11. 用“☆”定义一种新运算:对于任意有理数a和b,规定$a☆b= ab^{2}-2ab+b$.如:$2☆(-3)= 2×(-3)^{2}-2×2×(-3)+(-3)= 27$.
(1)求$(-4)☆7$的值;
(2)若$(1-3x)☆(-4)= 32$,求x的值.
(1)求$(-4)☆7$的值;
(2)若$(1-3x)☆(-4)= 32$,求x的值.
答案
(1)$(-4)☆7=(-4)×7^{2}-2×(-4)×7+7=-133$
(2)根据题意,得$(1-3x)×(-4)^{2}-2×(1-3x)×(-4)+(-4)=32$.整理,得$16(1-3x)+8(1-3x)-4=32$,解得$x=-\frac{1}{6}$
(2)根据题意,得$(1-3x)×(-4)^{2}-2×(1-3x)×(-4)+(-4)=32$.整理,得$16(1-3x)+8(1-3x)-4=32$,解得$x=-\frac{1}{6}$
12. (新考法·新定义题)阅读理解:a,b,c,d是有理数,我们把符号$\begin{vmatrix} a&b\\ c&d\end{vmatrix} $称为二阶行列式,并且规定:$\begin{vmatrix} a&b\\ c&d\end{vmatrix} =ad-bc$.例如$\begin{vmatrix} 2&3\\ 4&5\end{vmatrix} =2×5-3×4= 10-12= -2$.请你按照这种规定,解答下列各题:
(1)求$\begin{vmatrix} -1&-3\\ 2&7\end{vmatrix} $的值;
(2)求x的值,使得$\begin{vmatrix} \frac {x}{2}&\frac {x+1}{3}\\ 2&1\end{vmatrix} =-2$;
(3)求x的值,使得$\begin{vmatrix} x-1&-2\\ 3&-5\end{vmatrix} =\begin{vmatrix} x&4-2x\\ -3&7\end{vmatrix} $.
(1)求$\begin{vmatrix} -1&-3\\ 2&7\end{vmatrix} $的值;
(2)求x的值,使得$\begin{vmatrix} \frac {x}{2}&\frac {x+1}{3}\\ 2&1\end{vmatrix} =-2$;
(3)求x的值,使得$\begin{vmatrix} x-1&-2\\ 3&-5\end{vmatrix} =\begin{vmatrix} x&4-2x\\ -3&7\end{vmatrix} $.
答案
(1)$\begin{vmatrix}-1&-3\\2&7\end{vmatrix}=(-1)×7-(-3)×2=-7-(-6)=-1$
(2)根据题意,得$\frac{x}{2}×1-\frac{x+1}{3}×2=-2$.去分母,得$3x-4(x+1)=-12$.去括号,得$3x-4x-4=-12$.移项、合并同类项,得$-x=-8$.系数化为$1$,得$x=8$
(3)根据题意,得$-5(x-1)-(-2)×3=7x-(-3)(4-2x)$,即$-5(x-1)+6=7x+3(4-2x)$.去括号,得$-5x+5+6=7x+12-6x$.移项、合并同类项,得$-6x=1$.系数化为$1$,得$x=-\frac{1}{6}$
(2)根据题意,得$\frac{x}{2}×1-\frac{x+1}{3}×2=-2$.去分母,得$3x-4(x+1)=-12$.去括号,得$3x-4x-4=-12$.移项、合并同类项,得$-x=-8$.系数化为$1$,得$x=8$
(3)根据题意,得$-5(x-1)-(-2)×3=7x-(-3)(4-2x)$,即$-5(x-1)+6=7x+3(4-2x)$.去括号,得$-5x+5+6=7x+12-6x$.移项、合并同类项,得$-6x=1$.系数化为$1$,得$x=-\frac{1}{6}$
