已知x=(√3+√2)/(√3-√2),y=(√3-√2)/(√3+√2),则代数式x²-3xy+y²

问题描述:

已知x=(√3+√2)/(√3-√2),y=(√3-√2)/(√3+√2),则代数式x²-3xy+y²

x²-3xy+y²
=(x-y)²-xy
=(2√3)²-1
=11

不懂可以追问!

x=(√3+√2)÷(√3-√2)=(√3+√2)^2=5+2√6
y=(√3-√2)÷(√3+√2)=(√3-√2)^2=5-2√6
x+y=10
xy=1x²-3xy+y²=(x+y)²-5xy=10²-5x1=95

x=(√3+√2)/(√3-√2)=(√3+√2)²/(3-2)=(3+2√6+2)/1=5+2√6
y=(√3-√2)/(√3+√2))=(√3-√2)²/(3-2)=(3-2√6+2)/1=5-2√6
xy=(√3+√2)/(√3-√2)* (√3-√2)/(√3+√2)=1
x²-3xy+y²
=(x²+2xy+y²)-5xy
=(x+y)²-5*1
=(5+2√6+5-2√6)²-5
=10²-5
=95

先分母有理化得x=5+2√6,y=5-2√6
x+y=10,xy=1
x²-3xy+y²=(x+y)²-5xy=100-5=95