三道初一数学题(关于因式分解)
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三道初一数学题(关于因式分解)
1) (4a^2-1)(b^2-1)-8ab
2) x(x-1)(x+1)(x+2)-24
3) 已知a^2+2a=1,利用因式分解求多项式a^4+2a^3+2a-1的值
1) (4a^2-1)(b^2-1)-8ab
2) x(x-1)(x+1)(x+2)-24
3) 已知a^2+2a=1,利用因式分解求多项式a^4+2a^3+2a-1的值
![三道初一数学题(关于因式分解)](/uploads/image/z/16939893-21-3.jpg?t=%E4%B8%89%E9%81%93%E5%88%9D%E4%B8%80%E6%95%B0%E5%AD%A6%E9%A2%98%EF%BC%88%E5%85%B3%E4%BA%8E%E5%9B%A0%E5%BC%8F%E5%88%86%E8%A7%A3%EF%BC%89)
1)(4a^2-1)(b^2-1)-8ab
=4a²b²-4a²-b²+1-8ab
=4a²b²-4ab+1-4a²-b²-4ab
=(2ab-1)²-(2a+b)²
=(2ab-1+2a+b)(2ab-1-2a-b)
2) x(x-1)(x+1)(x+2)-24
=[(x-1)(x+2)][x(x+1)]-24
=(x²+x-2)(x²+x)-24
=(x²+x)²-2(x²+x)-24
=(x²+x+4)(x²+x-6)
=(x²+x+4)(x+3)(x-2)
3) 已知a^2+2a=1,利用因式分解求多项式a^4+2a^3+2a-1的值
a^2+2a=1,a^2+2a-1=0
a^4+2a^3+2a-1
=(a²+1)(a²-1)+2a(a²+1)
=(a²+2a-1)(a²+1)
=0
=4a²b²-4a²-b²+1-8ab
=4a²b²-4ab+1-4a²-b²-4ab
=(2ab-1)²-(2a+b)²
=(2ab-1+2a+b)(2ab-1-2a-b)
2) x(x-1)(x+1)(x+2)-24
=[(x-1)(x+2)][x(x+1)]-24
=(x²+x-2)(x²+x)-24
=(x²+x)²-2(x²+x)-24
=(x²+x+4)(x²+x-6)
=(x²+x+4)(x+3)(x-2)
3) 已知a^2+2a=1,利用因式分解求多项式a^4+2a^3+2a-1的值
a^2+2a=1,a^2+2a-1=0
a^4+2a^3+2a-1
=(a²+1)(a²-1)+2a(a²+1)
=(a²+2a-1)(a²+1)
=0