a的平方 b的平方 2a-6b+10=0则b的a次方的值为-搜答案-智慧作业帮

解:9a²-6a+1-b²=(9a²-6a+1)-b²=(3a-1)²-b²=(3a-1+b)(3a-1-b)

a^2-b^2-2b-1=a^2-(b^2+2b+1)=a^2-(b+1)^2=[a+(b+1)]*[a-(b+1)]=(a+b+1)*(a-b-1)

(a^2+b^2-1)^2-4a^2b^2=(a^2+b^2-1-2ab)(a^2+b^2-1+2ab)=[(a-b)^2-1][(a+b)^2-1]=(a-b+1)(a-b-1)(a+b+1)(a+

原式=(a+b)(a-b)/ab(a-b)÷(a+b)²/2ab=(a+b)/ab×2ab/(a+b)²=2/(a+b)∴原式=2/(5-√11-3+√11)=1

原式=2(a^2+b^2)(a+b)^2-(a+b)^2(a-b)^2=(a+b)^2(a^2+2ab+b^2)=(a+b)^4

解(a²+b²)²-2(a²+b²)=8令a²+b²=t则t²-2t-8=0∴(t-4)(t+2)=0∴t=4,或t=-

a平方-b平方+2a+1=(a²+2a+1)-b²=(a+1)²-b²=(a+1+b)(a+1-b)再问：是-2a再答：a²-b²-2a+1

解(a²+b²)×(a²+b²-1)=6令a²+b²=t则t(t-1)=6即t²-t-6=0(t-3)(t+2)=0∴t=3或t=

a的平方-b的平方+2a+1=a²+2a+1-b²=(a+1)²-b²=(a+1+b)(a+1-b)

[(a^2-b^2)/(a^2*b+ab^2)]/[(a^2+b^2)/(2ab)-1]=(a+b)(a-b)/[ab(a+b)]*2ab/(a^2+b^2-2ab)=2(a-b)/(a-b)^2=2

(1)（a+2b)/(a+b)+2b²/(a²-b²)=（a+2b)(a-b)/(a²-b²)+2b²/(a²-b²)=

能不能把原式再确认一下?感觉有问题.

=[(2a+b)+(a+2b)][(2a+b)-(a+2b)]=(3a+3b)(a-b)=3(a+b)(a-b)

(a²+b²-1)²=9(a²+b²-1)²-9=0(a²+b²-1-3)(a²+b²-1+3)=0

原式=[(a-b+2b)/(a-b)]²[(a+b-2b)/(a+b)]²=[(a+b)/(a-b)]²[(a-b)/(a+b)]²=[(a+b)/(a-b)×

(A^2-B^2+2B-1)/(A^2-B^2)+(A+B)=[A^2-(B-1)^2/(A+B)(A-B)]+A+B=(A+B-1)(A-B-1)/(A+B)(A-B+1)=A+B+1/(A+B)=

答案：2008/2009建议改下问题：已知a+b=2009,分子：a^2+b^2+2b-1分母：(a^2-b^2)+a+b求比值

原式=(a+b)(a-b)/(b+1)(b-1)×(b+1)/(a+b)=(a-b)/(b-1)a/(a²-b²)×a/(a-b)=a²/(a³-a²

[(a²-b²)/(a²b-ab²)]÷[1+(a²+b²)/2ab]=[(a+b)(a-b)/ab(a-b)]÷[(2ab+a²

原式=（3a-c）²-4b²=(3a-c+2b)(3a-c-2b)原式=(a²b-1/4b)²