七年级上册《字母表示数》的几道数学题
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七年级上册《字母表示数》的几道数学题
(1)已知|m+1|+|2n4|=0,求m²-3n+[m-5(m²-n)+3m]-4m的值(算式)
(2)今年捐款活动,某校初一2个班115名学生积极参加,已知甲班3分之一学生每人捐款10元,乙班5分之二学生每人捐10元,2班其余每人捐5元,设甲班有学生x人,使用代数式表示2班捐款的总额,并化简
(3)
(1)已知|m+1|+|2n4|=0,求m²-3n+[m-5(m²-n)+3m]-4m的值(算式)
(2)今年捐款活动,某校初一2个班115名学生积极参加,已知甲班3分之一学生每人捐款10元,乙班5分之二学生每人捐10元,2班其余每人捐5元,设甲班有学生x人,使用代数式表示2班捐款的总额,并化简
(3)
(1)已知|m+1|+|2n4|=0
由于绝对值大于等于0,则可知 m+1 = 2n4 = 0 ,即 m = -1 ,n = 0
那么m²-3n+[m-5(m²-n)+3m]-4m
= m² - 3n + ( m - 5m² + 5n + 3m ) - 4m
= m² - 3n + ( 4m - 5m² + 5n ) - 4m
= m² - 3n + 4m - 5m² + 5n - 4m
= ( m²-5m² ) + ( 4m - 4m ) +( - 3n + 5n )
= -4m² + 2n
= -4 * (-1)² + 2 * 0
= -4
(2)1/3 * x * 10 + (1 - 1/3) * x * 5 +(115 - x)* 2/5 * 10 +(115 - x) * ( 1 - 2/5) * 5
= 1/3 * 10x + 2/3 * 5x + (115 - x)* 4 +(115 - x)* 3/5 * 5
= 10x/3 + 10x/3 + 115 * 4 - 4x +(115 - x)* 3
= 20x/3 + 115 * 4 - 4x + 115 * 3 - 3x
= 20x/3 + 115 * (4 + 3) + (-4x - 3x)
= 20x/3 + 115 * 7 - 7x
= (20x/3 - 7x) + 805
= -1/3x + 805
由于绝对值大于等于0,则可知 m+1 = 2n4 = 0 ,即 m = -1 ,n = 0
那么m²-3n+[m-5(m²-n)+3m]-4m
= m² - 3n + ( m - 5m² + 5n + 3m ) - 4m
= m² - 3n + ( 4m - 5m² + 5n ) - 4m
= m² - 3n + 4m - 5m² + 5n - 4m
= ( m²-5m² ) + ( 4m - 4m ) +( - 3n + 5n )
= -4m² + 2n
= -4 * (-1)² + 2 * 0
= -4
(2)1/3 * x * 10 + (1 - 1/3) * x * 5 +(115 - x)* 2/5 * 10 +(115 - x) * ( 1 - 2/5) * 5
= 1/3 * 10x + 2/3 * 5x + (115 - x)* 4 +(115 - x)* 3/5 * 5
= 10x/3 + 10x/3 + 115 * 4 - 4x +(115 - x)* 3
= 20x/3 + 115 * 4 - 4x + 115 * 3 - 3x
= 20x/3 + 115 * (4 + 3) + (-4x - 3x)
= 20x/3 + 115 * 7 - 7x
= (20x/3 - 7x) + 805
= -1/3x + 805