(1)在复数范围内解方程x²+x+4=0(2)已知x∈C,解方程z-2/z/=-7+4i(3)已知实数x,y,满足x/1-i+y/1-2i=5/1-3i,求x,y的值
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/27 18:53:59
![(1)在复数范围内解方程x²+x+4=0(2)已知x∈C,解方程z-2/z/=-7+4i(3)已知实数x,y,满足x/1-i+y/1-2i=5/1-3i,求x,y的值](/uploads/image/z/5897311-7-1.jpg?t=%EF%BC%881%EF%BC%89%E5%9C%A8%E5%A4%8D%E6%95%B0%E8%8C%83%E5%9B%B4%E5%86%85%E8%A7%A3%E6%96%B9%E7%A8%8Bx%26%23178%3B%2Bx%2B4%3D0%282%29%E5%B7%B2%E7%9F%A5x%E2%88%88C%2C%E8%A7%A3%E6%96%B9%E7%A8%8Bz-2%2Fz%2F%3D-7%2B4i%283%29%E5%B7%B2%E7%9F%A5%E5%AE%9E%E6%95%B0x%2Cy%2C%E6%BB%A1%E8%B6%B3x%2F1-i%2By%2F1-2i%3D5%2F1-3i%2C%E6%B1%82x%2Cy%E7%9A%84%E5%80%BC)
x͒n@_R%fnKח>|AQ}P"*DT Q
(1)在复数范围内解方程x²+x+4=0(2)已知x∈C,解方程z-2/z/=-7+4i(3)已知实数x,y,满足x/1-i+y/1-2i=5/1-3i,求x,y的值
(1)在复数范围内解方程x²+x+4=0(2)已知x∈C,解方程z-2/z/=-7+4i(3)已知实数x,y,满足x/1-i+y/1-2i=5/1-3i,求x,y的值
(1)在复数范围内解方程x²+x+4=0(2)已知x∈C,解方程z-2/z/=-7+4i(3)已知实数x,y,满足x/1-i+y/1-2i=5/1-3i,求x,y的值
1)x12=[-1±√(1²-4*1*4)]/2=-1/2±√15i/2
∴x1=-1/2+√15i/2 x2=-1/2-√15i/2
2)设z=a+bi ∵a+bi-2√(a²+b²)=-7+4i ∴b=4 a-2√(a²+16)=-7 => 4a²+64=a²+14a+49
=>3a²-14a+15=0
解得:a1=5/3 a2=3
∴原方程的解为:z1=5/3+4i z2=3+4i
3)∵x(1+i)/2+y(1+2i)/5=5(1+3i)/10 => (5x+2y)+(5x+4y)i=5+15i
∴5x+2y=5 5x+4y=15 【相减】=> 2y=10 => y=5 => x=(5-2y)/5=-1
∴ x=-1 y=5