解:设4x2-3y2-6z2-7=0…① x2+2y2-7z2+1=0…② ①-②x4:-11y2+22z2-11=0 ∴2z2=y2+1 ∴y2+3=2z2+2…③ ①x2+②x3:11x2-33z2-11=0 ∴x2=3z2+1 ∴x2+2=3z2+3…④ 将③、④代入t ∴t=[5(3z2+3)2+2(2z2+2)2-(z2+1)2]/[2(3z2+3)2-3(2z2+2)2-10(z2+1)2] =52(z2+1)2/-4(z2+1)2=-13 |
解:设4x2-3y2-6z2-7=0…① x2+2y2-7z2+1=0…② ①-②x4:-11y2+22z2-11=0 ∴2z2=y2+1 ∴y2+3=2z2+2…③ ①x2+②x3:11x2-33z2-11=0 ∴x2=3z2+1 ∴x2+2=3z2+3…④ 将③、④代入t ∴t=[5(3z2+3)2+2(2z2+2)2-(z2+1)2]/[2(3z2+3)2-3(2z2+2)2-10(z2+1)2] =52(z2+1)2/-4(z2+1)2=-13 |