急求三道八上数学题1.已知x+y=0,x+3y=1,求3x²+12xy+13y²的值2.因式分解:4a²-4ab+b²-6a+3b-43.观察下列各式:1×2×3×4+1=5²;2×3×4×5+1=11²;3×4×5×6+1=19²;判断是否任意四个连续
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![急求三道八上数学题1.已知x+y=0,x+3y=1,求3x²+12xy+13y²的值2.因式分解:4a²-4ab+b²-6a+3b-43.观察下列各式:1×2×3×4+1=5²;2×3×4×5+1=11²;3×4×5×6+1=19²;判断是否任意四个连续](/uploads/image/z/1680802-34-2.jpg?t=%E6%80%A5%E6%B1%82%E4%B8%89%E9%81%93%E5%85%AB%E4%B8%8A%E6%95%B0%E5%AD%A6%E9%A2%981.%E5%B7%B2%E7%9F%A5x%2By%3D0%2Cx%2B3y%3D1%2C%E6%B1%823x%26sup2%3B%2B12xy%2B13y%26sup2%3B%E7%9A%84%E5%80%BC2.%E5%9B%A0%E5%BC%8F%E5%88%86%E8%A7%A3%EF%BC%9A4a%26sup2%3B-4ab%2Bb%26sup2%3B-6a%2B3b-43.%E8%A7%82%E5%AF%9F%E4%B8%8B%E5%88%97%E5%90%84%E5%BC%8F%EF%BC%9A1%C3%972%C3%973%C3%974%2B1%3D5%26sup2%3B%EF%BC%9B2%C3%973%C3%974%C3%975%2B1%3D11%26sup2%3B%EF%BC%9B3%C3%974%C3%975%C3%976%2B1%3D19%26sup2%3B%EF%BC%9B%E5%88%A4%E6%96%AD%E6%98%AF%E5%90%A6%E4%BB%BB%E6%84%8F%E5%9B%9B%E4%B8%AA%E8%BF%9E%E7%BB%AD)
急求三道八上数学题1.已知x+y=0,x+3y=1,求3x²+12xy+13y²的值2.因式分解:4a²-4ab+b²-6a+3b-43.观察下列各式:1×2×3×4+1=5²;2×3×4×5+1=11²;3×4×5×6+1=19²;判断是否任意四个连续
急求三道八上数学题
1.已知x+y=0,x+3y=1,求3x²+12xy+13y²的值
2.因式分解:4a²-4ab+b²-6a+3b-4
3.观察下列各式:1×2×3×4+1=5²;2×3×4×5+1=11²;3×4×5×6+1=19²;判断是否任意四个连续正整数之积与1的和都是某个正整数的平方,并说明理由.
答上追加
急求三道八上数学题1.已知x+y=0,x+3y=1,求3x²+12xy+13y²的值2.因式分解:4a²-4ab+b²-6a+3b-43.观察下列各式:1×2×3×4+1=5²;2×3×4×5+1=11²;3×4×5×6+1=19²;判断是否任意四个连续
1.3x²+12xy+13y²=(3x²+12xy+9y²)+4y²
=3(x+3y)(x+y)+y²
=0+4y²=4y².题目有问题吧
2.4a²-4ab+b²-6a+3b-4
=(2a-b)²+(-6a+3b)-4
=(2a-b+1)(2a-b-4).
3.任意四个连续正整数之积与1的和都是某个正整数的平方.理由如下:
设四个连续正整数为a,a+1,a+2,a+3,
则a(a+1)(a+2)(a+3)+1
=【a(a+3)】【(a+1)(a+2)】+1
=(a²+3a)(a²+3a+2)+1
=(a²+3a)²+2(a²+3a)+1
=(a²+3a+1)².
故任意四个连续正整数之积与1的和都是某个正整数的平方.