Question

已知ab是实数,且（根号2a+b）+ |b-根号2|=0 解关于x的方程（a+2)x+b的平方=a-1

Answer 1

∵（根号2a+b）+ |b-根号2|=0

∴2a+b=0 b-√2=0

a=-1/2√2 b=√2

（a+2)x+b的平方=a-1

(-1/2√2+2)x+(√2)²=-1/2√2-1

(-1/2√2+2)x=-1/2√2-5

x=(-1/2√2-5)/(-1/2√2+2)

x=(5+1/2√2)(2+1/2√2)/(2-1/2√2)(2+1/2√2)

x=(10+√2+5/2√2+1/2)/(4-1/2)

x=(21/2+7/2√2)/(7/2)

x=3+√2