9. 易错题 计算：
(1) $(-3)^2 - (1\dfrac{1}{2})^3 × \dfrac{2}{9} - 6 ÷ \left| -\dfrac{2}{3} \right|^3$.
(2) $(\dfrac{1}{3} - \dfrac{1}{5}) × (-15) ÷ \left| -\dfrac{1}{3} \right| + 1^2 - 2 × (-0.125) × 8$.
(3) $3 × [5^2 - 6 + (-8)^2 - 2 × (-2)^3 × \dfrac{1}{4}] ÷ (-3)^3$.
(4) $[-2^4 ÷ (-2\dfrac{2}{3})^2 + 5\dfrac{1}{2} × (-\dfrac{1}{6}) - \dfrac{1}{4}] ÷ \dfrac{1}{12}$.

(1) 原式$=9-\frac{27}{8}×\frac{2}{9}-6÷\frac{8}{27}=9-\frac{3}{4}-6×\frac{27}{8}=9-\frac{3}{4}-\frac{81}{4}=-12.$
(2) 原式$=(\frac{1}{5}×15-\frac{1}{3}×15)÷\frac{1}{3}+1-2×(-0.125×8)=(-2)×3+1-(-2)=-6+1+2=-3.$
(3) 原式$=3×[25-6+64-2×(-8)×\frac{1}{4}]×(-\frac{1}{27})=3×(25-6+64+4)×(-\frac{1}{27})=-\frac{29}{3}.$
(4) 原式$=(-16×\frac{9}{64}-\frac{11}{2}×\frac{1}{6}-\frac{1}{4})×12=(-\frac{9}{4}-\frac{11}{12}-\frac{1}{4})×12=-27-11-3=-41.$

10. 阅读材料：
求$1+2+2^2+2^3+2^4+\dots+2^{2025}$的值.
解:设$S=1+2+2^2+2^3+2^4+\dots+2^{2025}$①.
将等式两边同时乘2,得$2S=2+2^2+2^3+2^4+2^5+\dots+2^{2025}+2^{2026}$②.
由②$-$①,得$2S-S=2^{2026}-1$,即$S=2^{2026}-1$.
所以$1+2+2^2+2^3+2^4+\dots+2^{2025}=2^{2026}-1$.
请你仿照此方法计算:
(1) $1+2+2^2+2^3+\dots+2^9=$.
(2) $1+7+7^2+7^3+7^4+\dots+7^n$($n$为正整数).
(3) $1+2×2+3×2^2+4×2^3+\dots+9×2^8+10×2^9$.

(1) $2^{10}-1.$
(2) 设$S=1+7+7^2+7^3+7^4+\dots+7^n$①.将等式两边同时乘7,得$7S=7+7^2+7^3+7^4+7^5+\dots+7^n+7^{n+1}$②.由②$-$①,得$7S-S=7^{n+1}-1$,即$S=\frac{7^{n+1}-1}{6}.$所以$1+7+7^2+7^3+7^4+\dots+7^n=\frac{7^{n+1}-1}{6}.$
(3) 设$S=1+2×2+3×2^2+4×2^3+\dots+9×2^8+10×2^9$①.将等式两边同时乘2,得$2S=2+2×2^2+3×2^3+4×2^4+\dots+9×2^9+10×2^{10}$②.由①$-$②,得$S-2S=1+2+2^2+2^3+\dots+2^9-10×2^{10}.$由(1),得$1+2+2^2+2^3+\dots+2^9=2^{10}-1.$所以$-S=2^{10}-1-10×2^{10}$,即$S=9×2^{10}+1.$所以$1+2×2+3×2^2+4×2^3+\dots+9×2^8+10×2^9=9×2^{10}+1.$