早教吧作业答案频道 -->其他-->
2(x-1)的平方加上∣2y-1∣=－∣Z+1∣求X+y+Z的值
题目详情
2(x-1)的平方加上∣2y-1∣=－∣Z+1∣求X+y+Z的值
▼优质解答
答案和解析
∵2(x-1)²+∣2y-1∣=－∣Z+1∣,
∴2(x-1)²+∣2y-1∣+∣Z+1∣=0
又∵2(x-1)²≥0,∣2y-1|≥0,∣Z+1∣≥0,
∴x-1=0,x=1;
2y-1=0,y=1/2;
Z+1=0,z=-1.
∴x+y+z=1/2
∴2(x-1)²+∣2y-1∣+∣Z+1∣=0
又∵2(x-1)²≥0,∣2y-1|≥0,∣Z+1∣≥0,
∴x-1=0,x=1;
2y-1=0,y=1/2;
Z+1=0,z=-1.
∴x+y+z=1/2
看了2(x-1)的平方加上∣2y-...的网友还看了以下：
设x、y、z为实数,且(y-z)^2+(x-y)^2+(z-x)^2=(y+z-2x)^2+(x+ 2020-06-12 …
若z^2+z+1=0,求（1+z）*（1+z^2）*（1+z^4）*（1+z^8）…（1+z^1若 2020-06-30 …
方程组(xy+x)/(x+y+1)=2(xz+2x)/(z+y+2)=3(y+1)(z+2)/(z 2020-08-01 …
1..设x/a+y/b+z/c=1,a/x+b/y+c/z=0,求x*2/a*2+y*2/b*2+z 2020-10-30 …
x、y、z属于正实数,且xyz=1,求1/(x^2(y+1)+1)+1/(y^2(z+1)+1)+1 2020-10-31 …
x,y,z>0x^2/(1+x^2)+y^2/(1+y^2)+z^2/(1+z^2)=2求证x/(1 2020-10-31 …
1.x+y+z≠0且x/(y+z)=y/(x+y)=z/x+y,求x/(x+y+z)2.x+y+z= 2020-10-31 …
4x+8x^3+1+2x^2如何求得答案(4x+1)(2x-1)9(x-1)^2-25(x+1)^2 2020-10-31 …
计算∫∫e^z/√(x^2+y^2),其中Σ是z=√(x^2+y^2）在1原式=∫∫e^z/zdxd 2020-11-01 …
已知道2[√X+√(Y-1)+√(Z-2)]=X+Y+Z,求X,Y,Z2[√X+√(Y-1)+√(Z 2020-11-01 …