20%X 15%Y=1.25 X Y=7

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20%X 15%Y=1.25 X Y=7
已知x—xy=40,xy—y=—20,求代数式x—y和x+y—2xy的值

(x-xy)-(xy-y)=40-(-20)x-xy-xy+y=40+20x+y-2xy=60(x-xy)+(xy-y)=40+(-20)x-xy+xy-y=40-20x-y=20

24X15=(24+12)X10=( ) 36x15=(36+18)x10 =( ) 42x15=(42+21)x10=

我不知道诶,帮不了你的忙了!拜拜!

已知x-xy=8,xy-y=-9,求x+y-2xy的值

x-xy=8(1)xy-y=-9(2)则有(1)-(2):X-XY-XY+Y=X+Y-2XY=8-(-9)=17

有这样一道题,计算:(x-y)[(x+y)(x+y)-xy]-(x-y)[(x-y)(x-y)+xy]的值,其中x=20

原式=(x-y)(x²+2xy+y²-xy)-(x-y)(x²-2xy+y²+xy)=(x-y)(x²+xy+y²)-(x+y)(x&sup

.先化简,后计算.[2x(x²y-xy²)+xy(xy-x²)]/x²yx=20

[2x(x²y-xy²)+xy(xy-x²)]/x²y=[2x^3y-2x^2y^2+x^2y^2-x^3y]/x²y=x-y=4

求满足方程xy=20-3x+y的所有整数对(xy)

正确的解法如下:将已知变形为:xy+3x-y=20x(y+3)-(y+3)=17(x-1)(y+3)=17=1×17=-1×(-17)所以有四种情形:①(x-1)=1(y+3)=17得:x=2,y=1

已知xy^2=-2,求-xy(x^2y^5-xy^3-y)的值.

-xy(x^2y^5-xy^3-y)=-(xy^2)^3+(xy^2)^2+xy^2=-(-2)^3+4-2=8+4-2=10

xy'=y+xy的

xdy=(y+xy)dxdy/y=((1+x)/x)dxln|y|=ln|x|+x+cy=±e^(ln|x|+x+c)其中c是常数再问:真还不理解我们是选择题:y=cxe^xy=c+x-x^2y=cs

已知x-xy=40,xy-y=-20,求x-y和x+y-2xy的值.要有 过程,把你怎么想的写出来.

x-y=(x-xy)+(xy-y)=40+(-20)=20x+y-2xy=(x-xy)-(xy-y)=60希望满意!

x-xy=40,xy-y= -20,求代数式x-y和x+y-2xy

(x-xy)+(xy-y)=40-20x-xy+xy-y=20x-y=20(x-xy)-(xy-y)=40-(-20)x-xy-xy+y=60x+y-2xy=60

TMOD=0x15; TH0=0; TL0=0; TH1=(65536-2000)/256; TL1=(65536-20

这个TMOD=0x15;是高四位的M0=1低四位的M0=1,C/T=1;GATE\x05C/T\x05M1\x05M0\x05GATE\x05C/T\x05M1\x05M0M1M0工作方式计数器模式T

x^2+xy+x=36,y^2+xy+y=20,求x+y.

7或者-8再问:求过程^_^再答:两个等式两边相加

实数xy满足y>=1 y

答案:5.(用线性规划的知识解决)由y≥1,y≤2x-1作出可行域(∵直线x+y=m不确定,∴可行域暂时不确定,但不影响解题)∵目标函数z=x-y的最小值为-1∴y=x-z截距最大时,z最小,为-1,

已知x^2-xy=10,xy-y^2=20,求:

∵x^2-xy=10∵xy-y^2=20∴(x^2-xy)+(xy-y^2)=30∴x^2-y^2=30∵x^2-xy=10∵xy-y^2=20∴(x^2-xy)-(xy-y^2)=x^-2xy+y^

已知x-y=4xy,则2x+3xy-2yx-2xy-y

∵x-y=4xy,∴2x+3xy-2yx-2xy-y=2(x-y)+3xyx-y-2xy=8xy+3xy4xy-2xy=112.故答案为:112.

xy+x=20 xy+y=18

由题意得:X=Y+2.那么Y(Y+2)+Y+2=20(Y+2)×(Y+1)=20所以y=3那么x=5可待入xy+y=18就不对了.(Y+2)×(Y+1)=20,Y应该是-6,X是-4,答案就对了.X=

X²+2xy+y²/xy乘x²-2xy+y²/xy+y²=

X²+2xy+y²/xy乘x²-2xy+y²/xy+y²=(x+y)²/xy×(x-y)²/y(x+y)=(x+y)(x-y)&#

[(xy-2)(-xy-2)-4(xy-1)^2]除以(-xy),其中x=20,y=-25分之1

[(xy-2)(-xy-2)-4(xy-1)^2]除以(-xy)=[-x²y²+4-4(x²y²-2xy+1)]÷(-xy)=(-x²y²+

已知:x-xy=40,xy-y=-20,求代数式x-y和x+y-2xy的值.

x-y=(x-xy)+(xy-y)=40+(-20)=20x+y-2xy=(x-xy)-(xy-y)=60