2.

\(\frac{3n+9}{n-4}\in Z\)

\(\Rightarrow3n+9⋮n-4\)

\(\Rightarrow3n-12+21⋮n-4\)

\(\Rightarrow3\times\left(n-4\right)+21⋮n-4\)

\(\Rightarrow21⋮n-4\)

\(\Rightarrow n-4\inƯ\left(21\right)\)

\(\Rightarrow n-4\in\left\{-7;-3;-1;1;3;7\right\}\)

\(\Rightarrow n\in\left\{-3;1;3;5;7;11\right\}\)

\(B=\frac{6n+5}{2n-1}\in Z\)

\(\Rightarrow6n+5⋮2n-1\)

\(\Rightarrow6n-3+8⋮2n-1\)

\(\Rightarrow3\left(2n-1\right)+8⋮2n-1\)

\(\Rightarrow8⋮2n-1\)

\(\Rightarrow2n-1\inƯ\left(8\right)\)

\(\Rightarrow2n-1\in\left\{-8;-4;-2;-1;1;2;4;8\right\}\)

\(\Rightarrow2n\in\left\{-7;-3;-1;0;2;3;5;9\right\}\)

\(n\in Z\)

\(\Rightarrow n\in\left\{0;1\right\}\)

I don't now

mik ko biết 

sorry 

......................

1)\(4n+3⋮n-2\)

\(\Leftrightarrow4n+3=4\left(n-2\right)+11\)

\(\Rightarrow4\left(n-2\right)⋮n-2\)\(\Rightarrow n-2⋮n-2\)

\(\Rightarrow11⋮n-2\)

\(\Rightarrow n-2\in\left\{\pm1;\pm11\right\}\)

\(\Rightarrow n\in\left\{3;1;13;-9\right\}\)

2)\(xy+5x+y+10=0\)

\(\Leftrightarrow x\left(y+5\right)+y+5+5=0\)

\(\Leftrightarrow x\left(y+5\right)+\left(y+5\right)=-5\)

\(\Leftrightarrow\left(x+1\right).\left(y+5\right)=-5\)

3)

$\frac{1}{}$

$\frac{n+3}{2n-2}=\frac{n-4+7}{2n-2}=\frac{2(n-2)+7}{2n-2}=2+\frac{7}{2n-2}$

Để $\frac{n+3}{2n-2}$ thì $7⋮2n-2$

$\Rightarrow 2n-2\in Ư\left(7\right)\in \{\pm 1;\pm 7\}$

$2n-2=1\Rightarrow n=1,5$

$2n-2=-1\Rightarrow n=0,5$

$2n-2=7\Rightarrow n=4,5$

$2n-2=$

Các bạn giúp mình giải với nhé! Đúng thì mình k đúng nhé. Cảm ơn các bạn nhiều lắm. Yêu cả nhà.

\(1.\left(x-5\right)^{23}.\left(y+2\right)^7=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0\\\left(y+2\right)^7=0\end{cases}\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0^{23}\\\left(y+2\right)^7=0^7\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x-5=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0+5\\y=0-2\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x=5\\y=-2\end{cases}}\)

Vậy \(\left(x;y\right)=\left(5;-2\right)\)

2. \(A=\left(x-2\right)^2+|y+3|+7\)

Ta có :

\(\hept{\begin{cases}\left(x-2\right)^2\ge0\forall x\\|y+3|\ge0\forall y\end{cases}}\)

\(\Rightarrow\left(x-2\right)^2+|y+3|\ge0\forall x;y\)

\(\Rightarrow\left(x-2\right)^2+|y+3|+7\ge7\forall x;y\)

\(\Rightarrow A\ge7\forall x;y\)

Dấu bằng xảy ra

\(\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\|y+3|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-2=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)

Vậy GTNN của A là 7 khi \(\left(x;y\right)=\left(2;-3\right)\)

cau 1 :1,6

câu 2 : sai đề bài

cau 3 chua lam duoc 

cau 4 : chua lam duoc

cau 5 :101/10

1) 2n - 5 \(⋮\)n + 1

    2(n + 1) - 7 \(⋮\)n + 1

Do 2(n+1) \(⋮\)n+1 nên 7 \(⋮\)n+1 \(\Rightarrow\)n + 1 \(\in\)Ư(7) = { 1; -1; 7; -7}

Với n + 1 = 1 \(\Rightarrow\)n = 0

     n + 1 = -1 \(\Rightarrow\)n = -2

     n + 1 = 7 \(\Rightarrow\)n = 6

     n + 1 = -7 \(\Rightarrow\)n = -8

Vậy n = { 0; -2; 6; -8}