解方程2x^2-2根号2x+1 用因式分解法解

2个回答

  • 解;① (2x+3)^2=(5x-4)^2

    (2x+3+5x-4)(2x+3-5x+4)=0 x1=1/5 x2=7/3

    ..② (x-根号下2)^2-2(根号下2-x)=0

    (x-根号下2)(x-根号下2+2)=0 x1=根号2 x2=根号2-2

    ③ 4y^2+8y+4=0

    4(y-1)^2=0 y1=y2=1

    ..④ (x-5)(x+3)+x(x+6)=-17

    2x^2+4x+2=0 (x+1)^2=0 x1=x2=-1

    ..⑤ (3-x) ^2 =9-x ^2

    (3-x) ^2-(3-x)(3+x)=0

    (3-x)(3-x-3-x)=0 x1=3 x2=0

    ⑥(t-3)(t+4)=-12 t^2+t=0 t(t+2)=0 t1=0 t2=-2

    ..⑦ (x+5)(x+3)+x(x+6)=-17 2x^2+14x+32=0 x^2+7x+16=0 方程无解

    ..⑧x^2-2ax+a^2-b^2=0 [x-(a+b)][x-(a-b)]=0

    x1=a+b x2=a-