已知x,y满足y=8-2x 分之 根号 x2-16

2x-3y-1=0x-2y+2=0∴x=8y=5(3+y²)/(2x-5)是不是3+y²还是3y²=(3+25)/11=28/11

实数x,y满足√2x-3y-1+|x-2y+2|=0因为√2x-3y-1与|x-2y+2|都≥0所以,2x-3y-1=0x-2y+2=0y=5,x=84y/(4x-5)=20/(32-5)=20/27

2x-3y-1=0x-2y+2=02x-4y+4=0y-5=0y=55y=25x=82x-4=125y/(2x-4)=25/12

∵正数x、y，满足8x+1y=1，∴x+2y=(x+2y)(8x+1y)=10+xy+16yx≥10+2xy×16yx=18．当且仅当x＞0，y＞0，8x+1y＝1，xy＝16yx，解得x=12，y=

根号(x+y-8)+根号(8-x-y)=根号(3x-y-4)+根号(x-2y+7),根据二次根式有意义得：X+Y-8≥0,8-X-Y≥0,∴X+Y≥8,X+Y≤8,∴X+Y=8,左边为0,右边两个非负

已知x,y满足x^2+y^2+5/4=2x+y,求代数式xy/(x+y)x^2+y^2+5/4=2x+yx^2-2x+1+y^2-y+1/4=0(x-1)^2+(y-1/2)^2=0x-1=0,x=1

y＜√(x-1)+√(1-x)+1/2x-1≥0,1-x≥0,即x-1≤0∴x-1=0,x=1∴y＜√(x-1)+√(1-x)+1/2=1/2∴1-y＞0∴|1-y|/(y-1)=(1-y)/(y-1

x^2+y^2-4x-2y+5=0,即(x-2)^2+(y-1)^2=0所以x=2,y=1所以√(x+y)/√(8y-2x)=√3/2望采纳

已知x,y满足√（2x-3y-1）+√（x-2y+2）=0,则√（2x-3y-1）=0,且√（x-2y+2）=0∴{2x-3y-1=0x-2y+2=0解得{x=8y=5∴2x-4分之5y=12分之25

2分之3x+y=14分之5x-y=1,3x+y=2(1)5x-y=14(2)由(1)+(2)式得8x=16x=2y=-42x-3y+7=2*2-3*(-7)+7=32

有理数x,y满足1/2|x-2y|=-2|x-4|左右乘以2得：|x-2y|=-4|x-4|移项得：|x-2y|+4|x-4|=0绝对值的和相加为0,则必定是0+0=0即|x-2y|=0,4|x-4|

x平方+y平方-4x-2y+5=0(x-2)^2+(y-1)^2=0(x-2)^2=0(y-1)^2=0x=2y=1根号8y-2x分之（根号x）+y=(根号2)+1/根号(8*1-2*2)=[(根号2

∵分子x2-16≥0,16-x2≥0∴x2-16=0x=±4∵分母8-2x≠0x≠4∴x=-4y=（0+0-9）/【8-2*（-4）】=-9/16∴xy=-4*（-9/16）=9/4xy的平方根=±3

x^2+y^2+5/4=2x+y(x-1)^2+(y-1/2)^2=0x=1,y=1/2xy/(x+y)=1*(1/2)/(1+1/2)=1/3

y=√x^2-4+√4-x^2/(x-2)+3x^2-4>=04-x^2>=0x-2≠0解得：x=-2y=39x+8y=9×(-2)+8×3=6

因为-5|x-2y|≥08|y-1\2|≥0而-5|x-2y|+8|y-1\2|=0所以-5|x-2y|=08|y-1\2|=0x-2y=0y-1/2=0解得x=1y=1/2

将x^2+y^2+5/4=2x+y化为将x^2-2x+1+y^2-y+1/4=0即(x-1)^2+(y-1/2)^2=0上面的等式成立,则有x-1=0且y-1/2=0即x=1,y=1/2所以xy/(x

x^2+y^2-6x+8y+25=0x²-6x+9+y²+8y+16=0(x-3)²+(y+4)²=0x-3=0x=3y+4=0y=-4y^x=-4³

最小值0.5,1.5,-1最大值1,1,-1/3约束区域是一个三角形,把三角形的三个顶点代入.可以检验出最大值最小值.

等于0?(x+2y)^2-x-2y+1/4=0(x+2y-1/2)^2=0x+2y-1/2=0x+2y=1/2