因式分解,就是二次方程,急额.谢1,x^2-6x+9=02,(5x-3)^2=16(x-1)^23,(x+√2)（2x+√3）=√6注：√是根号
问题描述：
因式分解,就是二次方程,急额.谢
1,x^2-6x+9=0
2,(5x-3)^2=16(x-1)^2
3,(x+√2)（2x+√3）=√6
注：√是根号
答
1，x^2-6x+9=0,
(x-3)^2=0,
x1=x2=3;
2，(5x-3)^2=16(x-1)^2
(5x-3)^2-[4(x-1)]^2=0,
(5x-3+4x-4)(5x-3-4x+4)=0,
(9x-7)(x+1)=0,
x1=7/9,x2=-1;
3,(x+√2)（2x+√3）=√6
2x^2+(√3+2√2)x+√6=√6,
2x^2+(√3+2√2)x=0,
x(2x+√3+2√2)=0,
x1=0,x2=-(√3+2√2)/2.
答
1,x^2-6x+9=0 (x-3)^2=0x-3=0x=32,(5x-3)^2=16(x-1)^2 (5x-3)^2-[4(x-1)]^2=0(5x-3)^2-(4x-4)^2=0(5x-3+4x-4)(5x-3-4x+4)=0(9x-7)(x+1)=0x=7/9,x=-13,(x+√2)（2x+√3）=√6 2x^2+2√2x+√3x+√6=√62x^2+(2√2+√3)...