若y等于x加2分之根号x的平方减4加根号4减x的平方
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(x-3)^2+√(y+1)=0由非负数的性质:x=3,y=-1(x-1)^2/(y-2)=(3-1)^2/(-3)=-4/3
4.因为只有大于等于0的数才能被开根号,要使根号x-2和根号2-x都有意义,x必须等于2,当x等于2时,y等于0.所以x的平方加y的平方等于4.
解,x^2+根号2y=根号3;y^2+根号2x=根号3.因为x≠y,且x,y都为正数,所以x^2不等于y^2,即x^2=根号2x,y^2=根号2y,解得x,y一个等于0另一个等于2开三次方根,所以xy
即x=(√11+√7)/2——是这个吗?y=(√11-√7)/2x+y=√11xy=(√11+√7)(√11-√7)/4=(11-7)/4=1∴原式=(x+y)²-3xy=11-3=8
√x-√(1/x)=2两边平方x-2×√x×√(1/x)+1/x=2²x-2+1/x=4x+1/x=6所以(1+x²)/x=1/x+x²/x=1/x+x=6
y=2分之x的平方减7x加3
y=√2+√(x^2-2x+1/3)=√[(x-1)^2-2/3]+√2考虑到定义域要求:x^2-2x+1/3>=0所以x>=1+√6/3或者x
看看能理解吗
题有错吧,没负号吧
∵X²-2√2X+2+Y²+2√3X+3=0﹙X-√2﹚²+﹙Y+√3﹚²=0∴X=√2,Y=-√3∴ㄧX+Yㄧ=√2+√3
(x2-y2)/(x-y)2=(x+y)/(x-y)代入x、y得√3/6
x方+y方+5/4=2x+y(x方-2x+1)+(y方-y+1/4)=0(x-1)方+(y-1/2)方=0x=1,y=1/2xy/(x+y)=1/3
x²+y²-4x-2y+5=0x²-4x+4+y²-2y+1=0(x-2)²+(y-1)²=0x-2=0x=2y-1=0y=1√(x+y)/
根号下大于等于0x²-4>=0,x²>=44-x²>=0,x²
3∵x/y=1/2∴y=2x故(x^2+2xy+y^2)/(x^2-xy+y^2)=(x+y)^2/(x^2-xy+y^2)=(x+2x)^2/(x^2-x·2x+4x^2)=9x^2/(3x^2)=
y=√(2x-1)/(x²+x-2)=√(2x-1)/(x+2)(x-1)√(2x-1)>=0x>=1/2(x+2)(x-1)≠0x≠-2且x≠1综上x>=1/2且x≠1
证明:∵[√(x²+1)-1]²≥0.即x²+1-2√(x²+1)+1≥0.∴x²+2≥2√(x²+1).不等式两边同除以√(x²