《2025年运算升级卡七年级数学上册人教版》

1.$\frac {1}{4}ab^{2}-5a^{2}b-\frac {3}{4}a^{3}b+0.75ab^{2}$

答案： $ab^{2}-5a^{2}b-\frac{3}{4}a^{3}b$
解析：$\frac{1}{4}ab^{2}-5a^{2}b-\frac{3}{4}a^{3}b + 0.75ab^{2}=\frac{1}{4}ab^{2}+\frac{3}{4}ab^{2}-5a^{2}b-\frac{3}{4}a^{3}b=(\frac{1}{4}+\frac{3}{4})ab^{2}-5a^{2}b-\frac{3}{4}a^{3}b=ab^{2}-5a^{2}b-\frac{3}{4}a^{3}b$

2.$\frac {2}{3}a^{2}-8a-\frac {1}{2}+6a-\frac {2}{3}a^{2}+\frac {3}{2}$

答案： $-2a + 1$
解析：$\frac{2}{3}a^{2}-8a-\frac{1}{2}+6a-\frac{2}{3}a^{2}+\frac{3}{2}=(\frac{2}{3}a^{2}-\frac{2}{3}a^{2})+(-8a + 6a)+(-\frac{1}{2}+\frac{3}{2})=0 - 2a + 1=-2a + 1$

3.$(-\frac {1}{3}xy)+(-\frac {2}{5}x^{2})-\frac {1}{2}x^{2}-(-\frac {1}{6}xy)$

答案： $-\frac{1}{6}xy-\frac{9}{10}x^{2}$
解析：$(-\frac{1}{3}xy)+(-\frac{2}{5}x^{2})-\frac{1}{2}x^{2}-(-\frac{1}{6}xy)=-\frac{1}{3}xy-\frac{2}{5}x^{2}-\frac{1}{2}x^{2}+\frac{1}{6}xy=(-\frac{1}{3}xy+\frac{1}{6}xy)+(-\frac{2}{5}x^{2}-\frac{1}{2}x^{2})=(-\frac{2}{6}xy+\frac{1}{6}xy)+(-\frac{4}{10}x^{2}-\frac{5}{10}x^{2})=-\frac{1}{6}xy-\frac{9}{10}x^{2}$

4.$4(a - b)^{2}-2(a - b)+5(a - b)+3(a - b)^{2}$

答案： $7(a - b)^{2}+3(a - b)$
解析：$4(a - b)^{2}-2(a - b)+5(a - b)+3(a - b)^{2}=(4(a - b)^{2}+3(a - b)^{2})+(-2(a - b)+5(a - b))=7(a - b)^{2}+3(a - b)$

5.$3(a^{2}-2ab)-(-5ab + 3a^{2}-1)$

答案： $-ab + 1$
解析：$3(a^{2}-2ab)-(-5ab + 3a^{2}-1)=3a^{2}-6ab + 5ab - 3a^{2}+1=(3a^{2}-3a^{2})+(-6ab + 5ab)+1=0 - ab + 1=-ab + 1$