8.some

9.a

10.any

11.an

12.some

13.any

14.a

15.an

16.any

17.some

\(\dfrac{3c-4b}{2}=\dfrac{4a-2c}{3}=\dfrac{2b-3a}{4}\)

=\(\dfrac{2\left(3c-4b\right)}{4}=\dfrac{3\left(4a-2c\right)}{3}=\dfrac{4\left(2b-3a\right)}{16}\)

\(=\dfrac{6c-8b}{4}=\dfrac{12a-6c}{3}=\dfrac{8b-12a}{16}\)

\(=\dfrac{6c-8b+12a-6c+8b-12a}{4+3+16}=0\)

\(\left\{{}\begin{matrix}\dfrac{3c-4b}{2}=0\\\dfrac{4a-2c}{3}=0\\\dfrac{2b-3a}{4}=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}3c=4b\\4a=2c\\2b=3a\end{matrix}\right.\) => \(\left\{{}\begin{matrix}\dfrac{c}{4}=\dfrac{b}{3}\left(1\right)\\\dfrac{c}{4}=\dfrac{a}{2}\left(2\right)\\\dfrac{a}{2}=\dfrac{b}{3}\left(3\right)\end{matrix}\right.\)

Từ (1),(2) và (3) Suy ra : \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)

Theo tính chất dãy tỉ số bằng nhau ta có :

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{29}{4}=\dfrac{c+b-29}{4+3-4}=\dfrac{-27}{3}=-9\)

Từ \(\left\{{}\begin{matrix}\dfrac{a}{2}=-9\\\dfrac{b}{3}=-9\\\dfrac{c}{4}=-9\end{matrix}\right.=>\left\{{}\begin{matrix}a=-18\\b=-27\\c=-36\end{matrix}\right.\)

Vậy a = -18;b = -27;c = -36

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