帮忙解一下此方程(2k^2+14k)^2-4(k^2+1)(k^2+14k+24)=0,

2个回答

  • (2k^2+14k)^2-4(k^2+1)(k^2+14k+24)=0

    4(k^2+7k)^2-4(k^2+1)(k^2+14k+24)=0

    4(k^2+7k)^2-4(k^2+1)[(k^2+1)+(14k+23)]=0

    (k^2+7k)^2-4(k^2+1)^2-(k^2+1)(14k+23)=0

    [(k^2+7k)^2-(k^2+1)^2]-(k^2+1)(14k+23)=0

    [(k^2+7k-k^2-1)(k^2+7k+k^2+1)]-(k^2+1)(14k+23)=0

    (7k-1)(2k^2+7k+1)-(k^2+1)(14k+23)=0

    (7k-1)[(k^2+1)+(k^2+7k)]-(k^2+1)(14k+23)=0

    (7k-1)(k^2+1)+(7k-1)(k^2+7k)-(k^2+1)(14k+23)=0

    (k^2+1)[(7k-1)-(14k+23)]+(7k-1)(k^2+7k)=0

    (k^2+1)(7k-1-14k-23)+(7k-1)(k^2+7k)=0

    -(k^2+1)(7k+24)+(7k-1)(k^2+7k)=0

    -(k^2+1)(7k-1+25)+(7k-1)(k^2+7k)=0

    -(k^2+1)(7k-1)-25(k^2+1)+(7k-1)(k^2+7k)=0

    (7k-1)(k^2+7k-k^2-1)-25(k^2+1)=0

    (7k-1)(7k-1)-25(k^2+1)=0

    49k^2-14k+1-25k^2-25=0

    24k^2-14k-24=0

    12k^2-7k-12=0

    (3k-4)(4x+3)=0

    k=4/3或k=-3/4