Ta có: \(3a+11⋮a+2\)

<=> \(3\left(a+2\right)+5⋮a+2\)

<=> \(5⋮a+2\)

Vì a nguyên => \(a+2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Ta có bảng sau: 

Vậy \(a\in\left\{-1;-3;3;-7\right\}\)

Ezzz

ĐKXĐ: \(x\ne0;x\ne-2;x\ne\pm1\)

\(\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}=\dfrac{3}{4}\)

<=> \(\dfrac{1}{x-1}-\dfrac{1}{x}+\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}=\dfrac{3}{4}\)

<=> \(\dfrac{1}{x-1}-\dfrac{1}{x+2}=\dfrac{3}{4}\)

<=> \(\dfrac{x+2-x+1}{\left(x-1\right)\left(x+2\right)}=\dfrac{3}{4}\)

<=> \(\dfrac{3}{\left(x-1\right)\left(x+2\right)}=\dfrac{3}{4}\)

<=> \(\dfrac{12}{4\left(x-1\right)\left(x+2\right)}=\dfrac{3\left(x-1\right)\left(x+2\right)}{4\left(x-1\right)\left(x+2\right)}\)

<=> 12=3x²+3x-6

<=>3x²+3x-6-12=0

<=> 3x²+3x-18=0

<=> 3(x-2)(x+3)=0

<=> \(\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\) (thỏa mãn ĐKXĐ)

Vậy tập nghiệm của pt là S={2;-3}

Ta có: 4x-4xy-y=4

<=> 4x(1-y)-y-4=0

<=>4x(1-y)+(1-y)-5=0

<=>(1-y)(4x+1)=5

Vì \(x;y\in Z\Rightarrow1-y,4x+1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Ta có bảng sau:

Vì x, y là các số nguyên nên \(\left(x;y\right)\in\left\{\left(0;4\right),\left(1;0\right)\right\}\)

Vậy...

\(\sqrt{\dfrac{25}{9}}=\dfrac{5}{3}\) nhé bạn.

\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}:\sqrt{\dfrac{25}{9}}=\dfrac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}:\dfrac{5}{3}\)

\(=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}.\dfrac{5}{3}=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1-5\right)}.\dfrac{5}{3}=\dfrac{1-3}{1-5}.\dfrac{5}{3}=\dfrac{1}{2}.\dfrac{5}{3}=\dfrac{5}{6}\)

Để M nhận giá trị nguyên thì: 

\(n⋮n-1\)

=> \(n-1+1⋮n-1\)

=> 1\(⋮n-1\)

Vì \(n\in Z\Rightarrow n-1\inƯ\left(1\right)=\left\{\pm1\right\}\)

Ta có các trường hợp:

\(\left[{}\begin{matrix}n-1=1\\n-1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}n=2\\n=0\end{matrix}\right.\)

Vậy....

Theo đề bài, ta có: x-y=2 => x=2+y

=> P=xy+4=(2+y)y+4=2y+y²+4=y²+2y+1+3=(y+1)²+3

Ta có: \(\left(y+1\right)^2\ge0\) với mọi y

=> \(\left(y+1\right)^2+3\ge3\) với mọi y

Dấu "=" xảy ra khi y=-1, x=1

Vậy...