《2025年名校题库七年级数学上册》

3. (实外)
(1)先化简,再求值:$2(x^{2}y+xy)-3(x^{2}y-xy)-4x^{2}y$,其中$x= 1,y= -1;$
(2)已知$(a+2)^{2}+|a+b+5|= 0$,求$3a^{2}b-[2a^{2}b-(2ab-a^{2}b)-4a^{2}]-ab$的值.

答案： 解:
(1)原式=2x²y+2xy-3x²y+3xy-4x²y=-5x²y+5xy.当x=1,y=-1时,原式=-5×1²×(-1)+5×1×(-1)=0.
(2)原式=3a²b-(2a²b-2ab+a²b-4a²)-ab=3a²b-3a²b+2ab+4a²-ab=4a²+ab.由题意,得a=-2,b=-3,
∴原式=4×(-2)²+(-2)×(-3)=16+6=22.

4. (青羊区期末)先化简,再求值:$2x^{2}y+[8xy-2(3xy-2x^{2}y)-xy]$,其中$x= -1,y= 2.$

答案： 解:原式=2x²y+8xy-6xy+4x²y-xy=(2x²y+4x²y)+(8xy-xy-6xy)=6x²y+xy.当x=-1,y=2时,原式=6×(-1)²×2+(-1)×2=12-2=10.

5. (石室联中)先化简,再求值:$\frac {1}{3}x^{3}+(2x^{2}y+3xy^{2}-6)-3(\frac {1}{9}x^{3}+x^{2}y+xy^{2})$,其中$x,y满足|x+2|+(y-1)^{2}= 0.$

答案： 解:原式=$\frac{1}{3}x^{3}+2x^{2}y+3xy^{2}-6-\frac{1}{3}x^{3}-3x^{2}y-3xy^{2}=-x^{2}y-6$.
∵|x+2|+(y-1)²=0,
∴x+2=0,y-1=0,
∴x=-2,y=1,则原式=-(-2)²×1-6=-10.

6. (成外)按要求完成下列各题.
(1)已知$A= 4x^{2}-4xy+y^{2},B= x^{2}+xy-5y^{2}$,化简$A-2B;$
(2)先化简,再求值:$2(ab^{2}-2a^{2}b)-3(ab^{2}-a^{2}b)+(2ab^{2}-2a^{2}b)$,其中$a= 2,b= 1.$

答案： 解:
(1)A-2B=(4x²-4xy+y²)-2(x²+xy-5y²)=4x²-4xy+y²-2x²-2xy+10y²=4x²-2x²-4xy-2xy+10y²+y²=2x²-6xy+11y².
(2)原式=2ab²-4a²b-3ab²+3a²b+2ab²-2a²b=ab²+( -4a²b+3a²b-2a²b )=ab² -3a²b=ab²-3a²b,当a=8,b=9时,a=8,b=9原式=8×9²-8×8²×9=8×81-8×64×9=8×89-8×576=8×(89-72)=8×(81-576)=8×(-495)=-8960.当a=8,b=9时,a=8,b=9原式=8×9²-(8×8²)×9=8×81-(8×64)×9=8×81-8×(64×9)=8×81-8×
(576)=8×(81-576)=8×(-495)=-8960.

7. (龙泉驿区期中)已知$A= 2x^{2}-x+y-4xy,B= x^{2}-2x-y-xy+3$,若$(x+y-2)^{2}+|xy+1|= 0$,求$3A-2(A+B)$的值.

答案： 解:
∵(x+y-2)²+|xy+1|=0,
∴x+y-2=0,xy+1=0,
∴x+y=2,xy=-1.
∵A=2x²-x+y-4xy,B=x²-2x-y-xy+3,
∴3A-2(A+B)=3A-2A-2B=A-2B=2x²-x+y-4xy-2(x²-2x-y-xy+3)=2x²-x+y-4xy-2x²+4x+2y+2xy-6=3x+3y-2xy-6=3(x+y)-2xy-6=3×2-2×(-1)-6=6+2-6=2.