x y=420,30%x 40%y=160X80%-搜答案-大师作文网

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

1.2(Xy+Xy)-3(Xy-xy)-4Xy=2*2xy-0-4xy=4xy-4xy=02.1/2ab-5aC-(3acb)+(3aC-4aC)=1/2ab-5ac-3acb-ac=1/2ab-6a

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

Solution:x^2-2xy+3y^2=9(1)4x^2-5xy+6y^2=30(2)Eq(1)*5-Eq(2)*2,wehavex^2-y^2=5(3)Eq(2)-Eq(1),wegetx^2+

120%x+7=19x=1025+60%x40=265

X-Y=5,XY=3.XY是多少?

Y=X-5XY=X²-5X=3X²-5X-3=0X=(5±√37)/2Y=X-5X=(5-√37)/2,Y=(-5-√37)/2X=(5+√37)/2,Y=(-5+√37)/2

★★★1/7y是(1/7)y还是1/(7y)；电脑打出来的这种形式请多加一个括号.(11-y-1/7y)x40%+36%y+1/7yx35%=11x38.5%再问：7/1y都是（7/1）y再答：★白天

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

（1）∵xy+x=-1①，xy-y=-2②，∴①-②得x+y=1；（2）先把xy+x=-1，xy-y=-2的值代入代数式，得原式=-x-[2y-1+3x]+2[x+4]=-x-2y+1-3x+2x+8

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,

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

xy+1/xy>=2√（xy*1/xy）=2(当xy=1/xy即xy=1时取等号)x/y+y/x>=2√（x/y*y/x）=2(当x/y=y/x即x=y取等号)当x=y=1时可以同时满足两项的等号要求

原式可变为x(1+y)=30-y,显然y不等于-1,所以x=(30-y)/(y+1)=31/(y+1)-1,所以y+1能整除31,因此y+1可取1,31,-1,-31,代入可得x分别等于30,0,-3

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

89

(1-x)(200+400x)=224(x-1)(25+50x)=-2850x²+25x+3=0(5x+1)(10x+3)=0x=-1/2,x=-3/10

X+y=3 xy=-5 Xy=?

X+y=3xy=-5X-y=?(x-y)^2=(X+y)^2-4xy=9+20=29则x-y=±根号29

最大值14程序如下:p=0q=0for(i=1,i

x+2y+xy=30?x+2y≥2√(x×2y)=2√2√(xy)30=x＋2y＋xy≥2√2√(xy)+xy所以xy+2√2√(xy)-30≤0,令a=√(xy),显然a≥0：a²+(2√

原式=-xy²（x²y^4-xy²-1）∵xy²=-2原式=2（（-2)²-（-2）-1）=10