解下列一元二次方程 x^-49=0 2y^2=128 2x^-1/2=0 (3y-7)^2=3 7-2x^2=-15 t^2-45=0 (x+5)^2=16

题目

解下列一元二次方程 x^-49=0 2y^2=128 2x^-1/2=0 (3y-7)^2=3 7-2x^2=-15 t^2-45=0 (x+5)^2=16

答案

答：x^-49=0x²=49x²=7²x=7或者x=-72y^2=128y²=64y²=8²y=8或者y=-82x^-1/2=02x²=1/2x²=1/4x²=(1/2)²x=1/2或者x=-1/2 (3y-7)^2=33y-7=√3或者3y-7=-√3y=(7+√3)/3...