推荐回答(4个)
1.∵根号(x-3)中,x-3必须大于等于0,∴x≥3
∵根号(5-x)是分母,所以5-x>0,∴x<5
∴3≤x<5
2,这只有可能是0+0=0的形式,
∴x-2y-3=0,2x-3y-5=0
解得:x=1,y=-1
所以,根号(x-8y)=根号9=3
3.∵三角形的任意两边都大于第三边
∴a-b+c>0 故:根号((a-b+c)²)=|a-b+c|=a-b+c
又∵c-a-b<0,所以|c-a-b|=-(c-a-b)=a+b-c
∴a-b+c-(a+b-c)=a-b+c-a-b+c=2c-2b
4.∵根号(x-1)的取值范围:x≥1
根号(1-x)的取值范围:x≤1
∴x=1,y=-4
所以x的Y方的平方根为:1的(-4)次方的平方根,所以等于1
5.∵2a-根号(a²-4a+4)=2a-|a-2|
又∵a=根号3<2
∴原式=2a+a-2=3a-2=3根号3-2
6.移项得:a²-4a+4+根号(b-2) =0
∴式子为(a-2)²+根号(b-2)=0
同题二,为0+0=0形式
所以a=2,b=2,
∴ab=4
根号ab=2
1.若根号下X-3 加 根号下5-X 分之1 有意义,求X的取值范围
【解】:因为根号内部不小于零,并且分母不为零;
所以X-3>0 5-X>0 解得 X>3 且 x<5 所以3<X<5
2.已知X、Y满足:根号下X-2y-3 加 |2x-3y-5|=0.。求二次根式根号下x-8y的值
【解】因为根号里面和绝对值都是正数。 所以X-2Y=3 2X-3Y=5【这样的方程组会解吧】
解得 x=1 y=-1 所以x-8y=9
3.在△ABC中,a、b、c是三角形的三边长,化简 根号下(a-b+c)² 减 2|c-a-b|
【解】 因为两边之和大于第三边,两边之差小于第三边。 并且根号内不小于0.绝对值不小于零。
所以 根号下(a-b+c)² 化简为b+c-a . 2|c-a-b|化简为2(b+a-c)
整理得 b+c-a-2b-2a+2c
=c-b-3a
4.已知,根号下x-1 加 根号下1-x =y+4,求x的Y方的平方根
【解】 由已知的 x=1 所以y=-4
所以x的Y方的平方根正负1
5.化简求值:2a 减 根号下a²-4a+4 。其中a=根号3
【解】 a²-4a+4 化简为(a-2)² 所以原式=2a-a+2=a+2
把a=根号3 带入 得 原式=2+根号3
6.已知:a² 加 根号下b-2 =4a-4 。求根号下ab的值
【解】 通过整理 得 a² -4a+4=(-) 根号下b-2
(a-2)² =(-) 根号下b-2
所以 a=b=2 所以根号下ab的值为2.
1.要使根式有意义则x-3>=0 ,x>=3
分母有意义不能等于0 ,5-x>0 x<5
所以 3=2.这种情况只有可能是0+0=0的形式,
所以x-2y-3=0,2x-3y-5=0
解得:x=1,y=-1
所以,(x-8y)^½=9^½=3
3..因为三角形的任意两边都大于第三边
所以a-b+c>0 故:根号(a-b+c)²=a-b+c
又因为c-a-b<0,所以|c-a-b|=-(c-a-b)=a+b-c
所以a-b+c-2(a+b-c)=3c-a-3b
4.要使根式有意义,则x-1≥0且1-x≥0
所以x=1,y=-4
1^(-4)=1
1的平方根为1.
5.a²-4a+4 化简后为-(a-2)
所以原式为2a+(a-2)=3a-2=3倍根号3-2
6.a² -4a+4+(b-2)^½=0
(a-2)²+(b-2)^½=0
所以a-2=0 ,a=2
b-2=0 ,b=2
根号下ab=(ab)^½=2
1.x>=3,x不等于0
2。=3
3。=3c-a-3b
4.=1
5.=3根号下3减2
6。=2
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