两道初二分式数学题 急!
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两道初二分式数学题 急!
1.a²-b²/a²b+ab²÷(1- ( a²+b²/2ab)) 其中a=2+根号3 b=2-根号3
2.((x²-3/x-1)-2)÷1/x-1 其中x满足x²-2x-3=0
1.a²-b²/a²b+ab²÷(1- ( a²+b²/2ab)) 其中a=2+根号3 b=2-根号3
2.((x²-3/x-1)-2)÷1/x-1 其中x满足x²-2x-3=0
1.a=2+根号3 b=2-根号3
a²-b²/a²b+ab²÷(1- ( a²+b²/2ab))
=[(a+b)(a-b)]/[ab(a+b)]÷[-(a-b)²/2ab]
=-[(a-b)/ab]×[2ab/(a-b)²]=-2/(a-b)
=-2/(2+√3-2+√3)=-1/√3=-√3/3
2..x满足x²-2x-3=0,
((x²-3/x-1)-2)÷1/x-1
=[(x²-3-2x+2)]/(x-1)÷1/(x-1)
= x²-2x-3+2 =2,
a²-b²/a²b+ab²÷(1- ( a²+b²/2ab))
=[(a+b)(a-b)]/[ab(a+b)]÷[-(a-b)²/2ab]
=-[(a-b)/ab]×[2ab/(a-b)²]=-2/(a-b)
=-2/(2+√3-2+√3)=-1/√3=-√3/3
2..x满足x²-2x-3=0,
((x²-3/x-1)-2)÷1/x-1
=[(x²-3-2x+2)]/(x-1)÷1/(x-1)
= x²-2x-3+2 =2,