初一孩子的几道数学题,已做好,但不知对否,望会的网友告知一下,谢谢!
一.分解因式1.x^2(2x-1)+y^(1-2x)2.(m+4m+2)^2-43.已知a、b、c是△ABC三条边的长,且满足b^2+2ab=c^2+2ac,试判断△AB...
一.分解因式
1.x^2(2x-1)+y^(1-2x)
2.(m+4m+2)^2-4
3.已知a、b、c是△ABC三条边的长,且满足b^2+2ab=c^2+2ac,试判断△ABC的形状
4.已知a+b=2,ab=2,求1/2a^3b-a^2b^2+1/2ab^3的值 展开
1.x^2(2x-1)+y^(1-2x)
2.(m+4m+2)^2-4
3.已知a、b、c是△ABC三条边的长,且满足b^2+2ab=c^2+2ac,试判断△ABC的形状
4.已知a+b=2,ab=2,求1/2a^3b-a^2b^2+1/2ab^3的值 展开
1个回答
展开全部
1.x^2(2x-1)+y^2(1-2x)看看题是不是这样的,你的好像少了个平方
=x^2(2x-1)-y^2(2x-1)
=(x^2-y^2)(2x-1)
=(x+y)(x+y)(2x-1)
2..(m+4m+2)^2-4
=.(5m+2)^2-4
=(5m+2+2)(5m+2-2)
=5m(5m+4)
3. b^2+2ab=c^2+2ac
b^2-c^2+2ab-2ac=0
(b+c)(b-c)+2a(b-c)=0
(b-c)(b+c+2a)=0
b-c=0
b=c
△ABC为等腰三角形
4.1/2a^3b-a^2b^2+1/2ab^3
=1/2ab(a^2-2ab+b^2)
=1/2ab(a-b)^2
=1/2ab[(a+b)^2-4ab]
=1/2*2*[2^2-4*2]
=-4
=x^2(2x-1)-y^2(2x-1)
=(x^2-y^2)(2x-1)
=(x+y)(x+y)(2x-1)
2..(m+4m+2)^2-4
=.(5m+2)^2-4
=(5m+2+2)(5m+2-2)
=5m(5m+4)
3. b^2+2ab=c^2+2ac
b^2-c^2+2ab-2ac=0
(b+c)(b-c)+2a(b-c)=0
(b-c)(b+c+2a)=0
b-c=0
b=c
△ABC为等腰三角形
4.1/2a^3b-a^2b^2+1/2ab^3
=1/2ab(a^2-2ab+b^2)
=1/2ab(a-b)^2
=1/2ab[(a+b)^2-4ab]
=1/2*2*[2^2-4*2]
=-4
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