2. Tìm phân số \(\dfrac{a}{b}\):
\(\dfrac{5}{9}:\dfrac{a}{b}=\dfrac{2}{3}\) \(\dfrac{a}{b}-\dfrac{2}{3}=4\)
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a: \(1-\left(5\dfrac{4}{9}+a-7\dfrac{7}{18}\right):15\dfrac{3}{4}=0\)
=>\(\left(5+\dfrac{4}{9}+a-7-\dfrac{7}{18}\right):\dfrac{63}{4}=1\)
=>\(\left(a-2+\dfrac{1}{18}\right)=\dfrac{63}{4}\)
=>\(a-\dfrac{35}{18}=\dfrac{63}{4}\)
=>\(a=\dfrac{63}{4}+\dfrac{35}{18}=\dfrac{637}{36}\)
b: \(B=\left(\dfrac{2}{15}+\dfrac{5}{3}-\dfrac{3}{5}\right):\left(4\dfrac{2}{3}-2\dfrac{1}{2}\right)\)
\(=\dfrac{2+5\cdot5-3^2}{15}:\left(4+\dfrac{2}{3}-2-\dfrac{1}{2}\right)\)
\(=\dfrac{2+4^2}{15}:\left(2+\dfrac{2}{3}-\dfrac{1}{2}\right)\)
\(=\dfrac{18}{15}:\dfrac{13}{6}=\dfrac{6}{5}\cdot\dfrac{6}{13}=\dfrac{36}{65}\)
1: B là số nguyên
=>n-3 thuộc {1;-1;5;-5}
=>n thuộc {4;2;8;-2}
3:
a: -72/90=-4/5
b: 25*11/22*35
\(=\dfrac{25}{35}\cdot\dfrac{11}{22}=\dfrac{5}{7}\cdot\dfrac{1}{2}=\dfrac{5}{14}\)
c: \(\dfrac{6\cdot9-2\cdot17}{63\cdot3-119}=\dfrac{54-34}{189-119}=\dfrac{20}{70}=\dfrac{2}{7}\)
Bài 1:
a)
\(\dfrac{1}{2}=\dfrac{1\times6}{2\times6}=\dfrac{6}{12}\)
\(\dfrac{2}{3}=\dfrac{2\times4}{3\times4}=\dfrac{8}{12}\)
\(\dfrac{3}{4}=\dfrac{3\times3}{4\times3}=\dfrac{9}{12}\)
b)
\(\dfrac{1}{3}=\dfrac{1\times15}{3\times15}=\dfrac{15}{45}\)
\(\dfrac{2}{15}=\dfrac{2\times3}{15\times3}=\dfrac{6}{45}\)
\(\dfrac{4}{45}\) (giữ nguyên)
c)
\(\dfrac{1}{8}=\dfrac{1\times3}{8\times3}=\dfrac{3}{24}\)
\(\dfrac{2}{3}=\dfrac{2\times8}{3\times8}=\dfrac{16}{24}\)
\(\dfrac{5}{2}=\dfrac{5\times12}{2\times12}=\dfrac{60}{24}\)
d)
\(\dfrac{2}{7}=\dfrac{2\times4}{7\times4}=\dfrac{8}{28}\)
\(\dfrac{9}{4}=\dfrac{9\times7}{4\times7}=\dfrac{63}{28}\)
\(\dfrac{5}{28}\) (giữ nguyên)
Bài 2:
a)
\(4=\dfrac{4}{1}=\dfrac{4\times12}{1\times12}=\dfrac{48}{12}\)
\(\dfrac{9}{4}=\dfrac{9\times3}{4\times3}=\dfrac{27}{12}\)
b)
\(\dfrac{5}{8}=\dfrac{5\times30}{8\times30}=\dfrac{150}{240}\)
\(\dfrac{25}{30}=\dfrac{5}{6}=\dfrac{5\times40}{6\times40}=\dfrac{200}{240}\)
\(2=\dfrac{2}{1}=\dfrac{2\times240}{1\times240}=\dfrac{480}{240}\).
