出几道初二分式混合运算题
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2
a+1-a+3
a2-4a-5÷a2-9
a2-3a-10.
解
原式=[x+2
x(x-2)-x-1(x-2)2]•x
4-x
(括号内分式的分母中的多项式式分解因式.分式的除法法则)
=[(x+2)(x-2)x(x-2)2-x(x-1)x(x-2)2]•x4-x(异分母的分式减法的法则)
=x2-4-x2+x
x(x-2)2•x4-x
(整式运算)
=x-4x(x-2)2•x4-x
(合并同类项)
=x-4
x(x-2)2•(-xx-4)
(分式的符号法则)
=-1(x-2)2.
(分式的乘法法则)
计算x+y
x2-xy
+(x2-y2
x)2•(1
y-x)3.
解
原式=x+y
x(x-y)+(x+y)2(x-y)2x2•1(y-x)3
=x+y
x(x-y)-(x+y)2
x2(x-y)
=x2+xy-x2-2xy-y2
x2(x-y)
=-xy-y2
x2(x-y)=-xy+y2
x2(x-y).
x-y+4xy
x-y)(x+y-4xyx+y)
答案x2-y2
[1
(a+b)2-1(a-b)2]÷(1a+b-1a-b)
答案2a
(a+b)(a-b);
x
x-y•
y2
x+y-x4y
x4-y4÷x2
x2+y2
答案-xy
x+y
3x-2
x2-x-2+(1-1x+1)÷(1+1x-1)
答案x2
(x+1)(x-2);
(2x
x+1+2
x-1+4x
x2-1)×(2x
x+1+2
x-1-4x
x2-1).
答案4
(2m^2-4m)/(2-m)(m-1)-(1+m)/(1-m^2)
=2m(m-2)/(2-m)(m-1)-(1+m)/(1-m)(1+m)
=-2m/(m-1)-1/(1-m)
=(2m-1)/(1-m)
(-1)-a^2)/(a-1)-a
=(1-a-a^2-a^2+a)/(a-1)
=-(2a^2-1)/(a-1)
a+1-a+3
a2-4a-5÷a2-9
a2-3a-10.
解
原式=[x+2
x(x-2)-x-1(x-2)2]•x
4-x
(括号内分式的分母中的多项式式分解因式.分式的除法法则)
=[(x+2)(x-2)x(x-2)2-x(x-1)x(x-2)2]•x4-x(异分母的分式减法的法则)
=x2-4-x2+x
x(x-2)2•x4-x
(整式运算)
=x-4x(x-2)2•x4-x
(合并同类项)
=x-4
x(x-2)2•(-xx-4)
(分式的符号法则)
=-1(x-2)2.
(分式的乘法法则)
计算x+y
x2-xy
+(x2-y2
x)2•(1
y-x)3.
解
原式=x+y
x(x-y)+(x+y)2(x-y)2x2•1(y-x)3
=x+y
x(x-y)-(x+y)2
x2(x-y)
=x2+xy-x2-2xy-y2
x2(x-y)
=-xy-y2
x2(x-y)=-xy+y2
x2(x-y).
x-y+4xy
x-y)(x+y-4xyx+y)
答案x2-y2
[1
(a+b)2-1(a-b)2]÷(1a+b-1a-b)
答案2a
(a+b)(a-b);
x
x-y•
y2
x+y-x4y
x4-y4÷x2
x2+y2
答案-xy
x+y
3x-2
x2-x-2+(1-1x+1)÷(1+1x-1)
答案x2
(x+1)(x-2);
(2x
x+1+2
x-1+4x
x2-1)×(2x
x+1+2
x-1-4x
x2-1).
答案4
(2m^2-4m)/(2-m)(m-1)-(1+m)/(1-m^2)
=2m(m-2)/(2-m)(m-1)-(1+m)/(1-m)(1+m)
=-2m/(m-1)-1/(1-m)
=(2m-1)/(1-m)
(-1)-a^2)/(a-1)-a
=(1-a-a^2-a^2+a)/(a-1)
=-(2a^2-1)/(a-1)
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