方括号代表根号（1）[x+2][2x+3]-x=2(2)(x/x+1)平方-4x/x+1+3=0(3)解方程组:y-x=
2个回答
（1）[x+2][2x+3]-x=2
√(x+2)(2x+3)=2+x
等式两边平方,(x+2)(2x+3)=(2+x)²
2x²+7x+6=x²+4x+4
x²+3x+2=0
(x+2)(x+1)=0
x=-2,或x=-1
因为 x+2≥0,2x+3≥0.
所以,x≥-3/2
所以,x=-1
(2)(x/x+1)²-4x/x+1+3=0
等式两边同时乘以(x+1)²,原式变为：
x²-4x(x+1)+3(x+1)²=0
[x-3(x+1)][x-(x+1)]=0
即：x-3(x+1)=0
x=1/2
(3)解方程组:y-x=1,x²-xy-2y²=0
x²-xy-2y²=0
因式分解,有(x-2y)(x+y)=0
因为,y-x=1；即：x=y-1
所以：(x-2y)(x+y)=0
=>(y-1-2y)(y-1+y)=0
解之,y=-1,或y=1/2
当y=-1时,x=-1-1=-2
当y=1/2时,x=1/2-1=-1/2
(4)先化简,再求值：4/x²-4+2/x+2-1/x-2,其中x=√3-1
4/(x²-4)+2/(x+2)-1/(x-2)
=4/(x+2)(x-2)+2/(x+2)-1/(x-2)
=2[1/(x+2)-1/(x-2)]+2/(x+2)-1/(x-2)
=4/(x+2)-3/(x-2)
=[4(x-2)-3(x+2)]/(x+2)(x-2)
=1/(x+2)
将x=√3-1代入,有
原式=1/(x+2)
=1/(√3-1)
整理得=(√3+1)/2
(5)已知x=1/(√3-2),y=1/(√3+2),求x²y+xy²的值
x²y+xy²=xy（x+y）
=[1/(√3-2)+1/(√3+2)]/(√3-2)(√3+2)
=(√3+2)+(√3-2)
=2√3
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