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七年级数学填空题一般
题目
定义:形如关于xxyy的方程x+ky=bx+ky=bkx+y=bkx+y=b的两个方程互为共轭二元一次方程,其中k1k\neq 1;由这两个方程组成的方程组{x+ky=bkx+y=b\left\{\begin{array}{l}{x+ky=b}\\{kx+y=b}\end{array}\right.,叫做共轭方程组.
(1)(1)请写出方程4x+y=34x+y=3的共轭二元一次方程:______;
(2)(2)若方程x+ky=bx+ky=bxxyy的值满足表格:
xx1-122
yy2211
求这个方程的共轭二元一次方程;
(3)(3)若共轭方程组{x+ky=bkx+y=b\left\{\begin{array}{l}{x+ky=b}\\{kx+y=b}\end{array}\right.的解是{x=my=n\left\{\begin{array}{l}{x=m}\\{y=n}\end{array}\right.,请你求出mmnn的数量关系.
知识点:二元一次方程的解、二元一次方程组的解章节:未标注

答案与解析

答案

(1)方程4x+y=34x+y=3k=4k=4b=3b=3,所以它的共轭二元一次方程:x+4y=3x+4y=3
故答案为:x+4y=3x+4y=3
(2)(2)在方程x+ky=bx+ky=b中,当x=1x=-1时,y=2y=2;当x=2x=2时,y=1y=1
所以{1+2k=b2+k=b\left\{\begin{array}{l}{-1+2k=b}\\{2+k=b}\end{array}\right.
解得{k=3b=5\left\{\begin{array}{l}{k=3}\\{b=5}\end{array}\right.
所以方程x+ky=bx+ky=bx+3y=5x+3y=5,它的共轭二元一次方程3x+y=53x+y=5
(3)(3)若共轭方程组{x+ky=bkx+y=b\left\{\begin{array}{l}{x+ky=b}\\{kx+y=b}\end{array}\right.的解是{x=my=n\left\{\begin{array}{l}{x=m}\\{y=n}\end{array}\right.
{m+kn=bkm+n=b\left\{\begin{array}{l}{m+kn=b①}\\{km+n=b②}\end{array}\right.
把①代入②得,km+n=m+knkm+n=m+kn
(k1)m(k1)n=0\therefore \left(k-1\right)m-\left(k-1\right)n=0
(k1)(mn)=0\therefore \left(k-1\right)\left(m-n\right)=0
k1\because k\neq 1
mn=0\therefore m-n=0
m=nm=n.

解析

(1)方程4x+y=34x+y=3k=4k=4b=3b=3,所以它的共轭二元一次方程:x+4y=3x+4y=3
故答案为:x+4y=3x+4y=3
(2)(2)在方程x+ky=bx+ky=b中,当x=1x=-1时,y=2y=2;当x=2x=2时,y=1y=1
所以{1+2k=b2+k=b\left\{\begin{array}{l}{-1+2k=b}\\{2+k=b}\end{array}\right.
解得{k=3b=5\left\{\begin{array}{l}{k=3}\\{b=5}\end{array}\right.
所以方程x+ky=bx+ky=bx+3y=5x+3y=5,它的共轭二元一次方程3x+y=53x+y=5
(3)(3)若共轭方程组{x+ky=bkx+y=b\left\{\begin{array}{l}{x+ky=b}\\{kx+y=b}\end{array}\right.的解是{x=my=n\left\{\begin{array}{l}{x=m}\\{y=n}\end{array}\right.
{m+kn=bkm+n=b\left\{\begin{array}{l}{m+kn=b①}\\{km+n=b②}\end{array}\right.
把①代入②得,km+n=m+knkm+n=m+kn
(k1)m(k1)n=0\therefore \left(k-1\right)m-\left(k-1\right)n=0
(k1)(mn)=0\therefore \left(k-1\right)\left(m-n\right)=0
k1\because k\neq 1
mn=0\therefore m-n=0
m=nm=n.

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