tìm x thuộc Z để biểu thức sau thuộc Z
C = 42/(2x^2 - 5x + 7 )
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Cach lam ne:
C thuộc Z\Leftrightarrow42 chia het cho 2x^2 - 5x + 7 hay 2x^2-5x+7 la uoc cua 42
U(42)={1;2;3;6;7;21;14;42;-1;-2;-3;-14;-6;-7;-42;-21}
Neu 2x^2 - 5x + 7=1 thi 2x^2 - 5x+6=0ma2x^2 - 5x+6>0(loai)
Neu 2x^2 - 5x + 7=-1thi 2x^2 - 5x+8=0 ma2x^2 - 5x+8>0 (loai)
Neu 2x^2 - 5x + 7=2 thi 2x^2 - 5x+5=0 ma2x^2 - 5x+5>0(loai)
Neu 2x^2 - 5x + 7=-2thi 2x^2 - 5x+9=0ma2x^2 - 5x+9>0(loai)
Neu 2x^2 - 5x + 7=3 thi 2x^2 - 5x+4=0ma2x^2 - 5x+4>0(loai)
Neu 2x^2 - 5x + 7=-3thi 2x^2 - 5x+10=0ma2x^2 - 5x+10>0(loai)
Neu 2x^2 - 5x + 7=-6 thi 2x^2 - 5x+13=0ma2x^2 - 5x+13>0(loai)
Neu 2x^2 - 5x + 7=-7thi 2x^2 - 5x+14=0ma2x^2 - 5x+14>0(loai)
Neu 2x^2 - 5x + 7=-14 thi 2x^2 - 5x+21=0ma2x^2 - 5x+21>0(loai)
Neu 2x^2 - 5x + 7=-21thi 2x^2 - 5x+28=0ma2x^2 - 5x+28>0loai)
Neu 2x^2 - 5x + 7=-42 thi 2x^2 - 5x+49=0ma2x^2 - 5x+49>0(loai)
Neu 2x^2 - 5x + 7=4thi 2x^2 - 5x+3=0\Rightarrowx=1
Neu 2x^2 - 5x + 7=14thi 2x^2 - 5x-7=0\Rightarrowx=-1
Neu 2x^2 - 5x + 7=21thi 2x^2 - 5x-14=0\Rightarrowx kg co nghiem nguyen
Neu 2x^2 - 5x + 7=42thi 2x^2 - 5x-35=0\Rightarrowx kg co nghiem nguyen
Vay voi x=1hoac x=-1 thi C nguyen
Dung thi nho thanks nha