1.
设t=2x+1,则x=(t-1)/2
f(t)=[(t-1)/2]^2-2*(t-1)/2
=t^2/4-t+1/4-t+1
=t^2/4-2t+1
f(x)=x^2/4-2x+1
f((√2￣ )=(√2￣)^2/4-2(√2￣)+1=3/2-2√2￣
2.
设f(x)=ax^2+bx+c
f(2)=f(4)说明二次函数f(x)关于x=(2+4)/2=3对称,x=3为其对称轴
又f(2)=f(4)＞f(3),说明函数图像开口向上,左半边递减,
f(5)=f(1)