1.先化简再求值(½ x²y+1/5 xy²-¼xy)·(-½xy
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/10/06 16:09:23
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﹚
(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
(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.
(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.
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