笔触:

【24-25学年闵行区六下期末多校联考券1 第27题】对于有理数xxyy,定义新运算:xy=ax+byx*y = ax + byxy=axbyx \otimes y = ax - by,其中aabb是常数。 例如:32=3a+2b3*2 = 3a + 2b21=2ab2 \otimes 1 = 2a - b

已知32=13*2 = -121=42 \otimes 1 = 4,则根据定义可以得到:

{3a+2b=12ab=4\begin{cases} 3a + 2b = -1 \\ 2a - b = 4 \end{cases}

(1) a=a = \underline{\quad\quad}b=b = \underline{\quad\quad}

(2) 若xy+x2y=6x*y + x \otimes 2y = 6,求x+yx + y的值;

(3) 若关于xxyy的方程组

{xy=2m4xy=8m\begin{cases} x*y = 2m - 4 \\ x \otimes y = 8m \end{cases}

的解也满足方程xy=4x - y = 4,求mm的值;

(4) 若关于xxyy的方程组

{a1xb1y=c1a2xb2y=c2\begin{cases} a_1x * b_1y = c_1 \\ a_2x \otimes b_2y = c_2 \end{cases}

的解为{x=6y=15\begin{cases} x = 6 \\ y = 15 \end{cases},则关于xxyy的方程组

{3a1(2xy)5b1(x+2y)=c13a2(2xy)5b2(x+2y)=c2\begin{cases} 3a_1(2x - y) * 5b_1(x + 2y) = c_1 \\ 3a_2(2x - y) \otimes 5b_2(x + 2y) = c_2 \end{cases}

的解为\underline{\quad\quad}

解:

{a=1b=2,\because \begin{cases} a=1 \\ b=-2 \end{cases},

xy=x2y, xy=x+2y,\therefore x^{*}y = x - 2y,\ x \otimes y = x + 2y,

xy=x2y, x2y=x+4y,\therefore x^{*}y = x - 2y,\ x \otimes 2y = x + 4y,

xy+x2y=6,\because x^{*}y + x \otimes 2y = 6,

x2y+x+4y=6,\therefore x - 2y + x + 4y = 6,

解得

x+y=3;x + y = 3;

解:

{x2y=2m4x+2y=8m,\therefore \begin{cases} x - 2y = 2m - 4 ①\\ x + 2y = 8m ② \end{cases},

x=y+4x = y+4代入得到

{y+4=2m43y+4=8m\therefore \begin{cases} -y + 4 = 2m - 4 ③\\ 3y + 4 = 8m ④ \end{cases}

×3+③ \times 3 + ④得到16=14m1216 = 14m - 12,得到m=2m = 2

×3+① \times 3 + ②得到4(xy)=14m124(x-y) = 14m - 12

{3(2xy)=65(x+2y)=15\begin{cases} 3(2x-y) = 6 \\ 5(x+2y)=15 \end{cases}

解得:

{x=75y=45\begin{cases} x = \dfrac{7}{5} \\ y = \dfrac{4}{5} \end{cases}