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242/363 + 1616/ 2121= 5/7 x y
\(\frac{2}{3}+\frac{16}{21}=\frac{5}{7}\)x y
\(\frac{14}{21}\)+\(\frac{16}{21}\)= \(\frac{5}{7}\)x y
\(\frac{10}{7}\)=\(\frac{5}{7}\)x y
\(\frac{10}{7}\): \(\frac{5}{7}\)=y
\(\frac{10}{7}\)x \(\frac{7}{5}\)=y
2=y
vậy y=2
\(\frac{242}{363}+\frac{1616}{2121}=\frac{5}{7}\times y\)
\(\frac{121\times2}{121\times3}+\frac{101\times16}{101\times21}=\frac{5}{7}\times y\)
\(\frac{2}{3}+\frac{16}{21}=\frac{5}{7}\times y\)
\(\frac{10}{7}=\frac{5}{7}\times y\)
\(\Rightarrow y=\frac{10}{7}:\frac{5}{7}\)
\(y=2\)
\(\dfrac{242}{363}+\dfrac{1616}{2121}=\dfrac{2}{7}y\)
\(\dfrac{2\cdot121}{3\cdot121}+\dfrac{16\cdot101}{21\cdot101}=\dfrac{2}{7}y\)
\(\dfrac{2}{7}y=\dfrac{2}{3}+\dfrac{16}{21}\)
\(\dfrac{2}{7}y=\dfrac{10}{7}\)
\(y=\dfrac{10}{7}\div\dfrac{2}{7}=5\)
\(\left(\dfrac{242}{363}+\dfrac{1616}{2121}\right)=\dfrac{2}{7}.x\)
\(\Leftrightarrow\dfrac{242}{363}\left(1+\dfrac{8}{7}\right)=\dfrac{2}{7}.x\)
\(\Leftrightarrow\dfrac{242}{363}.\dfrac{15}{7}=\dfrac{2}{7}.x\)
\(\Leftrightarrow\dfrac{10}{7}=\dfrac{2}{7}.x\)
\(\Leftrightarrow x=\dfrac{10}{7}:\dfrac{2}{7}\)
\(\Leftrightarrow x=\dfrac{10}{7}.\dfrac{7}{2}\)
\(\Leftrightarrow x=5\)
` 242/363 + 1616/2121 = 2/7 xxy`
`2/7 xxy= 2/3 + 16/21`
`2/7 xxy= 14/21 +16/21`
`2/7 xxy= 30/21`
`y=10/7 : 2/7`
`y=10/7 xx 7/2`
`y=70/14`
`y=5`
__
` (y + 1/4) + (y + 1/16) + (y + 1/16) =2`
`(y+y+y)+(1/4 + 1/16+1/16)=2`
`3y + (4/16 +1/16 +1/16)=2`
`3y + 6/16=2`
`3y=2-6/16`
`3y= 32/16-6/16`
`3y= 26/16`
`y=26/16 : 3`
`y=26/48`
`y=13/24`
\(a,\dfrac{242}{363}+\dfrac{1616}{2121}=\dfrac{2}{7}\times y\)
\(\dfrac{2}{7}\times y=\dfrac{2\times121}{3\times121}+\dfrac{16\times101}{21\times101}\)
\(\dfrac{2}{7}\times y=\dfrac{2}{3}+\dfrac{16}{21}\)
\(\dfrac{2}{7}\times y=\dfrac{14}{21}+\dfrac{16}{21}\)
\(\dfrac{2}{7}\times y=\dfrac{30}{21}\)
\(\dfrac{2}{7}\times y=\dfrac{10}{7}\)
\(y=\dfrac{10}{7}:\dfrac{2}{7}\)
\(y=\dfrac{10}{7}\times\dfrac{7}{2}\)
\(y=5\)
\(---\)
\(b,\left(y+\dfrac{1}{4}\right)+\left(y+\dfrac{1}{16}\right)+\left(y+\dfrac{1}{16}\right)=2\)
\(\left(y+y+y\right)+\left(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{16}\right)=2\)
\(3\times y+\left(\dfrac{4}{16}+\dfrac{2}{16}\right)=2\)
\(3\times y+\dfrac{6}{16}=2\)
\(3\times y+\dfrac{3}{8}=2\)
\(3\times y=2-\dfrac{3}{8}\)
\(3\times y=\dfrac{16}{8}-\dfrac{3}{8}\)
\(3\times y=\dfrac{13}{8}\)
\(y=\dfrac{13}{8}:3\)
\(y=\dfrac{13}{8}\times\dfrac{1}{3}\)
\(y=\dfrac{13}{24}\)
#\(Toru\)
Câu b:
\(\frac{21}{8}:\frac{5}{6}+\frac{1}{2}:\frac{5}{6}\)
