f（x+1）=x平方减3x加2,求f（庚号2）的值.

令x+1=√2
则x=√2-1
所以f(√2)=(√2-1)²-3(√2-1)+2
=(3-2√2)-(3√2-3)+2
=8-5√2

a²+c²-b²=6/5ac
cosB=(a²+c²-b²)/2ac=6/5)ac/2ac=3/5
00
sin²B+cos²B=1
所以 sinB=4/5
sin2B=2sinBcosB=24/25
2sin²[(A+C)/2]+sin2B
=2*[1-cos(A+C)]/2+sin2B
=1-cos(180-B)+sin2B
=1+cosB+sin2B
=1+3/5+24/25
=64/25