=-17
c: \(\dfrac{-8}{15}< \dfrac{x}{40}< \dfrac{-7}{15}\)
\(\Leftrightarrow-64< 3x< -56\)
hay \(x\in\left\{-21;-20;-19\right\}\)
d: \(\dfrac{-1}{2}< \dfrac{x}{18}< \dfrac{-1}{3}\)
=>-9<x<-6
hay \(x\in\left\{-8;-7\right\}\)
a,Vì \(\dfrac{5}{6}\)=\(\dfrac{x}{24}\) nên ta có: 5.24:6= 20 \(\Rightarrow\)x =20
Mấy câu sau làm tương tự như vậy.
b,x =21
c,x =9
d,x = -5
Bài 1: Tính ( hợp lý nếu có thể )
\(A=\dfrac{-3}{8}+\dfrac{12}{25}+\dfrac{5}{-8}+\dfrac{2}{-5}+\dfrac{13}{25}\)
\(=\left(\dfrac{-3}{8}+\dfrac{5}{-8}\right)+\left(\dfrac{12}{25}+\dfrac{13}{25}\right)+\dfrac{2}{-5}\)
\(=-1+1+\dfrac{2}{-5}\)
\(=0+\dfrac{2}{-5}\)
\(=\dfrac{2}{-5}\)
\(B=\dfrac{-3}{15}+\left(\dfrac{2}{3}+\dfrac{3}{15}\right)\)
\(=\left(\dfrac{-3}{15}+\dfrac{3}{15}\right)+\dfrac{2}{3}\)
\(=0+\dfrac{2}{3}\)
\(=\dfrac{2}{3}\)
\(C=\dfrac{-5}{21}+\left(\dfrac{-16}{21}+1\right)\)
\(=\left(\dfrac{-5}{21}+\dfrac{-16}{21}\right)+1\)
\(=-1+1\)
\(=0\)
\(D=\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)
\(=\left(\dfrac{5}{-12}+\dfrac{7}{12}\right)+\dfrac{-1}{6}\)
\(=\dfrac{1}{6}+\dfrac{-1}{6}\)
\(=0\)
Bài 2: Tìm x,biết:
a) \(x+\dfrac{2}{3}=\dfrac{4}{5}\)
\(x=\dfrac{4}{5}-\dfrac{2}{3}\)
\(x=\dfrac{2}{15}\)
Vậy \(x=\dfrac{2}{15}\)
b) \(x-\dfrac{2}{3}=\dfrac{7}{21}\)
\(\Rightarrow x-\dfrac{2}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{2}{3}\)
\(x=\dfrac{3}{3}=1\)
Vậy \(x=1\)
c) sai đề hay sao ấy bạn.bỏ dấu - ở x thì đúng đề.mk giải luôn nha!
\(x-\dfrac{3}{4}=\dfrac{-8}{11}\)
\(x=\dfrac{-8}{11}+\dfrac{3}{4}\)
\(x=\dfrac{1}{44}\)
Vậy \(x=\dfrac{1}{44}\)
d) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}-\dfrac{2}{5}\)
\(x=-\dfrac{3}{20}\)
Vậy \(x=-\dfrac{3}{20}\)
a) \(\dfrac{7}{13}\)\(\times\)\(\dfrac{7}{15}\)-\(\dfrac{5}{12}\)\(\times\)\(\dfrac{21}{39}+\dfrac{49}{91}\)\(\times\)\(\dfrac{8}{15}\)
= \(\dfrac{7}{13}\)\(\times\)\(\dfrac{7}{15}\)-\(\dfrac{5}{12}\times\dfrac{7}{13}+\dfrac{7}{13}\times\dfrac{8}{15}\)
= \(\dfrac{7}{13}\left(\dfrac{7}{15}-\dfrac{5}{12}+\dfrac{8}{15}\right)\)
= \(\dfrac{7}{13}\times\dfrac{7}{12}\)
= \(\dfrac{49}{156}\)
b) \(\left(\dfrac{12}{199}+\dfrac{23}{200}-\dfrac{34}{201}\right)\times\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
= \(\left(\dfrac{12}{199}+\dfrac{23}{200}-\dfrac{34}{201}\right)\times0\)
= 0
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
` (x+721) /2020+(x+21) /700+ (x+721)/2021=-1`
`<=>(x+721)/2020+(x+21) /700+ 1+ (x+721)/2021=0`
`<=>(x+721)/2020+(x+721) /700+ (x+721)/2021=0 `
`<=>(x +721) (1/2020+1/700+1/2021 )= 0`
Vì `1/2020+1/700+1/2021>0`
`=>x+721 =0 <=>x=-721`
Vậy `S={-721}`
Ta có: \(\dfrac{x+721}{2020}+\dfrac{x+21}{700}+\dfrac{x+721}{2021}=-1\)
\(\Leftrightarrow\dfrac{x+721}{2020}+\dfrac{x+721}{700}+\dfrac{x+721}{2021}=0\)
\(\Leftrightarrow\left(x+721\right)\left(\dfrac{1}{2020}+\dfrac{1}{700}+\dfrac{1}{2021}\right)=0\)
mà \(\dfrac{1}{2020}+\dfrac{1}{700}+\dfrac{1}{2021}>0\)
nên x+721=0
hay x=-721
Vậy: S={-721}