已知：x^2+y^2=1.z^+w^2=1.xz+yw=0.xy+zw=0.求x^2+z^2的值與y^2+z^2的值．

熱心網友

樓上好象作的有問題:請看:(x+z)^2+(y+w)^2-2(xz+yw)=2,由已知xz+yw=0,知(x+z)^2+(y+w)^2=0,應該(x+z)^2+(y+w)^2=2。解:∵x^+y^=1。。。。。。。。(1) z^+w^=1。。。。。。(2)xz+yw=0。。。。。(3) xy+zw=0。。。。。。(4)∴(1)-(2)得: x^+y^-( z^+w^)=0 (x-z)＾+2xz-{(y+w)＾-2yw}=0由(3)知: (x-z)＾=(y+w)＾ x＾+z＾-2xz=y＾+w＾+2yw x＾+z＾=y＾+w＾。。。。。。。。(5)(1)+(2)得: x^+y^+ z^+w^=2 x＾+z＾=2-(y＾+w＾)。。。。(6)(5)+(6)得: 2(x＾+z＾)=2 x＾+z＾=1將 x＾+z＾=1帶入(6)得 y＾+w＾=1。

熱心網友

x^2+z^2=1,y^2+z^2=1由己知得x^2+y^2+z^2+w^2=2, (x+z)^2+(y+w)^2-2(xz+yw)=2,由已知xz+yw=0,知(x+z)^2+(y+w)^2=0,故知x+z=0,y+w=0,z=-x,w=-y同理(x+y)^2+(z+w)^2=0,得y=-x,w=-z,因而有x^2=y^2=z^2=w^2故x^2+z^2=x^2+y^2=1,同理y^2+z^2=1.發現計算過程有錯,返回準備修改時,樓下己有了答案,我就不再修正了.