Question

4道初二数学题求解，要过程，谢谢

3x^2-7x+3=0(配方法)

x^2+2(√2 -1)x+3-2√2=0

x^2-x-k^2+k=0
(x-1)^2+2(1-x)=15

Answer 1

3x^2-7x+3=0(配方法)

x²-7x/3=-1

x²-7x/3+(7/6)²=-1+49/36

(x-7/6)²=13/36

x-7/6=±√13/6

x1=(7+√13)/6     x2=(7-√13)/6     

x^2+2(√2 -1)x+3-2√2=0

(x+√2-1)²=0

x+√2-1=0

x=1-√2

x²-x-k²+k=0

a=1   b=-1   c=-k²+k

Δ=1-4(-k²+k)=4k²-4k+1=(2k-1)²

代入求根公式得

x=[1±√(2k-1)²]/2=[1±(2k-1)]/2

x1=k    x2=1-k

(x-1)^2+2(1-x)=15

(x-1)²+2(1-x)-15=0

(x-1+5)(x-1-3）=0

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

x1=-4   x2=4