《2025年通城学典活页检测七年级数学下册北师大版》

3. 先化简,再求值:$(x - 2y)(x + y)-(x - 2y)(x + 2y)-x(x - y)$,其中$x=\frac{1}{3}$,$y = -1$.

答案：
$(x - 2y)(x + y)-(x - 2y)(x + 2y)-x(x - y)=x^{2}+xy - 2xy - 2y^{2}-(x^{2}-4y^{2})-(x^{2}-xy)=x^{2}+xy - 2xy - 2y^{2}-x^{2}+4y^{2}-x^{2}+xy=-x^{2}+2y^{2}$.当$x=\frac{1}{3}$,$y = - 1$时,原式$=-(\frac{1}{3})^{2}+2\times(-1)^{2}=-\frac{1}{9}+2=\frac{17}{9}$

4. ★已知$a = (\frac{1}{4})^{-1}$,$b = -(-\frac{1}{4})^{2}$,求代数式$a(a + 2b)-(a + 1)^{2}+2a$的值.

答案： $a(a + 2b)-(a + 1)^{2}+2a=a^{2}+2ab - a^{2}-2a - 1+2a=2ab - 1$.当$a = (\frac{1}{4})^{-1}=4$,$b=-(-\frac{1}{4})^{2}=-\frac{1}{16}$时,原式$=2\times4\times(-\frac{1}{16})-1=-\frac{1}{2}-1=-\frac{3}{2}$

5. ★先化简,再求值:$[(2x - y)^{2}-(2x - y)(y + 2x)-4xy]\div2y$,其中$x$,$y$满足$(x + 1)^{2}+|y - 2| = 0$.

答案： $[(2x - y)^{2}-(2x - y)(y + 2x)-4xy]\div2y=[(4x^{2}-4xy + y^{2})-(4x^{2}-y^{2})-4xy]\div2y=[4x^{2}-4xy + y^{2}-4x^{2}+y^{2}-4xy]\div2y=(2y^{2}-8xy)\div2y=y - 4x$.因为$x$,$y$满足$(x + 1)^{2}+\vert y - 2\vert = 0$,所以$x = -1$,$y = 2$.所以原式$=2 - 4\times(-1)=6$

6. 试说明代数式$[(x - y)^{2}-(x + y)(x - y)]\div(-2y)+y$的值与$y$的值无关.

答案： $[(x - y)^{2}-(x + y)(x - y)]\div(-2y)+y=[x^{2}-2xy + y^{2}-(x^{2}-y^{2})]\div(-2y)+y=(x^{2}-2xy + y^{2}-x^{2}+y^{2})\div(-2y)+y=(-2xy + 2y^{2})\div(-2y)+y=x - y + y=x$.所以代数式$[(x - y)^{2}-(x + y)(x - y)]\div(-2y)+y$的值与$y$的值无关

7. ★小亮在做“先化简,再求值:$(2x + k)(3x + 2)-6x(x + 3)+5x + 16$,其中$x = 2$”一题时,错将$x = 2$看成$x = -2$,但结果却和正确答案一样,由此,你能推算出$k$值吗？

答案： $(2x + k)(3x + 2)-6x(x + 3)+5x + 16=6x^{2}+4x + 3kx + 2k-6x^{2}-18x + 5x + 16=(3k - 9)x + 2k + 16$.由题意,得$3k - 9 = 0$,解得$k = 3$