= \(\frac{63}{20}+\frac{3}{5}\)
= \(\frac{15}{4}\)
\(\left(\frac{21}{8}+\frac{1}{2}\right):\frac{5}{6}\)
\(\frac{25}{8}:\frac{5}{6}\)
\(\frac{25}{8}.\frac{6}{5}\)
\(\frac{30}{8}\)
\(\left(X+\frac{1}{1.3}\right)+\left(X+\frac{1}{3.5}\right)+...+\left(X+\frac{1}{23.25}\right)=11.X+\)\(\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(\Leftrightarrow12X+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)+11X\)\(+\frac{\left(1+\frac{1}{3}+...+\frac{1}{81}\right)-\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)}{2}\)
\(\Leftrightarrow X+\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\right)=\frac{242}{243}:2\)
\(\Leftrightarrow X+\frac{12}{25}=\frac{121}{243}\)
\(\Leftrightarrow X=\frac{109}{6075}\)
Vậy X=109/6075
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Đặt:
\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{23.25}=\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{25-23}{23.25}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}=1-\frac{1}{25}=\frac{24}{25}\)
=> \(A=\frac{12}{25}\)
Đặt \(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)
=> \(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)=1-\frac{1}{3^5}=\frac{242}{243}\)
=> \(2B=\frac{242}{243}\Rightarrow B=\frac{121}{243}\)
Giải phương trình:
\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}\right)+...+\left(x+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)\)
\(12x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{242}\right)\)
\(12x+\frac{12}{25}=11x+\frac{121}{243}\)
\(12x-11x=\frac{121}{243}-\frac{12}{25}\)
\(x=\frac{109}{6075}\)
= \(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{19}{20}=\frac{1}{20}\)
=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{19}{20}\)
= đến đây bn xem trong thống kê hỏi đáp nhé
= :))
(1-1/2).(1-1/3).(1-1/4).(1-1/5)=b/100
=> 1/2.2/3.3/4.4/5=b/100
=> 1/5=b/100
=> b=100:5=20
\(\frac{242}{363}+\frac{1616}{2121}=\frac{2}{7}.\frac{2015}{A}\)
\(\frac{11.11.2}{11.11.3}+\frac{101.16}{101.21}=\frac{2}{7}.\frac{2015}{A}\)
\(\frac{2}{3}+\frac{16}{21}=\frac{2}{7}.\frac{2015}{A}\)
\(\Rightarrow\frac{2}{7}.\frac{2015}{A}=\frac{10}{7}\)
\(\frac{2015}{A}=\frac{10}{7}:\frac{2}{7}\)
\(\frac{2015}{A}=5\)
\(A=2015:5\)
\(A=403\)
\(\left(\frac{242}{363}+\frac{1616}{2121}\right)=\frac{2}{7}.\frac{2015}{A}\)
\(\Leftrightarrow\left(\frac{2}{3}+\frac{16}{21}\right)=\frac{2}{7}.\frac{2015}{A}\)
\(\Leftrightarrow\frac{10}{7}=\frac{2}{7}.\frac{2015}{A}\)
\(\Leftrightarrow5=\frac{2015}{A}\)
\(\Leftrightarrow A=403\)