Tìm x;y ϵ Z sao cho :
\(\dfrac{x}{2}=\dfrac{2}{y}-\dfrac{11}{6}\)
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\(\dfrac{4}{x}-\dfrac{y}{2}=\dfrac{1}{4}\Leftrightarrow\dfrac{8-xy}{2x}=\dfrac{1}{4}\Leftrightarrow\dfrac{16-2xy}{4x}=\dfrac{x}{4x}\)
\(\Rightarrow16-2xy=x\Leftrightarrow x+2xy=16\Leftrightarrow x\left(1+2y\right)=16\)
\(\Rightarrow x;1+2y\inƯ\left(16\right)=\left\{\pm1;\pm2;\pm4;\pm8;\pm16\right\}\)
x | 1 | -1 | 2 | -2 | 4 | -4 | 8 | -8 | 16 | -16 |
2y + 1 | 16 | -16 | 8 | -8 | 4 | -4 | 2 | -2 | 1 | -1 |
y | 15/2 ( ktm ) | -17/2 ( ktm ) | 7/2 ( ktm ) | -9/2 ( ktm ) | 3/2 ( ktm ) | -5/2 ( ktm ) | 1/2 ( ktm ) | -3 / 2 ( ktm ) | 0 | -1 |
Lời giải:
$\frac{5}{x}-\frac{y}{3}=\frac{1}{6}$
$\Rightarrow \frac{15-xy}{3x}=\frac{1}{6}$
$\Rightarrow \frac{2(15-xy)}{6x}=\frac{x}{6x}$
$\Rightarrow 2(15-xy)=x$
$\Rightarrow 30=2xy+x$
$\Rightarrow 30=x(2y+1)$
$\Rightarrow x=\frac{30}{2y+1}$
Vì $x$ nguyên nên $\frac{30}{2y+1}$ nguyên
$\Rightarrow 2y+1$ là ước của $30$
Vì $2y+1$ lẻ nên $2y+1\in\left\{\pm 1; \pm 3; \pm 5; \pm 15\right\}$
$\Rightarrow y\in\left\{-1; 0; -2; 1; -3; 2; -8; 7\right\}$
Tương ứng với các giá trị $y$ trên ta có: $x\in\left\{-30; 30; -10; 10; -6; 6; -2;2\right\}$
3/ Ta có:
\(A=\dfrac{1-2x}{x+3}\)
\(A=\dfrac{-2x+1}{x+3}\)
\(A=\dfrac{-2x-6+7}{x+3}\)
\(A=\dfrac{-2\left(x+3\right)+7}{x+3}\)
\(A=-2+\dfrac{7}{x+3}\)
A nguyên khi \(\dfrac{7}{x+3}\) nguyên
⇒ 7 ⋮ \(x+3\)
\(\Rightarrow x+3\inƯ\left(7\right)\)
\(\Rightarrow x+3\in\left\{1;-1;7;-7\right\}\)
\(\Rightarrow x\in\left\{-2;-4;4;-10\right\}\)
Bài 3 :
\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)
\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)
\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)
\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)
.....
\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)
\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)
a) Để y nguyên thì \(6x-4⋮2x+3\)
\(\Leftrightarrow-13⋮2x+3\)
\(\Leftrightarrow2x+3\in\left\{1;-1;13;-13\right\}\)
\(\Leftrightarrow2x\in\left\{-2;-4;10;-16\right\}\)
hay \(x\in\left\{-1;-2;5;-8\right\}\)
\(\Rightarrow3xy=12-11y\Leftrightarrow3xy+11y=12\)
\(\Leftrightarrow y\left(3x+11\right)=12\Rightarrow y;3x+11\inƯ\left(12\right)=12\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm\right\}\)
-> bạn tự lập bảng