三道初二数学题 要过程啊!!快,高手进
1/a+1/b=1/(a+b)求b/a+a/b的值化简求值(2a)/(a²-4)+1/(2-a)a=1/2(x+2y)/(x+y)+(2y²)/(x&...
1/a+1/b=1/(a+b) 求b/a+a/b的值
化简求值 (2a)/(a²-4) +1/( 2-a) a=1/2
(x+2y)/(x+y)+(2y²)/(x²-y²) x=-2 y=1/3 展开
化简求值 (2a)/(a²-4) +1/( 2-a) a=1/2
(x+2y)/(x+y)+(2y²)/(x²-y²) x=-2 y=1/3 展开
5个回答
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因为1/a+1/b=1/(a+b)
(a+b)/ab=1/(a+b)
(a+b)^2=ab
所以b/a+a/b=(a^2+b^2)/ab
=〔(a+b)^2-2ab〕/ab=-1
此题主要考查你对(a+b)^2,ab,a^2+b^2三者关系的理解情况,依据完全平方公式可推导出这三者之间的关系。在中考中常用到
(2a)/(a²-4) +1/( 2-a)
=(2a)/(a²-4) -(a+2)/( a²-4)
=(a-2)/(a²-4)
=1/(a+2)
把a=1/2代入上式可得
1/(a+2)=2/5
此题考查了对a+b,a-b,a^2-b^2这三者的关系的掌握情况,依据为平方差公式,因此要熟练掌握平方差公式。
(x+2y)/(x+y)+(2y²)/(x²-y²)
=(x+2y)(x-y)/(x²-y²)+(2y²)/(x²-y²)
=x(x+y)/(x²-y²)
=x/(x-y)
把x=-2 y=1/3代入上式可得
x/(x-y)=6/7
此题中考查了平方差公式和利用提公因式法分解因式,要对a+b,a-b,a^2-b^2这三者的关系熟练掌握,依据为平方差公式。
(a+b)/ab=1/(a+b)
(a+b)^2=ab
所以b/a+a/b=(a^2+b^2)/ab
=〔(a+b)^2-2ab〕/ab=-1
此题主要考查你对(a+b)^2,ab,a^2+b^2三者关系的理解情况,依据完全平方公式可推导出这三者之间的关系。在中考中常用到
(2a)/(a²-4) +1/( 2-a)
=(2a)/(a²-4) -(a+2)/( a²-4)
=(a-2)/(a²-4)
=1/(a+2)
把a=1/2代入上式可得
1/(a+2)=2/5
此题考查了对a+b,a-b,a^2-b^2这三者的关系的掌握情况,依据为平方差公式,因此要熟练掌握平方差公式。
(x+2y)/(x+y)+(2y²)/(x²-y²)
=(x+2y)(x-y)/(x²-y²)+(2y²)/(x²-y²)
=x(x+y)/(x²-y²)
=x/(x-y)
把x=-2 y=1/3代入上式可得
x/(x-y)=6/7
此题中考查了平方差公式和利用提公因式法分解因式,要对a+b,a-b,a^2-b^2这三者的关系熟练掌握,依据为平方差公式。
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1、∵1/a+1/b=(a+b)/ab,由条件1/a+1/b=1/(a+b)
可得(a+b)/ab=1/(a+b) ,∴(a+b)²=ab
b/a+a/b=(b²+a²)/(ab)=[(b+a)²-2ab]/(ab),将(a+b)²=ab代入可得
原式=(ab-2ab)/(ab)=(-ab)/(ab)=-1。
2、(2a)/(a²-4) +1/( 2-a)=(2a)/[(a+2)(a-2)]-(a+2)/[(a+2)(a-2)]
=(2a-a-2)/[(a+2)(a-2)]
=(a-2)/[(a+2)(a-2)]
=1/(a+2)
当a=1/2时,原式=1/[1+(1/2)]=2/3
3、(x+2y)/(x+y)+(2y²)/(x²-y²) =[(x+2y)(x-y)]/[(x+y)(x-y)]+(2y²)/[(x+y)(x-y)]
=[(x²+xy-2y²)+2y²]/[(x+y)(x-y)]
=(x²+xy)/[(x+y)(x-y)]
=[x(x+y)]/[(x+y)(x-y)]
=x/(x-y)
当x=-2 y=1/3时,原式=-2/(-2-1/3)=6/7.
可得(a+b)/ab=1/(a+b) ,∴(a+b)²=ab
b/a+a/b=(b²+a²)/(ab)=[(b+a)²-2ab]/(ab),将(a+b)²=ab代入可得
原式=(ab-2ab)/(ab)=(-ab)/(ab)=-1。
2、(2a)/(a²-4) +1/( 2-a)=(2a)/[(a+2)(a-2)]-(a+2)/[(a+2)(a-2)]
=(2a-a-2)/[(a+2)(a-2)]
=(a-2)/[(a+2)(a-2)]
=1/(a+2)
当a=1/2时,原式=1/[1+(1/2)]=2/3
3、(x+2y)/(x+y)+(2y²)/(x²-y²) =[(x+2y)(x-y)]/[(x+y)(x-y)]+(2y²)/[(x+y)(x-y)]
=[(x²+xy-2y²)+2y²]/[(x+y)(x-y)]
=(x²+xy)/[(x+y)(x-y)]
=[x(x+y)]/[(x+y)(x-y)]
=x/(x-y)
当x=-2 y=1/3时,原式=-2/(-2-1/3)=6/7.
