用变形补码计算X Y,X-Y,并指出结果是否溢出X=0.1001
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3xy[2xy-x(y-2)+x-1]=3xy(2xy-xy+2x+x-1)=3xy(xy+3x-1)=3x^2y^2+9x^2y-3xy
3xy[2xy-x(y-2)+x]=3xy[2xy-xy+2x+x]=3xy(xy+3x)=3x²y²+9x²y
这题少括号了吧.
【x²+xy/(x-y)】/【xy/(x-y)】=【x²(x-y)/(x-y)+xy/(x-y)】/【xy/(x-y)】={【x²(x-y)+xy】/(x-y)}/【xy
-3[y-(3x²-3xy)]-[y+2(4x²-4xy)]=-3[y-3x²+3xy]-[y+8x²-8xy]=-3y+9x²-9xy-y-8x
(xy-y^2)÷x-y/xy=y(x-y)×[xy/(x-y)]=xy²
x+y=1.10010x-y=0.10111+1.00101=1.11100
两个正数相加,两个负数相加,两个符号不相同的数相减,都有可能产生溢出.
[x]补=1101010+[y]补=1111111————————————[x+y]补=1101001
没区别,两位乘法复杂点.具体两位乘法忘了.因为没必要纠结这个.要考试的话一般都是考存储和指令子类的,一位的话给你发个例题吧.原码一位乘法中,符号位与数值位是分开进行计算的.运算结果的数值部分是乘数与被
(xy-x²)乘(x-y)分之xy=-x(x-y)乘[(x-y)分之xy]=-x²y
[X+Y]补=X补+Y补=00.1011+11.0111=00.0010(无溢出)[X-Y]补=[X]补+[-Y]补=00.1011+00.1001=01.0100(溢出,结果错误)
变形补码就是双符号位补码(1)取补:[x]补=00.11011[y]补=00.00011[-y]补=[[y]补]变补=11.11101[x+y]补=[x]补+[y]补=00.11110x+y=[[x+
Y补码:-Y=逐位取反(Y)+1=0110001+1=0110010然后X-Y=X+(-Y)=1011110+0110010=0010000.即为所得
我就做一道,剩下的我把要领讲一下:(1)[X]原=00001111〔Y〕原=10000001〔X〕补=00001111〔Y〕补=11111111〔X+Y〕补=[X]补+[Y]补=000001111+1
题目多,奖分少,仅选其中(1)、(2)两题.(1)X=0.11100,Y=-0.11111.-Y=+0.11111..[X]补=00.11100.[-Y]补=00.11111(+-----------
解(x-y)/(xy-y²)=(x-y)/y(x-y)=1/y
用双符号位?[X]补=00.10011[-Y]补=11.00111[X-Y]补=[X]补+[-Y]补=11.11010(符号位是11,无溢出)X-Y=-0.00110
X原码=-106Y原码=74则X-Y=-180X-Y原码=10110100(符号位溢出)X-Y补码=11001100
x【原】=11110x【补】=10010y【原】=10101y【补】=11011x【补】+y【补】=01101结果有溢出