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1. (2025·苏州期中)我们规定一种新运算“$\otimes$”:对于任意有理数$m$和$n$,规定$m\otimes n = mn^2 - mn + n$,如:$1\otimes 3 = 1× 3^2 - 1× 3 + 3 = 9$.
(1)求$3\otimes (-2)$的值;
(2)化简$(1 - x)\otimes (-1)$;
(3)若$A = 3\otimes x$,$B = (1 - x)\otimes (-1) - 2$,比较$A$与$B$的大小,并说明理由.
(1)求$3\otimes (-2)$的值;
(2)化简$(1 - x)\otimes (-1)$;
(3)若$A = 3\otimes x$,$B = (1 - x)\otimes (-1) - 2$,比较$A$与$B$的大小,并说明理由.
答案
(1) $3\otimes(-2) = 3×(-2)^2 - 3×(-2) - 2 = 3×4 + 6 - 2 = 12+6-2=16$.
(2) $(1-x)\otimes(-1) = (1-x)×(-1)^2 - (1-x)×(-1) -1 = 1-x+1-x-1=-2x+1$.
(3) $A＞B$,理由:$A=3\otimes x=3x^2-3x+x=3x^2-2x$,$B=(1-x)\otimes(-1)-2=-2x+1-2=-2x-1$,$A-B=3x^2-2x-(-2x-1)=3x^2+1＞0$,故$A＞B$.
(2) $(1-x)\otimes(-1) = (1-x)×(-1)^2 - (1-x)×(-1) -1 = 1-x+1-x-1=-2x+1$.
(3) $A＞B$,理由:$A=3\otimes x=3x^2-3x+x=3x^2-2x$,$B=(1-x)\otimes(-1)-2=-2x+1-2=-2x-1$,$A-B=3x^2-2x-(-2x-1)=3x^2+1＞0$,故$A＞B$.
2. (2025·宿迁校级期中)数学上,我们把$\begin{vmatrix} a&b \\ c&d \end{vmatrix}$称作二阶行列式,规定它的运算法则为$\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc$,例如$\begin{vmatrix} 2&3 \\ 4&5 \end{vmatrix}=2×5-3×4=-2$,请根据上述材料解答下列各题:
(1)$\begin{vmatrix} -6&4 \\ 3&\dfrac{1}{2} \end{vmatrix}=$ ______ ;
(2)计算:$\begin{vmatrix} 1&2 \\ 3&4 \end{vmatrix}+\begin{vmatrix} 5&6 \\ 7&8 \end{vmatrix}+\dots+\begin{vmatrix} 97&98 \\ 99&100 \end{vmatrix}$;
(3)已知$a,b$满足行列式$\begin{vmatrix} a&-1 \\ -a^2+b&a-1 \end{vmatrix}=5$,求代数式$\dfrac{5a-3b-4ab}{2}+2ab-b+2$的值.
(1)$\begin{vmatrix} -6&4 \\ 3&\dfrac{1}{2} \end{vmatrix}=$ ______ ;
(2)计算:$\begin{vmatrix} 1&2 \\ 3&4 \end{vmatrix}+\begin{vmatrix} 5&6 \\ 7&8 \end{vmatrix}+\dots+\begin{vmatrix} 97&98 \\ 99&100 \end{vmatrix}$;
(3)已知$a,b$满足行列式$\begin{vmatrix} a&-1 \\ -a^2+b&a-1 \end{vmatrix}=5$,求代数式$\dfrac{5a-3b-4ab}{2}+2ab-b+2$的值.
答案
(1) -15
(2) $\begin{vmatrix} 1&2 \\ 3&4 \end{vmatrix}+\begin{vmatrix} 5&6 \\ 7&8 \end{vmatrix}+\dots+\begin{vmatrix} 97&98 \\ 99&100 \end{vmatrix}=(1×4-2×3)+(5×8-6×7)+\dots+(97×100-98×99)=(-2)+(-2)+\dots+(-2)=(-2)×25=-50$.
(3) 因为$\begin{vmatrix} a&-1 \\ -a^2+b&a-1 \end{vmatrix}=5$,所以$a(a-1)-(-1)×(-a^2+b)=5$,即$b-a=5$,所以$\frac{5a-3b-4ab}{2}+2ab-b+2=\frac{5a-3b-4ab}{2}+\frac{4ab-2b+4}{2}=\frac{5a-5b+4}{2}=\frac{5}{2}(a-b)+2=-\frac{5}{2}(b-a)+2=-\frac{5}{2}×5+2=-\frac{21}{2}$.
(2) $\begin{vmatrix} 1&2 \\ 3&4 \end{vmatrix}+\begin{vmatrix} 5&6 \\ 7&8 \end{vmatrix}+\dots+\begin{vmatrix} 97&98 \\ 99&100 \end{vmatrix}=(1×4-2×3)+(5×8-6×7)+\dots+(97×100-98×99)=(-2)+(-2)+\dots+(-2)=(-2)×25=-50$.
(3) 因为$\begin{vmatrix} a&-1 \\ -a^2+b&a-1 \end{vmatrix}=5$,所以$a(a-1)-(-1)×(-a^2+b)=5$,即$b-a=5$,所以$\frac{5a-3b-4ab}{2}+2ab-b+2=\frac{5a-3b-4ab}{2}+\frac{4ab-2b+4}{2}=\frac{5a-5b+4}{2}=\frac{5}{2}(a-b)+2=-\frac{5}{2}(b-a)+2=-\frac{5}{2}×5+2=-\frac{21}{2}$.