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第一题
1/a+1/b=1/(a+b)化简求职有:
(b+a)/ab=1/(a+b)
则有ab=(a+b)²
要求的式子为b/a+a/b化简得
(b²+a²)/ab将ab=(a+b)²代入上式化简后得到-1.
第二题
(2a)/(a+2)(a-2)+1/( 2-a)
=[(2a)-(2+a)]/(a+2)(a-2)
=1/(a+2)
将a=1/2代入
得到2/5
第三题
(x+2y)/(x+y)+(2y²)/(x²-y²)化简得
[(x+2y(x-y)+2y²]/(x+y)(x-y)
=(x²+xy)/(x+y)(x-y)
=x/(x-y)
将x=-2,y=1/3代入得结果6/7
1/a+1/b=1/(a+b)化简求职有:
(b+a)/ab=1/(a+b)
则有ab=(a+b)²
要求的式子为b/a+a/b化简得
(b²+a²)/ab将ab=(a+b)²代入上式化简后得到-1.
第二题
(2a)/(a+2)(a-2)+1/( 2-a)
=[(2a)-(2+a)]/(a+2)(a-2)
=1/(a+2)
将a=1/2代入
得到2/5
第三题
(x+2y)/(x+y)+(2y²)/(x²-y²)化简得
[(x+2y(x-y)+2y²]/(x+y)(x-y)
=(x²+xy)/(x+y)(x-y)
=x/(x-y)
将x=-2,y=1/3代入得结果6/7
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1/a+1/b=1/(a+b)
so (a+b)/ab=(a+b)
即(a+b)^2=ab
a^2+b^2=-ab
b/a+a/b=(b^2+a^2)/ab=-1
2. (2a)/(a²-4) +1/( 2-a)=2a/[(a+2)(a-2)]-1/(a-2)=(2a)/[(a+2)(a-2)]-(a+2)/[(a+2)(a-2)]=1/(a+2)=1/(1/2+2)=2/5
3.(x+2y)/(x+y)+(2y²)/(x²-y²)
=(x+2y)(x-y)/[(x+y)(x-y)+2y²/[(x+y)(x-y)]
=[(x+2y)(x-y)+2y²]/[(x+y)(x-y)]
=x(x+y)/[(x+y)(x-y)]
=x/(x-y)
=-2/(-2-1/3)
=6/7
so (a+b)/ab=(a+b)
即(a+b)^2=ab
a^2+b^2=-ab
b/a+a/b=(b^2+a^2)/ab=-1
2. (2a)/(a²-4) +1/( 2-a)=2a/[(a+2)(a-2)]-1/(a-2)=(2a)/[(a+2)(a-2)]-(a+2)/[(a+2)(a-2)]=1/(a+2)=1/(1/2+2)=2/5
3.(x+2y)/(x+y)+(2y²)/(x²-y²)
=(x+2y)(x-y)/[(x+y)(x-y)+2y²/[(x+y)(x-y)]
=[(x+2y)(x-y)+2y²]/[(x+y)(x-y)]
=x(x+y)/[(x+y)(x-y)]
=x/(x-y)
=-2/(-2-1/3)
=6/7
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1. 1/a+1/b=1/(a+b)
(b+a)/ab=1/(a+b)
(a+b)²=ab
a²+ab+b²=0
由题知a,b,a+b都不等于0,两边同时除以ab,
a/b+1+b/a=0->b/a+a/b=-1.
2. (2a)/(a²-4) +1/( 2-a)
=(2a)/((a+2)(a-2))+1/(2-a)
=(2a-a-2)/((a+2)(a-2))
=(a-2)/((a+2)(a-2))
=1/(a+2)=1/(1/2+2)=5/2
3. (x+2y)/(x+y)+(2y²)/(x²-y²)
=((x+2y)(x-y)+2y²)/((x+y)(x-y))
=(x²+xy)/((x+y)(x-y))
=x/(x-y)
=-2/(-2-1/3)=6/7
(b+a)/ab=1/(a+b)
(a+b)²=ab
a²+ab+b²=0
由题知a,b,a+b都不等于0,两边同时除以ab,
a/b+1+b/a=0->b/a+a/b=-1.
2. (2a)/(a²-4) +1/( 2-a)
=(2a)/((a+2)(a-2))+1/(2-a)
=(2a-a-2)/((a+2)(a-2))
=(a-2)/((a+2)(a-2))
=1/(a+2)=1/(1/2+2)=5/2
3. (x+2y)/(x+y)+(2y²)/(x²-y²)
=((x+2y)(x-y)+2y²)/((x+y)(x-y))
=(x²+xy)/((x+y)(x-y))
=x/(x-y)
=-2/(-2-1/3)=6/7
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