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Gọi \(ƯC\left(12n+1;30n+2\right)=d\)
\(\Rightarrow12n+1⋮d\Rightarrow60n+5⋮d\)
và \(30n+2⋮d\Rightarrow60n+ 4⋮d\)
Do đó \(60n+5-60n-4⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=1\)
Vậy \(\dfrac{12n+1}{30n+2}\) là phân số tối giản.
Gọi (12n+1),(30n+2) là d (1)
=>30n+2 \(⋮\) d
=> 2(30n + 2) \(⋮\) d hay 60n +4 \(⋮\) d
Tương tự ta chưng minh:
12n + 1 \(⋮\)d (2)
=> 5(12n+1) \(⋮\) d hay 60n +5 \(⋮\)d
Do đó (60n + 5) - ( 60n +4 ) \(⋮\)d hay 1 \(⋮\) d
=> d = 1 hoặc -1
Từ (1) và(2) ta có( 12n+1 ;30n+2) =1
=> P/s 12n + 1 /30n+2 là ps tối giản
a)\(123-5:\left(x+4\right)=38\)
\(5:\left(x+4\right)=123-38\)
\(5:\left(x+4\right)=85\)
\(x+4=5:85\)
\(x=\dfrac{1}{17}-4\)
\(x=-\dfrac{67}{17}\)
b)\(70-5.\left(x-3\right)=45\)
\(5.\left(x-3\right)=70-45\)
\(5.\left(x-3\right)=35\)
\(x-3=35:5\)
\(x-3=7\)
\(x=7+3\)
\(x=10\)
Câu 14)
\(a,\\ =-\dfrac{3}{8}+\dfrac{8}{17}+\dfrac{-5}{8}-\dfrac{3}{5}+\dfrac{9}{17}\\ =\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\left(\dfrac{8}{17}+\dfrac{9}{17}\right)-\dfrac{3}{5}\\ =\left(-1\right)+1-\dfrac{3}{5}=0-\dfrac{3}{5}=\dfrac{-3}{5}\\ b,\\ =\dfrac{7}{15}.\dfrac{-15}{14}+\left(\dfrac{27}{16}-\dfrac{1}{8}\right):\dfrac{5}{8}\)
\(=\dfrac{-1}{2}+\dfrac{25}{16}.\dfrac{8}{5}=\dfrac{-1}{2}+\dfrac{5}{2}=2\\ c,\\ =\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+.....+\dfrac{2}{99}-\dfrac{2}{100}\\ =1-\dfrac{1}{50}=\dfrac{49}{50}\)
Câu 15
\(a,2x+\dfrac{-1}{4}=\dfrac{3}{2}\\ 2x=\dfrac{3}{2}-\dfrac{-1}{4}=\dfrac{7}{4}\\ x=\dfrac{7}{4}:2=\dfrac{7}{8}\\ b,\dfrac{15}{x}=\dfrac{-3}{4}\\ x=\dfrac{15.4}{-3}=-20\)
A=\(\frac{3n+9}{n+2}=3+\frac{3}{n+2}\)
muốn A nguyên thì n+2 thuocj Ư(3)=(-1,-3,1,3)
giải từng TH ra là được :
n+2=-1=>n=-3
n+2=-3=>n=-5
n+2=1=>n=-1
n+2=3=>n=1
vậy n=( -1,-3,-5,1) thì A nguyên
\(\dfrac{\dfrac{2}{5}+\dfrac{2}{7}-\dfrac{2}{9}-\dfrac{2}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{9}-\dfrac{4}{11}}=\dfrac{2.\left[\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right]}{4.\left[\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right]}\)\(=\dfrac{2}{4}=\dfrac{1}{2}\)
\(B=\dfrac{\dfrac{2}{5}+\dfrac{2}{7}-\dfrac{2}{9}-\dfrac{2}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{9}-\dfrac{4}{11}}=\dfrac{2.\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4.\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{1}{2}\)
a) Ư(60):{ 1;2;3;4;5;6;10;12;15;20;30;60}
Ư(84):{ 1;2;4;6;7;12;14;21;42;84}
Ư(120):{ 1;2;3;4;5;6;8;10;12;15;20;24;30;40;60;120}
ƯC(60;84;120):{ 2;4;6;12}
nhưng vì x_> 6 nên x = 2,4,6
Đề thi Hậu Lộc năm 2016 2017 r nhé