第1个回答 2022-10-19
1.(x²+2x)(x²+2x-2)-3 =(x²+2x)[(x²+2x)-2]-3 =(x²+2x)^2-2(x²+2x)-3 =(x²+2x-3)(x²+2x+1) =(x-1)(x+3)(x+1)² 2.(a+b)²-4(a+b-1)=(a+b)²-4(a+b)+4=(a+b-2)² 3.-4x²+12xy-9y²=-(4x²-12xy+9y²)=-(2x-3y)² 4.a³+b³+c³-3abc=(a+b)³-3(a²b+ab²)+c³-3abc=(a+b+c)(a²-ab+b²)-3ab(a+b+c) =(a+b+c)(a²-4ab+b²) 5.8x²y-8xy+2y =2y(4x²-4x+1)=2y(2x-1)² 6.-9x²+6x-1 =-(9x²-6x+1)=-(3x-1)² 7.(2x-3y)³+(3x-2y)³-125(x-y)³ =(2x-3y+3x-2y)[(2x-3y)²-(2x-3y)(3x-2y)+(3x-2y)²]-125(x-y)³ =5(x-y)(7x²-11xy+7y²)-125(x-y)³=5(x-y)(7x²-11xy+7y²-25x²+50xy-25y²) =-5(x-y)(18x²-39xy+18y²) 8.x²-4y²-9z²-12yz=x²-(4y²+9z²+12yz)=x²-(2y+3z)²=(x+2y+3z)(x-2y-3z) 9.(ab+1)(a+1)(b+1)+ab =(ab+1)(a+1)(b+1)+(ab+1)-1 =(ab+1)[(a+1)(b+1)-1]=(ab+1)(ab+a+b-1) 10.(a²+b²)xy+ab(x²+y²)=a²xy+b²xy+abx²+aby²=(a²xy+abx²)+(b²xy+aby²) =ax(ay+bx)+by(bx+ay)=(ax+by)(ay+bx) ∵(x+y)(a+b)=ax+bx+ay+by=(ax+by)+(ay+bx)=2×2=4 ∴ay+bx=4/(ax+by)=4/5 ∴(a²+b²)xy+ab(x²+y²)=5×4/5=4 11.∵x³+ax²+bx+8有两个因式x+1,x+2 ∴第三个因式为一次式,且系数为1,常数项为4,即x+4 ∴x³+ax²+bx+8=(x+1)(x+2)(x+4)=x³+7x²+11x+8 ∴a=7,b=14 ∴a+b=21 故选 D 12.2x³+x²-13x+6=(2x³+x²-1)-(12x-6)=x(x+1)(2x-1)-6(2x-1) =(2x-1)(x²+x-6)=(2x-1)(x-2)(x+3) 故选 A 13.x³-7x+6=(x³-x)-(6x-6)=x(x+1)(x-1)-6(x-1)=(x-1)(x²+x-6) =(x-1)(x-2)(x+3)