Bài 13: Bội và ước của một số nguyên

a,(x+1)(xy-1)=3

có 3=1.3=3.1=-1.-3=-3.-1

x+1 1 3 -3 -1

x 0 2 -4 -2

xy-1 3 1 -1 -3

y Φ 1 0 1

vậy x;y là (2;1),(-4;0),(-2;1)

b, Làm tương tự

a/ \(19⋮3-n\)

\(\Leftrightarrow3-n\inƯ\left(19\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3-n=1\\3-n=19\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}n=2\left(tm\right)\\n=-22\left(loại\right)\end{matrix}\right.\)

Vậy ...

b/ \(21⋮1-2n\)

\(\Leftrightarrow1-2n\inƯ\left(21\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}1-2n=1\\1-2n=21\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}n=0\\n=-10\left(loại\right)\end{matrix}\right.\)

Vậy ...

c/ \(9⋮\left|n-1\right|\)

\(\Leftrightarrow\left|n-1\right|\inƯ\left(9\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|n-1\right|=9\\\left|n-1\right|=3\\\left|n-1\right|=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}n-1=9\\n-1=-9\end{matrix}\right.\\\left[{}\begin{matrix}n-1=3\\n-1=-3\end{matrix}\right.\\\left[{}\begin{matrix}n-1=1\\n-1=-1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}n=10\\n=-8\left(lọai\right)\end{matrix}\right.\\\left[{}\begin{matrix}n=4\\n=-2\left(loại\right)\end{matrix}\right.\\\left[{}\begin{matrix}n=2\\n=0\end{matrix}\right.\end{matrix}\right.\)

Vậy ...

d/ \(-15⋮3-2n\)

\(\Leftrightarrow3-2n\inƯ\left(-15\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3-2n=1\\3-2n=15\\3-2n=3\\3-2n=5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}n=1\\n=-6\left(loại\right)\\n=0\\n=-1\left(loại\right)\end{matrix}\right.\)

Vậy...

a: \(8⋮x\)

mà x<0

nên \(x\in\left\{-1;-2;-4;-8\right\}\)

b: \(\left\{{}\begin{matrix}-8⋮x\\12⋮x\end{matrix}\right.\Leftrightarrow x\inƯC\left(-8;12\right)\)

hay \(x\in\left\{1;-1;2;-2;4;-4\right\}\)

\(4x+3⋮x-2\)

Mà \(x-2⋮x-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x+3⋮x-2\\4x-8⋮x-2\end{matrix}\right.\)

\(\Leftrightarrow11⋮x-2\)

\(\Leftrightarrow x-2\inƯ\left(11\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=11\\x-2=-1\\x-2=-11\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=13\\x=1\\x=-9\end{matrix}\right.\)

Vậy ..

Nguyễn Thanh Hằng CTV23 tháng 1 2018 lúc 17:19

$4x+3⋮x-2$

Mà $x-2⋮x-2$

$\iff \{\begin{array}{c}4x+3⋮x-2\\ 4x-8⋮x-2\end{array}$

$\iff 11⋮x-2$

$\iff x-2\in Ư\left(11\right)$

$\iff [\begin{array}{c}x-2=1\\ x-2=11\\ x-2=-1\\ x-2=-11\end{array}$ $\iff [\begin{array}{c}x=3\\ x=13\\ x=1\\ x=-9\end{array}$

Vậy ..

a/ \(-7⋮2n+1\)

\(\Leftrightarrow2n+1\inƯ\left(-7\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2n+1=1\\2n+1=7\\2n+1=-1\\2n+1=-7\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}n=0\\n=3\\n=-1\\n=-3\end{matrix}\right.\)

Vậy ...

b/ \(-11⋮2n+5\)

\(\Leftrightarrow2n+5\inƯ\left(-11\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2n+5=1\\2n+5=11\\2n+5=-1\\2n+5=-11\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}n=-2\\n=3\\n=-3\\n=-8\end{matrix}\right.\)

Vậy ..

(x+7+1) ⋮(x+7)

Vì (x+7) ⋮ (x+7) nên 1⋮ (x+7)

=> x+7ϵ Ư(1)={1;-1}

.x+7=1 .x+7=-1

x=1-7 x=(-1)-7

x=-6 x=-8

Vậy x ϵ{-6;-8}

thua

Bạn Kuroko no Basket gần đúng, bài này phải kẻ bảng

Gọi (a,b) = d

Vì \(\dfrac{a}{b}=\dfrac{4}{5}\) mà (4,5) = 1\(\Rightarrow a=4d,b=5d\)

Ta có: \(ab=\left(a,b\right)\cdot\left[a,b\right]\Rightarrow\left[a,b\right]=\dfrac{ab}{\left(a,b\right)}=\dfrac{4d.5d}{d}=4.5.d=20.d=140\)

\(\Rightarrow d=7\Rightarrow\left\{{}\begin{matrix}a=4.7=28\\b=5.7=35\end{matrix}\right.\)

gọi (a,b) = d

Vì $\frac{a}{b}=\frac{4}{5}$ mà (4,5) = 1$\Rightarrow a=4d,b=5d$

Ta có: $ab=\left(a,b\right)\cdot \left[a,b\right]\Rightarrow \left[a,b\right]=\frac{ab}{\left(a,b\right)}=\frac{4d.5d}{d}=\mathrm{4.5.}d=20.d=140$

$\Rightarrow d=7\Rightarrow \{\begin{array}{c}a=4.7=28\\ b=5.7=35\end{array}$

\(n^2+3n-13⋮n+3\)

Mà \(n+3⋮n+3\)

\(\Leftrightarrow\left\{{}\begin{matrix}n^2+3n-13⋮n+3\\n^2+3n⋮n+3\end{matrix}\right.\)

\(\Leftrightarrow13⋮n+3\)

\(\Leftrightarrow n+3\inƯ\left(13\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}n+3=1\\n+3=13\\n+3=-1\\n+3=-13\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}n=-2\\n=10\\n=-4\\n=-16\end{matrix}\right.\)

Vậy ..

$n^2+3n-13⋮n+3$

Mà $n+3⋮n+3$

$\iff \{\begin{array}{c}{n}^2+3n-13⋮n+3\\ n^2+3n⋮n+3\end{array}$

$\iff 13⋮n+3$

$\iff n+3\in Ư\left(13\right)$

$\iff [\begin{array}{c}n+3=1\\ n+3=13\\ n+3=-1\\ n+3=-13\end{array}$

$\iff [\begin{array}{c}n=-2\\ n=10\\ n=-4\\ n=-16\end{array}$

Vậy ..

2x(y+1)+(y+1)=-5

=>(2x+1)(y+1)=-5

=> 2x+1 va y+1 ∈U(-5)=(1,5,-1,-5)

Ta co bang :

a)3x-5=4x+6

=> 3x-4x=5+6

=> -1x=11

x=11:(-1)

x=-11

b)2x+15=4x-23

=> 2x-4x=15-23

=>-2x=-8

x=-8:(-2)

x=4

c)|x+1|-12=17+(-15)

=> |x+1|-12=17-15

=>|x+1|-12=2

=>|x+1|=2+12

=>|x+1|=14

Trường hợp 1: x+1=14.

x=14-1.

x=13.

Trường hợp 2: x+1=-14

x=-14-1

x=-15

Vậy x=13; x=-15.