Question

已知:x=√3+√2/√3-√2,y=√3-√2/√3+√2,求x三次方-xy二次方/x四次方y+2x三次方y二次方+x二次方y三次

方

Answer 1

x=(√3+√2)/(√3-√2)
=(√3+√2)²/[(√3+√2)(√3-√2)]
=5+2√6
y=(√3-√2)/(√3+√2)
=(√3-√2)²/[(√3-√2)(√3+√2)]
=5-2√6
∴x-y=4√6
x+y=10
xy=1

(x³-xy²)/(x^4y+2x³y²+x²y³)
=[x(x²-y²)]/[x²y(x²+2xy+y²)
=[x(x+y)(x-y)]/[x²y(x+y)²]
=(x-y)/[xy(x+y)]
=(4√6)/(1×10)
=(2√6)/5

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Answer 2

已知:x=(√3+√2)/(√3-√2),y=(√3-√2)/(√3+√2),求(x三次方-xy二次方)/(x四次方y+2x三次方y二次方+x二次方y三次)?
x=(√3+√2)/(√3-√2)=[(√3+√2)(√3+√2)]/[(√3-√2)(√3+√2)]=(3+2√6+2)/(3-2)=5+2√6;
y=(√3-√2)/(√3+√2)=[(√3-√2)(√3-√2)]/[(√3+√2)(√3-√2)]=(3-2√6+2)/(3-2)=5-2√6;
x+y=10
x-y=4√6
xy=(5+2√6)(5-2√6)=5²-(2√6)²=25-24=1

(x三次方-xy二次方)/(x四次方y+2x三次方y二次方+x二次方y三次)
=x(x²-y²)/[x²y(x²+2xy+y²)]
=(x+y)(x-y)/[xy(x+y)²]
=(x-y)/[xy(x+y)]
=4√6/[1*(10)]
=(2√6)/5