a, ĐKXĐ: xe1;x≥0
b, C=x+x−23x+9x−3−x+2x+1+1−xx+2
⇔C=(x+2)(x−1)3x+3x−3−x+2x+1−x−1x+2
⇔C=(x+2)(x−1)3(x+x−1)−(x+2)(x−1)(x+1)(x−1)−(x+2)(x−1)(x+2)(x+2)
⇔C=(x+2)(x−1)3x+3x−3−x+1−x−4x−4
⇔C=(x+2)(x−1)x−x−6=(x+2)(x−1)(x+2)(x−3)=x−1x−3
c, C=x−1x−3=x−1x−1−2=1−x−12
Để C nguyên thì x−12 cũng nguyên
⇒(x−1)∈Ư(2)={1;−1;2;−2}
Do: x≥0 nên x−1≥−1
⇒x−1∈{1;−1;2}
⇒x∈{4;0;9}(đều thỏa mãn ĐKXĐ và x∈Z )