√3(x²-x)-√2(x²+x)=0
√3x(x-1)-√2x(x+1)=0
x(√3x-√3-√2x-√2)=0
x[(√3-√2)x-(√3+√2)]=0
所以
x1=0
x2=(√3+√2)/(√3-√2)
=(√3+√2)²/[(√3-√2)(√3+√2)]
=(5+2√6)
√3(x²-x)-√2(x²+x)=0
√3x(x-1)-√2x(x+1)=0
x(√3x-√3-√2x-√2)=0
x[(√3-√2)x-(√3+√2)]=0
所以
x1=0
x2=(√3+√2)/(√3-√2)
=(√3+√2)²/[(√3-√2)(√3+√2)]
=(5+2√6)