Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-...+\frac{1}{19}-\frac{1}{21}\right).462-\left[2,04:\left(x+1,05\right)\right]:0,12=19\)
\(\left(\frac{1}{11}-\frac{1}{21}\right).462-\left[2,04:\left(x+1,05\right)\right]:0,12=19\)
\(\frac{10}{231}.462-\left[2,04:\left(x+1,05\right)\right]:0,12=19\)
\(20-\left[2,04:\left(x+1,05\right)\right]:0,12=19\)
\(\left[2,04:\left(x+1,05\right)\right]:0,12=20-19=1\)
\(2,04:\left(x+1,05\right)=0,12\)
\(x+1,05=2,04:0,12\)
x+1,05=17
x=17-1,05=15,95
Bài 1:
\(\Leftrightarrow\left(\dfrac{1}{11}-\dfrac{1}{21}\right)\cdot462-\left[2.04:\left(x+1.05\right)\right]:0.12=19\)
\(\Leftrightarrow\left[2.04:\left(x+1.05\right)\right]:0.12=1\)
\(\Leftrightarrow2.04:\left(x+1.05\right)=0.12\)
\(\Leftrightarrow x+1.05=17\)
hay x=15,85
\(\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{19\cdot21}\right)\cdot462-\left[0,04:\left(x+1,05\right)\right]:0,12=19\)\(\Leftrightarrow\dfrac{2}{2}\cdot\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\cdot462-\left[\dfrac{1}{25}:\left(x+\dfrac{21}{20}\right)\right]:\dfrac{3}{25}=19\)\(\Leftrightarrow\left(\dfrac{1}{11}-\dfrac{1}{21}\right)\cdot462-\left[\dfrac{1}{25}:\left(x+\dfrac{21}{20}\right)\right]:\dfrac{3}{25}=19\)
\(\Leftrightarrow20-\left[\dfrac{1}{25}:\left(x+\dfrac{21}{20}\right)\right]:\dfrac{3}{25}=19\)
\(\Leftrightarrow\left[\dfrac{1}{25}:\left(x+\dfrac{21}{20}\right)\right]:\dfrac{3}{25}=1\)
\(\rightarrow\dfrac{1}{25}:\left(x+\dfrac{21}{20}\right)=\dfrac{3}{25}\)
\(\Leftrightarrow x+\dfrac{21}{20}=\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{1}{3}-\dfrac{21}{20}=-\dfrac{43}{60}\)
Vậy \(x=-\dfrac{43}{60}\)
2,04 : ( x + 1,05 ) : 1,02 = 0,2
2,04 : ( x + 1,05 ) = 0,2 x 1 ,02
2,04 : ( x + 1,05 ) = 0,204
( x + 1,05 ) = 2,04 : 0 ,204
( x + 1,05 ) = 10
x = 10 - 1,05
x = 8,95
\(2,04:\left(x+1,05\right):1,02=0,2\)
=> \(2,04:1,02:\left(x+1,05\right)=0,2\)
=> \(2:\left(x+1,05\right)=0,2\)
=> \(x+1,05=2:0,2\)
=> \(x+1,05=10\)
=> \(x=10-1,05\)
=> \(x=8,95\)