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`a in ZZ`
`=>6n-4 vdots 2n+1`
`=>3(2n+1)-7 vdots 2n+1`
`=>7 vdots 2n+1`
`=>2n+1 in Ư(7)={+-1,+-7}`
`=>2n in {0,-2,6,-8}`
`=>n in {0,-1,3,-4}`
`b in ZZ`
`=>3n+2 vdots 4n-4`
`=>12n+8 vdots 4n-4`
`=>3(4n-4)+20 vdots 4n-4`
`=>20 vdots 4n-4`
`=>4n-4 in Ư(20)={+-1,+-2,+-4,+-5,+-10,+-20}`
`=>4n-4 in {+-4,+-20}`
`=>n-1 in {+-1,+-5}`
`=>n in {0,2,6,-4}`
`c in ZZ`
`=>4n-1 vdots 3-2n`
`=>2(3-2n)-7 vdots 3-2n`
`=>7 vdots 3-2n`
`=>3-2n in Ư(7)={+-1,+-7}`
`=>2n in {4,0,-4,10}`
`=>n in {2,0,-2,5}`
a) đk: \(n\ne\dfrac{-1}{2}\)
Để \(\dfrac{6n-4}{2n+1}\) nguyên
<=> \(\dfrac{3\left(2n+1\right)-7}{2n+1}\) nguyên
<=> \(3-\dfrac{7}{2n+1}\) nguyên
<=> \(7⋮2n+1\)
Ta có bảng
2n+1 | 1 | -1 | 7 | -7 |
n | 0 | -1 | 3 | -4 |
tm | tm | tm | tm |
b)đk: \(n\ne1\)
Để \(\dfrac{3n+2}{4n-4}\) nguyên
=> \(\dfrac{3n+2}{n-1}\) nguyên
<=> \(\dfrac{3\left(n-1\right)+5}{n-1}\) nguyên
<=> \(3+\dfrac{5}{n-1}\) nguyên
<=> \(5⋮n-1\)
Ta có bảng:
n-1 | 1 | -1 | 5 | -5 |
n | 2 | 0 | 6 | -4 |
Thử lại | tm | loại | tm | loại |
c) đk: \(n\ne\dfrac{3}{2}\)
Để \(\dfrac{4n-1}{3-2n}\) nguyên
<=> \(\dfrac{4n-1}{2n-3}\) nguyên
<=> \(\dfrac{2\left(2n-3\right)+5}{2n-3}\) nguyên
<=> \(2+\dfrac{5}{2n-3}\) nguyên
<=> \(5⋮2n-3\)
Ta có bảng:
2n-3 | 1 | -1 | 5 | -5 |
n | 2 | 1 | 4 | -1 |
tm | tm | tm | tm |
\(a,\dfrac{x}{5}=-\dfrac{3}{y}\Rightarrow xy=-15\\ \Rightarrow xy=-1\cdot15=-15\cdot1=-5\cdot3=-3\cdot5\\ \Rightarrow\left(x;y\right)=\left\{\left(-1;-15\right);\left(1;-15\right);\left(15;-1\right);\left(-15;1\right);\left(3;-5\right);\left(-5;3\right);\left(5;-3\right);\left(-3;5\right)\right\}\)\(g,-\dfrac{11}{x}=\dfrac{y}{3}\\ \Rightarrow xy=-33\\ \Rightarrow xy=-3\cdot11=-11\cdot3=-1\cdot33=-33\cdot1\\ \Rightarrow\left(x;y\right)=\left\{\left(-3;11\right);\left(11;-3\right);\left(-11;3\right);\left(3;-11\right);\left(-1;33\right);\left(33;-1\right);\left(-33;1\right);\left(1;-33\right)\right\}\)
a, 9.27n=3n
32.33n=3n
32+3n=3n
2+3n=n
n-3n=2
-2n=2
n=-1
bạn nhớ k cho mk nha
b, (23:4).2n=4
(23:22).2n=22
21.2n=22
21+n = 22
1+n=2
n=1
bạn nhớ k cho minh nha