Question

数学问题:因式分解

这个内容不太理解,望指点!
分解因式: x^2(x^2-1)+4(1-x^2)
(50又1/11)^2-(49又10/11)^2

Answer 1

x^2(x^2-1)+4(1-x^2) =x^2(x^2-1) - 4(x^2-1)
提取公因式=(x^2-1)(x^2-4)
平方差公式=(x+1)(x-1)(x+2)(x-2)

(50又1/11)^2 - (49又10/11)^2
平方差公式=[(50又1/11)+(49又10/11)]*[(50又1/11)-(49又10/11)]
=100*(2/11)
=200/11

Answer 2

1、原式=x²(x²-1)-4(x²-1)
=(x²-1)(x²-4)=(x+1)(x-1)(x+2)(x-2)
2、原式=(50+1/11)²-(49+10/11)²
=(50+1/11+49+10/11)(50+1/11-49-10/11)
=100×2/11
=200/11

Answer 3

x^2(x^2-1)+4(1-x^2)
= x^2(x^2-1)-4(x^2-1)
= (x^2-1)(x^2-4)
=(x+1)(x-1)(x+2)(x-2)

(50又1/11)^2-(49又10/11)^2
= { (50+1/11)+(49+10/11) } { (50+1/11)-(49+10/11) }
= 100*2/11
= 200/11

Answer 4

希望对你有帮助，祝你周末玩的开心
(*^__^*) 嘻嘻……
（1）原式=x^2(x+1)(x-1)+4(1-x)(1+x)
=x^2(x+1)(x-1)-4(x-1)(x+1)
=(x+1)(x-1)(x^2-4)
=(x+1)(x-1)(x+2)(x-2)
(2)原式=(50又1/11+49又10/11)(50又1/11-49又10/11)
=100*2/11
=200/11

Answer 5

x^2(x^2-1)+4(1-x^2)
(50又1/11)^2-(49又10/11)^2=x^4-x^2+(4-4x^2)(551/11)^2-(539/11)^2=x^4-x^2+(303601/121)*4-(303601/121)*4x^2-290521/121=x^4+(1214283/121)x^2+923883/121相乘的两个括号内每一项对应相乘即可,如(a+b)(c+d)=(a+b)c+(a+b)d=ac+ad+bc+bd,e(a+b)(c+d)=eac+ead+ebc+ebd.不知我表述的是否清楚了