已知a,b满足|a+1/5|+(a+b+7/10)^2=0.求(4a-b)^2-[8a(1-a)-(3a-b)(3a+b)]的值.

来源:学生作业帮助网 编辑:作业帮 时间:2024/10/03 06:45:37
已知a,b满足|a+1/5|+(a+b+7/10)^2=0.求(4a-b)^2-[8a(1-a)-(3a-b)(3a+b)]的值.
x){}Ku^bDmC}mD$ms}C8#[g4Lui="}_`gC>l[̋募h UHQF`qFҩ~OH=jʆ@CDM`6Rm,PeQdv% l Ag`1/!A~OZ @G '?m]w]o\

已知a,b满足|a+1/5|+(a+b+7/10)^2=0.求(4a-b)^2-[8a(1-a)-(3a-b)(3a+b)]的值.
已知a,b满足|a+1/5|+(a+b+7/10)^2=0.求(4a-b)^2-[8a(1-a)-(3a-b)(3a+b)]的值.

已知a,b满足|a+1/5|+(a+b+7/10)^2=0.求(4a-b)^2-[8a(1-a)-(3a-b)(3a+b)]的值.
由a+1/5=0 a+b+7/10=0
解得a=-1/5 b=-1/2
(4a-b)^2-[8a(1-a)-(3a-b)(3a+b)]
=[4×(-1/5)-(-1/2)]²-[8×(-1/5)×(1+1/5)-(3×(-1/5)-(-1/2))(3×(-1/5)+(-1/2))]
=131/50

|a+1/5|+(a+b+7/10)^2=0得a+1/5=0,a+b+7/10=0
a=-1/5,b=-1/2
代入即可求解