电子课本网 第86页

第86页

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解:因为关于​$x,y$​的二元一次方程组
​$\begin {cases}4x - y = 5\\ax + by = - 1\end {cases}$​与​$\begin {cases}3x + y = 9\\3ax - 4by = 18\end {cases}$​有公共解,
所以​$\begin {cases}4x - y = 5\\3x + y = 9\end {cases}$​的解即为两个方程组的公共解。
解得​$\begin {cases}x = 2\\y = 3\end {cases}。$​
​$ $​把​$\begin {cases}x = 2\\y = 3\end {cases}$​代入​$\begin {cases}ax + by = - 1\\3ax - 4by = 18\end {cases}$​得
​$\begin {cases}2a + 3b = - 1\\6a - 12b = 18\end {cases},$​解得​$\begin {cases}{a = 1}\\{b=-1}\end {cases}$​
​$ $​所以​$a^2+b^2-2ab=1 + 1-2×1×(-1)=4。$​
解:因为关于​$x,y$​的方程组​$\begin {cases}4x + my = 16\\2x + y = 4n + 2\end {cases}$​和
​$\begin {cases}3x + my = 13\\2x - 3y = - 6\end {cases}$​的解相同,
所以​$\begin {cases}4x + my = 16\\3x + my = 13\end {cases},$​
两式相减得:​$x = 3。$​
把​$x = 3$​代入​$2x - 3y = - 6$​得:​$6 - 3y = - 6,$​
解得​$y = 4。$​
​$ $​把​$\begin {cases}x = 3\\y = 4\end {cases}$​代入​$\begin {cases}2x + y = 4n + 2\\4x + my = 16\end {cases}$​
得​$\begin {cases}6 + 4 = 4n + 2\\12 + 4m = 16\end {cases},$​解得​$\begin {cases}{m = 1}\\{n = 2}\end {cases}$​
解:因为甲同学看错​$a$​的符号,
所以把​$x = 3,y = 2$​代入​$x + cy = 4$​得:
​$3 + 2c = 4,$​解得​$c=\frac {1}{2}。$​
​$ $​且由方程组的第一个方程可得​$-3a + 2b = 6。$​
​$ $​因为乙同学看漏​$c,$​
所以把​$x = 6,y = - 2$​代入​$ax + by = 6$​得
​$6a-2b = 6。$​
​$ $​联立​$\begin {cases}-3a + 2b = 6\\6a - 2b = 6\end {cases},$​
两式相加得:​$3a = 12,$​解得​$a = 4。$​
​$ $​把​$a = 4$​代入​$-3a + 2b = 6$​得:​$-12 + 2b = 6,$​
解得​$b = 9。$​
​$ $​所以​$2a - b + 4c=2×4-9 + 4×\frac {1}{2}=8 - 9 + 2 = 1。$​
解:​$ (1)\begin {cases}3x + 5y = 2m&①\\x + y = m - 1&②\end {cases}$​
由① - ②得:​$2(x + 2y)=m + 1。$​
​$ $​因为​$x + 2y = 2,$​
所以​$2×2=m + 1,$​解得​$m = 3。$​
​$ (2)$​因为​$a\geqslant m,$​即​$a\geqslant 3,$​
所以​$a + 1>0,$​​$2 - a<0。$​
​$ $​所以​$\vert a + 1\vert -\vert 2 - a\vert =a + 1-(a - 2)$​
​$=a + 1 - a + 2 = 3。$​