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a) 2(x+1)-4 = 5x+7
2x + 2 - 4 - 5x=7
-3x = 7 - 2 + 4
-3x = 9
x = 9 : (-3)
x = -3
b) 10 = 10 + 9 + 8 + ... + x
=> 9 + 8 + ... + x = 0
=> x = -9
1) \(xy-2x-y=-6\Rightarrow x\left(y-2\right)-y=-6\Rightarrow x\left(y-2\right)-y+2=-6+2\)
\(\Rightarrow x\left(y-2\right)-\left(y-2\right)=-4\Rightarrow\left(y-2\right)\left(x-1\right)=-4\)
\(\Rightarrow x-1\inƯ\left(-4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
Ta có bảng sau:
x - 1 | 1 | -1 | 2 | -2 | 4 | -4 |
y - 2 | -4 | 4 | -2 | 2 | -1 | 1 |
Suy ra ta có các cặp (x,y) sau:
x | 2 | 0 | 3 | -1 | 5 | -3 |
y | -2 | 6 | 0 | 4 | 1 | 3 |
2) \(|x+1|+|x-2|+|x+7|=5x-1\)
Ta thấy: \(|x+1|\ge0,|x-2|\ge0,|x+7|\ge0\) với \(\forall x\inℤ\)
Mà \(|x+1|+|x-2|+|x+7|=5x-10\Rightarrow5x-10\ge0\Rightarrow5x\ge10\Rightarrow x\ge2>0\)
\(\Rightarrow|x+1|=x+1,|x-2|=x-2,|x+7|=x+7\)
\(\Rightarrow|x+1|+|x-2|+|x+7|=x+1+x-2+x+7=5x-10\)
\(\Rightarrow\left(x+x+x\right)+\left(1-2+7\right)=5x-10\Rightarrow3x+6=5x-10\)
\(\Rightarrow3x-5x=-10-6\Rightarrow-2x=-16\Rightarrow x=\frac{-16}{-2}=8\)
Với mọi x thì /x+1/>=0
/x-2/>=0
/x+7/>=0
\(\Rightarrow\)5x-10>=0
Nên x>=2
\(\Rightarrow\)x+1+x-2+x+7=3x+6=5x-10
\(\Rightarrow\)2x=16
\(\Rightarrow\)x=8
Ta có |x+1|\(\ge\)0
|x-2|\(\ge\)0
|x+7|\(\ge\)0
\(\Rightarrow5x-10\ge0\)
\(\Rightarrow x+1+x-2+x+7=5x-10\)
3x+(1-2+7)=5x-10
3x+6=5x-10
6+10=5x-3x
16=2x
x=8
d) Ta có: \(n^2+5n+9⋮n+3\)
\(\Leftrightarrow n^2+3n+2n+6+3⋮n+3\)
\(\Leftrightarrow n\left(n+3\right)+2\left(n+3\right)+3⋮n+3\)
mà \(n\left(n+3\right)+2\left(n+3\right)⋮n+3\)
nên \(3⋮n+3\)
\(\Leftrightarrow n+3\inƯ\left(3\right)\)
\(\Leftrightarrow n+3\in\left\{1;-1;3;-3\right\}\)
hay \(n\in\left\{-2;-4;0;-6\right\}\)
Vậy: \(n\in\left\{-2;-4;0;-6\right\}\)
d) Ta có: n2+5n+9⋮n+3n2+5n+9⋮n+3
⇔n2+3n+2n+6+3⋮n+3⇔n2+3n+2n+6+3⋮n+3
⇔n(n+3)+2(n+3)+3⋮n+3⇔n(n+3)+2(n+3)+3⋮n+3
mà n(n+3)+2(n+3)⋮n+3n(n+3)+2(n+3)⋮n+3
nên 3⋮n+33⋮n+3
⇔n+3∈Ư(3)⇔n+3∈Ư(3)
⇔n+3∈{1;−1;3;−3}
Ta có: \(\hept{\begin{cases}GTTDx+1\ge0\\GTTDx-2\ge0\\GTTDx+7\ge0\end{cases}}\)với mọi x \(\Rightarrow\)/x+1/+/x-2/+/x+7/ \(\ge\)0 với mọi x hay 5x-10\(\ge\)0 \(\Rightarrow5x\ge10\Rightarrow x\ge2\)
Với \(x\ge2\), ta có: /x+1/+/x-2/+/x+7/=x+1+x-2+x+7=5x-10 hay 3x+6=5x-10 \(\Rightarrow\)3x+16=5x \(\Rightarrow\)2x=16 \(\Rightarrow\)x=8
Vậy x=8