Ta có :\(A=\frac{x^2+3x+1}{x+2}=\frac{x^2+2x+x+2-1}{x+2}=\frac{x\left(x+2\right)+x+2-1}{x+2}=\frac{\left(x+1\right)\left(x+2\right)-1}{x+2}\)

\(=x+1-\frac{1}{x+2}\)

Để A nguyên => \(\frac{1}{x+2}\inℤ\Rightarrow1⋮x+2\Rightarrow x+2\inƯ\left(1\right)\)

=> \(x+2\in\left\{-1;1\right\}\)

=> x \(\in\left\{-3;-1\right\}\)

Vậy  x \(\in\left\{-3;-1\right\}\)thì A nguyên 

Thank You!

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 \(\left(x;y\right)=\left(2;-3\right)\)

ĐKXĐ: \(x\ne\pm3\)

a

Khi x = 1:

\(A=\dfrac{3.1+2}{1-3}=\dfrac{5}{-2}=-2,5\)

Khi x = 2:

\(A=\dfrac{3.2+2}{2-3}=-8\)

Khi x = \(\dfrac{5}{2}:\)

\(A=\dfrac{3.2,5+2}{2,5-3}=\dfrac{9,5}{-0,5}=-19\)

b

Để A nguyên => \(\dfrac{3x+2}{x-3}\) nguyên

\(\Leftrightarrow3x+2⋮\left(x-3\right)\\3\left(x-3\right)+11⋮\left(x-3\right) \)

Vì \(3\left(x-3\right)⋮\left(x-3\right)\) nên \(11⋮\left(x-3\right)\)

\(\Rightarrow\left(x-3\right)\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\\ \Rightarrow x\left\{4;2;-8;14\right\}\)

c

Để B nguyên => \(\dfrac{x^2+3x-7}{x+3}\) nguyên

\(\Rightarrow x\left(x+3\right)-7⋮\left(x+3\right)\)

\(\Rightarrow-7⋮\left(x+3\right)\\ \Rightarrow x+3\inƯ\left\{\pm1;\pm7\right\}\)

\(\Rightarrow x=\left\{-4;-11;-2;4\right\}\)

d

\(\left\{{}\begin{matrix}A.nguyên.\Leftrightarrow x=\left\{-8;2;4;14\right\}\\B.nguyên\Leftrightarrow x=\left\{-11;-4;-2;4\right\}\end{matrix}\right.\)

=> Để A, B cùng là số nguyên thì x = 4.

Ta có: ĐK \(x\ne-1\)

\(A=\frac{x^2+2x}{x+1}=\frac{x^2+2x+1-1}{x+1}=\frac{\left(x+1\right)^2-1}{x+1}=x+1-\frac{1}{x+1}\)

Để A là số nguyên thì ta có \(x+1\inƯ\left(1\right)\)

Ta có bảng sau:

Vậy \(x\in\left\{0;-2\right\}\)

1: Để A nguyên thì x+3-4 chia hết cho x+3

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

=>\(x\in\left\{-2;-4;-1;-5;1;-7\right\}\)

2: Để B nguyên thì 2x+4-9 chia hết cho x+2

=>\(x+2\in\left\{1;-1;3;-3;9;-9\right\}\)

=>\(x\in\left\{-1;-3;1;-5;7;-11\right\}\)

\(A=\frac{3x-1}{x-1}=\frac{3\left(x-1\right)+2}{x-1}=3+\frac{2}{x-1}\)

\(B=\frac{2x^2+x-1}{x+2}=\frac{\left(x+2\right)\left(2x-3\right)+5}{x+2}=2x-3+\frac{5}{x+2}\)

Để A,B đều là số nguyên thì \(x-1\in\left\{1;2;-1;-2\right\}\) và \(x+2\in\left\{1;5;-1;-5\right\}\)

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