1.已知x=根号5+2,y=根号5-2求代数式x平方+5xy+y平方的值。
1.已知x=根号5+2,y=根号5-2求代数式x平方+5xy+y平方的值。2.先化简,再求值。x-1分之x-2÷(x+1-x-1分之3),其中x=根号3-23.若a=1-...
1.已知x=根号5+2,y=根号5-2求代数式x平方+5xy+y平方的值。2.先化简,再求值。
x-1分之x-2÷(x+1-x-1分之3),其中x=根号3-2
3.若a=1-根号2,先化简,再求a平方+a分之a平方-1+a平方-a分之根号a平方-2a+1的值 展开
x-1分之x-2÷(x+1-x-1分之3),其中x=根号3-2
3.若a=1-根号2,先化简,再求a平方+a分之a平方-1+a平方-a分之根号a平方-2a+1的值 展开
2个回答
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1、x²+5xy+y
=(√5+2)²+5(√5+2)(√5-2)+√5-2
=(9+4√5)+(25-20)+√5-2
=12+5√5
2、[(x-2)/(x-1)]/[(x+1)-3/(x-1)]
=[(x-2)/(x-1)]/[(x²-1-3)/(x-1)]
=[(x-2)(x-1)]/[(x-1)(x²-2²)]
=(x-2)/[(x+2)(x-2)]
=1/(x+2)
=1/(√3-2+2)
=√3/3 a平方+a分之a平方-1+a平方-a分之根号a平方-2a+1
3、[(a²-1)/(a²+a)]+[√(a²-2a+1)/(a²-a)] (题目叙述的太不规范)
=[(a²-1)/a(a+1)]+[(a-1)/a(a-1)]
=[(a-1)/a]+(1/a)
=1
或
a²+(a²/a)-1+a²-(√a²/a)-2a+1
=a²+a-1+a²-(1/a)-2a+1
=2a²-a-(1/a)
=2(1-√2)²-(1-√2)-1/(1-√2)
=2-8+4-1+√2-(1+√2)/(1-2)
=-3+√2+1+√2
=2√2-2
=2(√2-1)
=(√5+2)²+5(√5+2)(√5-2)+√5-2
=(9+4√5)+(25-20)+√5-2
=12+5√5
2、[(x-2)/(x-1)]/[(x+1)-3/(x-1)]
=[(x-2)/(x-1)]/[(x²-1-3)/(x-1)]
=[(x-2)(x-1)]/[(x-1)(x²-2²)]
=(x-2)/[(x+2)(x-2)]
=1/(x+2)
=1/(√3-2+2)
=√3/3 a平方+a分之a平方-1+a平方-a分之根号a平方-2a+1
3、[(a²-1)/(a²+a)]+[√(a²-2a+1)/(a²-a)] (题目叙述的太不规范)
=[(a²-1)/a(a+1)]+[(a-1)/a(a-1)]
=[(a-1)/a]+(1/a)
=1
或
a²+(a²/a)-1+a²-(√a²/a)-2a+1
=a²+a-1+a²-(1/a)-2a+1
=2a²-a-(1/a)
=2(1-√2)²-(1-√2)-1/(1-√2)
=2-8+4-1+√2-(1+√2)/(1-2)
=-3+√2+1+√2
=2√2-2
=2(√2-1)
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