Tìm x :
\(\left(\frac{2}{11\times13}+\frac{2}{13\times15}+...+\frac{2}{19\times21}\right)\times462-2,04\div\left(x+15\right)+11,02=30\)
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Ta có : \(\left(\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{19.21}\right)642-\left[0,04:\left(x+1,05\right)\right]:0,12=19\)
=> \(\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}\right)642-0,04:\left(x+1,05\right):0,12=19\)
=>
(nãy bấm nhầm) tiếp nà :
=> \(\left(\frac{1}{11}-\frac{1}{21}\right)462-0,04:\left(x+1,05\right):0,12=19\)
=> \(\frac{10}{231}.462-0,04:\left(x+1,05\right):0,12=19\)
=> \(20-0,04:\left(x+1,05\right):0,12=19\)
=> 0,04 : (x + 1,05) : 0,12 = 1
=> 0,04 : (x + 1,05) = 0,12
=> \(x+1,05=\frac{1}{3}\)
=> \(x=\frac{1}{3}-1,05=...\)
\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(x-10\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)=\frac{3}{11}\)
\(x-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10.\frac{4}{55}=\frac{3}{11}\)
\(x-\frac{40}{55}=\frac{3}{11}\)
\(x=\frac{3}{11}+\frac{40}{55}\)
\(x=\frac{55}{55}=1\)
nha.
\(a,\)\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=-\frac{37}{45}\)
\(x+\left(\frac{9-5}{5.9}+\frac{13-9}{9.13}+\frac{17-13}{13.17}+...+\frac{45-41}{41.45}\right)=-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+....+\frac{1}{41}-\frac{1}{45}\right)-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(x+\frac{8}{45}=-\frac{37}{45}\)
\(x=-\frac{37}{45}-\frac{8}{45}\)
\(x=-1\)
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
2/
a) \(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\)
\(=\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+....+\frac{1}{17}-\frac{1}{21}\right)\)
\(=1-\frac{1}{21}=\frac{20}{21}\)
b) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot..\cdot\frac{2016}{2017}\)
\(=\frac{1}{2017}\)
c) \(A=2000-5-5-5-..-5\)(có 200 số 5)
\(A=2000-\left(5\cdot200\right)\)
\(A=2000-1000\)
\(A=1000\)
\(\left(\frac{2}{11\times13}+\frac{2}{13\times15}+...+\frac{2}{19\times21}\right)\times462-2,04\div\left(x+15\right)+11,02=30\)
\(\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}\right)\times462-2,04\div\left(x+15\right)+11,02=30\)
\(\left(\frac{1}{11}-\frac{1}{21}\right)\times462-2,04\div\left(x+15\right)=30-11,02\)
\(\frac{10}{231}\times462-2,04\div\left(x+15\right)=18,98\)
\(20-2,04\div\left(x+15\right)=18,98\)
\(2,04\div\left(x+15\right)=20-18,98\)
\(2,04\div\left(x+15\right)=1,02\)
\(x+15=2,04\div1,02\)
\(x+15=2\)
\(x=2-15\)
\(x=-13\)