a-b分之a 减去a-b 分之2b乘a-2b分之ab
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(a+b/(a-b)=2取倒数(a-b)/(a+b)=1/2原式=(3/4)[(a+b/(a-b)]-(1/2)[(a-b)/(a+b)]=3/2-1/4=4分之5
原式=我只写分子,分母是ab是不变的(a-b)^2-(a+b)^2这里可用平方差公式a^2-b^2=(a+b)(a-b)所以原式=(a-b+a+b)(a-b-a-b)=-4ab分母为ab所以相除的-4
(a/b)-(b/a)-(a²+b²)/ab=(a²-b²-a²-b²)/ab=-2b²/ab=-2b/a9a²-4b&
=[(2a-b)(a-b)/(a+b)(a-b)-b(a+b)/(a+b)(a-b)]×(a+b)/(a-2b)=(2a²-3ab+b²-ab-b²)/(a+b)(a-b
(a-b分之a-b+3)+(b-a分之b-a+a)-(a-b分之a-b+b+a)=1+(a-b分之3)+1+(b-a分之a)-1-(a-b分之b+a)=1+(a-b分之3)-(a-b分之a)-(a-b
=(a+b)/(a+b)(a-b)-(a-b)/(a+b)(a-b)+2b/(a²+b²)=2b/(a²-b²)+2b/(a²+b²)=2b
由条件得:b^2-a^2=ab左右同时除以a^2或b^2,可以解得到a/b和b/a的值,带入即可.
原式=[a/(a²-b²)-1/(a+b)]÷[b/(b-a)]=[a/(a²-b²)-(a-b)/(a²-b²)]×[-(a-b)/b]=
解原式=[(a-b)/2(a+b)]-[(a²+b²)/(a-b)(a+b)]=(a-b)²/2(a+b)(a-b)-2(a²+b²)/(a-b)(a
解:b分之a减去a分之b减去ab分之a²+b²=a/b-b/a-(a²+b²)/(ab)=(a²-b²)/(ab)-(a²+b
分之a
1/2a-1/(a-b)*[(a-b)/2a-a^2+b^2]=1/2a-1/(a-b)*[(a-b)/2a-(a^2-b^2)]=1/2a-1/(a-b)*(a-b)/2a+1/(a-b)*(a^2
解析1/a+1/b=3a/b(a-b)-b/a(a-b)=a²/ab(a-b)-b²/ab(a-b)=(a²-b²)/ab(a-b)=(a-b)(a+b)/ab
1-(a+2b)/(a+b)=(a+b)/(a+b)-(a+2b)/(a+b)=(a+b-a-2b)/(a+b)=-b/(a+b)再问:��3��10�ĸ����η������Ĵη�������3��
(a-ab)/(a+2b+b-a)/(b-a-b)/2a答:(1)(a-ab)/3b/(-a)/2a(2)/-1/2(3)*-2(4)(a/3b-a/3)*-2(5)-2a/3b+2a/3
=a(a+b)/(a+b)(a-b)+b(a-b)/(a+b)(a-b)-2ab/(a+b)(a-b)=(a²+ab+ab-b²-2ab)/(a+b)(a-b)=(a²-
原式=(a-b)²/2²-(a²-b²)/4=(a²-2ab+b²)/4-(a²-b²)/4=(a²-2ab+
(a-b)分之1-(a+b)分之1+(a^2+b^2)分之2b=(a+b)/(a²-b²)-(a-b)/(a²-b²)+2b/(a²+b²)
原式=2b/(a²+b²)+(a+b-a+b)/(a²-b²)=2b/(a²+b²)+2b/(a²-b²)=2b[(a&
(a-b)分之1-(a+b)分之1+(a平方-b平方)分之2b=(a平方-b平方)分之2b+(a平方-b平方)分之2b=(a平方-b平方)分之4