a^4+b^4+c^4+d^4-4abcd=0
(a^4-2a^2b^2+b^4)+(c^4-2c^2d^2+d^4)+(2a^2b^2-4abcd+2c^2d^2)=0
(a^4-2a^2b^2+b^4)+(c^4-2c^2d^2+d^4)+2(a^2b^2-2abcd+c^2d^2)=0
(a^2-b^2)^2+(c^2-d^2)^2+2(ab-cd)^2=0
由于平方数都大于或等于0,且a、b、c、d>0,所以由上式可知:
(a^2-b^2)^2=0,可得:a^2=b^2,即a=b,
(c^2-d^2)^2=0,可得:c^2=d^2,即c=d,
2(ab-cd)^2=0,可得:ab=cd;
由a=b,c=d,ab=cd可得:a=b=c=d,
(3x-y+4)(x+2y-1)
用双十字相乘法做
3x-y4
x2y-1
或用待定系数法:
由于3x2+5xy-2y2=(3x-y)(x+2y),故可设
3x2+5xy-2y2+x+9y-4
=(3x-y+a)(x+2y+b)
=3x2+5xy-2y2+(a+3b)x+(2a-b)y+ab.
由于3x2+5xy-2y2=(3x-y)(x+2y),故可设
3x2+5xy-2y2+x+9y-4
=(3x-y+a)(x+2y+b)
=3x2+5xy-2y2+(a+3b)x+(2a-b)y+ab.
比较两边系数最后解得a=4b=1
∴原式=(3x-y+4)(x+2y-1).
1111=11*101,111111=111*1001
显然1111.1111=11..11*10..01(注意,11..11是n位数,10..01是n+1位数)
左面是2n个1,右边是n个1,10..01中间n-1个0
1111.1111-22..22=11..11*10..01-2*11..11
=11..11*(10..01-2)
=11..11*(99..99)
=11..11*11..11*9=(3*11..11)^2=33..33^2