a,a+1/4=2 3/4-1 1/2
a+1/2=5/4
a=5/4-1/2
a=3/4
b,a-7/4=13/4-7/9
a-7/4=89/36
a= 89/36+7/4
a=152/36
c,3/2-a=17/6-1/6
3/2-a=8/3
a= 3/2-8/3
a= -7/6
a, \(\dfrac{3}{4}\) = \(\dfrac{3\times5}{4\times5}\) = \(\dfrac{15}{20}\) \(\dfrac{5}{7}\) = \(\dfrac{5\times3}{7\times3}\) = \(\dfrac{15}{21}\)
Vì \(\dfrac{15}{20}\) > \(\dfrac{15}{21}\) nên \(\dfrac{3}{4}\) > \(\dfrac{5}{7}\)
b, \(\dfrac{2}{7}\) = \(\dfrac{2\times2}{7\times2}\) = \(\dfrac{4}{14}\) < \(\dfrac{4}{9}\)
Vậy \(\dfrac{2}{7}\) < \(\dfrac{4}{9}\)
c, \(\dfrac{5}{8}\) < 1 < \(\dfrac{8}{5}\)
Vậy \(\dfrac{5}{8}\) < \(\dfrac{8}{5}\)
`A=(8 2/7-4 2/7)-3 4/9`
`=8+2/7-4-2/7-3-4/9`
`=4-3-4/9`
`=1-4/9=5/9`
`B=(10 2/9-6 2/9)+2 3/5`
`=10+2/9-6-2/9+2+3/5`
`=4+2+3/5`
`=6+3/5=33/5`
Bài 2:
`a)5 1/2*3 1/4`
`=11/2*13/4`
`=143/8`
`b)6 1/3:4 2/9`
`=19/3:38/9`
`=19/3*9/38=3/2`
`c)4 3/7*2`
`=31/7*2`
`=62/7`
Bài 1:
\(A=\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\)
\(A=\left(\dfrac{58}{7}-\dfrac{30}{7}\right)-\dfrac{31}{9}\)
\(A=4-\dfrac{31}{9}\)
\(A=\dfrac{5}{9}\)
\(B=\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\)
\(B=\left(\dfrac{92}{9}-\dfrac{56}{9}\right)+\dfrac{13}{5}\)
\(B=4+\dfrac{13}{5}\)
\(B=\dfrac{33}{5}\)
Bài 2:
a) Ta có: \(A=\dfrac{4}{n-1}+\dfrac{6}{n-1}-\dfrac{3}{n-1}\)
\(=\dfrac{4+6-3}{n-1}\)
\(=\dfrac{7}{n-1}\)
Để A là số tự nhiên thì \(7⋮n-1\)
\(\Leftrightarrow n-1\inƯ\left(7\right)\)
\(\Leftrightarrow n-1\in\left\{1;7\right\}\)
hay \(n\in\left\{2;8\right\}\)
Vậy: \(n\in\left\{2;8\right\}\)
ta có B=2n+9/n+2-3n+5n+1/n+2=4n+10/n+2 Để B là STN thì 4n+10⋮n+2 4n+8+2⋮n+2 4n+8⋮n+2 ⇒2⋮n+2 n+2∈Ư(2) Ư(2)={1;2} Vậy n=0
\(\sqrt{\dfrac{1}{4}+\dfrac{1}{\left(2n-1\right)^2}+\dfrac{1}{\left(2n+1\right)^2}}=\sqrt{\dfrac{\left(2n-1\right)^2\left(2n+1\right)^2+4\left(2n-1\right)^2+4\left(2n+1\right)^2}{4\left(2n-1\right)^2\left(2n+1\right)^2}}\)
\(=\sqrt{\dfrac{\left(4n^2-1\right)^2+4\left(4n^2-4n+1\right)+4\left(4n^2+4n+1\right)}{4\left(2n-1\right)^2\left(2n+1\right)^2}}\)
\(=\sqrt{\dfrac{16n^4+24n^2+9}{4\left(2n-1\right)^2\left(2n+1\right)^2}}=\sqrt{\dfrac{\left(4n^2+3\right)^2}{4\left(2n-1\right)^2\left(2n+1\right)^2}}=\dfrac{4n^2+3}{2\left(2n-1\right)\left(2n+1\right)}\)
\(=\dfrac{\left(4n^2-1\right)+4}{2\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)
\(=\dfrac{1}{2}+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\)
Do đó:
\(P=\left(\dfrac{1}{2}+\dfrac{1}{1}-\dfrac{1}{3}\right)+\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}\right)+...+\left(\dfrac{1}{2}-\dfrac{1}{399}-\dfrac{1}{401}\right)\)
\(=\dfrac{1}{2}.200+1-\dfrac{1}{401}=\dfrac{40500}{401}\)
\(\Rightarrow Q=400\)
a) \(\dfrac{2}{5}\times?=\dfrac{3}{10}\)
\(?=\dfrac{3}{10}:\dfrac{2}{5}=\dfrac{3}{4}\)
b) \(\dfrac{1}{8}:?=\dfrac{1}{5}\)
\(?=\dfrac{1}{8}:\dfrac{1}{5}=\dfrac{5}{8}\)
a: Phân số cần tìm là: \(\dfrac{3}{10}:\dfrac{2}{5}=\dfrac{3}{10}\cdot\dfrac{5}{2}=\dfrac{15}{20}=\dfrac{3}{4}\)
b: Phân số cần tìm là \(\dfrac{1}{8}:\dfrac{1}{5}=\dfrac{5}{8}\)
\(\dfrac{a}{b}=\dfrac{5}{9}:\dfrac{2}{3}\)
\(\dfrac{a}{b}=\dfrac{5}{6}\)
\(\dfrac{a}{b}=4+\dfrac{2}{3}\)
\(\dfrac{a}{b}=\dfrac{14}{3}\)
a)\(\dfrac{a}{b}=\dfrac{5}{9}:\dfrac{2}{3}=\dfrac{5}{9}\times\dfrac{3}{2}=\dfrac{5}{6}\)
b)\(\dfrac{a}{b}=4+\dfrac{2}{3}=\dfrac{12}{3}+\dfrac{2}{3}=\dfrac{14}{3}\)