1.先化简再求值(½ x²y+1/5 xy²-¼xy)·(-½xy²),其中x=-2 y=-½[3xy(1-x)-6xy﹙x-½)]·2x·(-xy)²,其中x=﹣1 y=23xy²(-1/3x²y+4x²y²)-4x²y(﹣¾xy²
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![1.先化简再求值(½ x²y+1/5 xy²-¼xy)·(-½xy²),其中x=-2 y=-½[3xy(1-x)-6xy﹙x-½)]·2x·(-xy)²,其中x=﹣1 y=23xy²(-1/3x²y+4x²y²)-4x²y(﹣¾xy²](/uploads/image/z/1682742-30-2.jpg?t=1.%E5%85%88%E5%8C%96%E7%AE%80%E5%86%8D%E6%B1%82%E5%80%BC%EF%BC%88%26%23189%3B+x%26%23178%3By%EF%BC%8B1%EF%BC%8F5+xy%26%23178%3B%EF%BC%8D%26%23188%3Bxy%29%C2%B7%28-%26%23189%3Bxy%26%23178%3B%29%2C%E5%85%B6%E4%B8%ADx%3D-2+y%3D-%26%23189%3B%5B3xy%281%EF%BC%8Dx%29%EF%BC%8D6xy%EF%B9%99x%EF%BC%8D%26%23189%3B%29%5D%C2%B72x%C2%B7%28%EF%BC%8Dxy%29%26%23178%3B%2C%E5%85%B6%E4%B8%ADx%3D%EF%B9%A31+y%3D23xy%26%23178%3B%28%EF%BC%8D1%EF%BC%8F3x%26%23178%3By%EF%BC%8B4x%26%23178%3By%26%23178%3B%29%EF%BC%8D4x%26%23178%3By%28%EF%B9%A3%26%23190%3Bxy%26%23178%3B)
1.先化简再求值(½ x²y+1/5 xy²-¼xy)·(-½xy²),其中x=-2 y=-½[3xy(1-x)-6xy﹙x-½)]·2x·(-xy)²,其中x=﹣1 y=23xy²(-1/3x²y+4x²y²)-4x²y(﹣¾xy²
1.先化简再求值
(½ x²y+1/5 xy²-¼xy)·(-½xy²),其中x=-2 y=-½
[3xy(1-x)-6xy﹙x-½)]·2x·(-xy)²,其中x=﹣1 y=2
3xy²(-1/3x²y+4x²y²)-4x²y(﹣¾xy²)·(﹣4y) 其中x=-3,y=1/3
简答题.
已知(m-x)·﹙﹣x﹚+n(x+m﹚=x²+5x-6对于任意数x都成立,求
m(n-1)+n(m+1)的值
计算:
(x+1﹚﹙x²+x+1﹚-﹙x-1﹚﹙x²-x+1﹚≥﹙4x+3﹚﹙x-2﹚
1.先化简再求值(½ x²y+1/5 xy²-¼xy)·(-½xy²),其中x=-2 y=-½[3xy(1-x)-6xy﹙x-½)]·2x·(-xy)²,其中x=﹣1 y=23xy²(-1/3x²y+4x²y²)-4x²y(﹣¾xy²
1.先化简再求值
(½ x²y+1/5 xy²-¼xy)·(-½xy²),其中x=-2 y=-½
(½ x²y+1/5 xy²-¼xy)·(-½xy²)
=(1/20)xy(10x+4 y-5)·(-½xy²)
=(-1/40)x²·y^3(10x+4y-5)
=(-1/40)(-2)²x(-1/2)^3(10x(-2)+4x(-1/2)-5)
=(-1/40)x4x(-1/8)x(-20-2-5)
=1/80x(-27)
=-27/80
[3xy(1-x)-6xy﹙x-½)]·2x·(-xy)²,其中x=﹣1 y=2
[3xy(1-x)-6xy﹙x-½)]·2x·(-xy)²
=3xy[(1-x)-2(x-1/2)]·2x·x²y²
=6x^4y^3(2-3x)
=6(-1)^4(2)^3(2-3x(-1))
=48x5=240
3xy²(-1/3x²y+4x²y²)-4x²y(﹣¾xy²)·(﹣4y) 其中x=-3,y=1/3
3xy²(-1/3x²y+4x²y²)-4x²y(﹣¾xy²)·(﹣4y)
=3xy²x²y(-1/3+4y)-12x^3y^4
=3x^3y^3(-1/3+4y)-12x^3y^4
=3x^3y^3[(-1/3+4y)-4y]
=(-1/3)3x^3y^3
=-x^3y^3
=-(xy)^3
=-(-3x1/3)^3
=-(-1)
=1.
简答题.
已知(m-x)·﹙﹣x﹚+n(x+m﹚=x²+5x-6对于任意数x都成立,求
m(n-1)+n(m+1)的值
(m-x)·(-X)+n(x+m)
=x²-mx+nx+mn
=x²+(n-m)x+mn
=x²+5X-6
所以 n-m=5 mn=6
m(n-1)+n(m+1)
=mn-m+mn+n
=2mn+(n-m)
=12+5
=17
计算:
(x+1﹚﹙x²+x+1﹚-﹙x-1﹚﹙x²-x+1﹚≥﹙4x+3﹚﹙x-2﹚
由(x+1﹚﹙x²+x+1﹚-﹙x-1﹚﹙x²-x+1﹚=4x²+2
﹙4x+3﹚﹙x-2﹚=4x²-5x-6,
原不等式(x+1﹚﹙x²+x+1﹚-﹙x-1﹚﹙x²-x+1﹚≥﹙4x+3﹚﹙x-2﹚
可化为4x²+2≥4x²-5x-6,
5x≥-8
x≥-8/5,
即原不等式的解为:
x大于或等于负一又五分之三.
1 提取公因式 1/20xy,
2 =(3xy-3x²y-6x²y+3xy)(2x³y²)
=12x^4y³-18x^5y³
=96+144
=240
3 (m-x)*(-X)+n(x+m)
=x^2-mx+nx+mn
=x^2+(n-m)x+mn
=x²+5X...
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1 提取公因式 1/20xy,
2 =(3xy-3x²y-6x²y+3xy)(2x³y²)
=12x^4y³-18x^5y³
=96+144
=240
3 (m-x)*(-X)+n(x+m)
=x^2-mx+nx+mn
=x^2+(n-m)x+mn
=x²+5X-6
所以 n-m=5 mn=6
m(n-1)+n(m+1)
=mn-m+mn+n
=2mn+(n-m)
=12+5
=17
这题应该与(x+1)^3有关吧,(x+1﹚﹙x²+x+1﹚=(x+1)^3-x^2-x
因为(X+1)^3=x^3+3x^2+3x+1
自己算吧
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