求代数式的值时,如果代数式中含有同类项,通常先
合并同类项
,再进行计算.
答案：合并同类项
1. 若多项式$x^{2}-3kxy+6xy-8化简后不含xy$项,则$k$的值为 (
A
)
A.2
B.-2
C.0
D.3
答案：A
解析：
解：原式$=x^{2}+(-3k + 6)xy - 8$
因为化简后不含$xy$项，所以$-3k + 6 = 0$
解得$k = 2$
答案：A
2. 若把$x-y$看成一项,合并$2(x-y)^{2}+3(x-y)+5(y-x)^{2}+3(y-x)$得 (
A
)
A.$7(x-y)^{2}$
B.$-3(x-y)^{2}$
C.$-3(x+y)^{2}+6(x-y)$
D.$(y-x)^{2}$
答案：A
解析：
解：
$\begin{aligned}&2(x-y)^{2}+3(x-y)+5(y-x)^{2}+3(y-x)\\=&2(x-y)^{2}+3(x-y)+5(x-y)^{2}-3(x-y)\\=&[2(x-y)^{2}+5(x-y)^{2}]+[3(x-y)-3(x-y)]\\=&7(x-y)^{2}+0\\=&7(x-y)^{2}\end{aligned}$
答案：A
3. 合并同类项:
(1)$\frac{1}{2}x-\frac{1}{3}x-\frac{2}{3}x+1$;
(2)$7m^{2}n-3mn^{2}+5m^{2}n+mn^{2}$.
答案：(1)$-\frac{1}{2}x+1$ (2)$12m^{2}n-2mn^{2}$
解析：
(1)解：原式$=(\frac{1}{2}-\frac{1}{3}-\frac{2}{3})x+1$
$=(\frac{1}{2}-1)x+1$
$=-\frac{1}{2}x+1$
(2)解：原式$=(7m^{2}n+5m^{2}n)+(-3mn^{2}+mn^{2})$
$=12m^{2}n-2mn^{2}$
4. 求下列各式的值:
(1)$4xy-3x^{2}-xy+y^{2}+x^{2}-3xy-2y+2x^{2}$,其中$x= 1\frac{13}{15},y= -1$;
(2)$\frac{1}{2}x^{2}-\frac{1}{4}x+0.2x^{3}+0.25x-0.5x^{2}-\frac{1}{5}x^{3}$,其中$x= \frac{12}{13}$.
答案：(1)3 (2)0
解析：
(1)解：原式$=(4xy - xy - 3xy) + (-3x^2 + x^2 + 2x^2) + y^2 - 2y$
$=0 + 0 + y^2 - 2y$
$=y^2 - 2y$
当$x=1\frac{13}{15}$，$y=-1$时，
原式$=(-1)^2 - 2×(-1)$
$=1 + 2$
$=3$
(2)解：原式$=0.2x^3 - \frac{1}{5}x^3 + \frac{1}{2}x^2 - 0.5x^2 + (-\frac{1}{4}x + 0.25x)$
$=(0.2 - 0.2)x^3 + (\frac{1}{2} - \frac{1}{2})x^2 + (-\frac{1}{4} + \frac{1}{4})x$
$=0 + 0 + 0$
$=0$
5. 把$(a+b)与(x-y)$各当作一个整体,合并下列各式中的同类项:
(1)$-7(a+b)-7(a+b)+6(a+b)$;
(2)$3(x-y)^{2}-(x-y)+(x-y)^{2}+6(x-y)$.
答案：(1)$-8(a+b)$ (2)$4(x-y)^{2}+5(x-y)$
解析：
(1) $-7(a+b)-7(a+b)+6(a+b)$
$=(-7-7+6)(a+b)$
$=-8(a+b)$
(2) $3(x-y)^{2}-(x-y)+(x-y)^{2}+6(x-y)$
$=(3+1)(x-y)^{2}+(-1+6)(x-y)$
$=4(x-y)^{2}+5(x-y)$
