百度作业网拜托各位高手 超难数学题
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拜托各位高手 超难数学题
1．二次函数f(x)=px2+qx+r中实数p,q,r满足p/(m+2)+q/(m+1)+r/m=0,其中m>0,求证：
（1）pf(m/(m+1))
1．二次函数f(x)=px2+qx+r中实数p,q,r满足p/(m+2)+q/(m+1)+r/m=0,其中m>0,求证：
（1）pf(m/(m+1))
1．二次函数f(x)=px2+qx+r中实数p,q,r满足p/(m+2)+q/(m+1)+r/m=0,其中m>0,求证：
（1）pf(m/(m+1))r=-pm/(m+2)-qm/(m+1)
pf(m/(m+1))=p²m²/(m+1)²+qmp/(m+1)-p²m/(m+2)-qpm/(m+1)=-p²m/(m+2)(m+1)²0
g(0)=p²[-m/(m+2)-qm/p(m+1)]q/p≤-2(m+1)/(m+2)
显然q/p≤-2(m+1)/(m+2),与-(m+1)/(m+2)≤q/p矛盾
则当q/pf(x)=f(y)+f(x/y)
当x->y
[f(x)-f(y)]/(x-y)=[f(x/y)-f(1)]/y(x/y-1)
=>f'(x)=f'(1)/x
f(x)=ln|x|f'(1)+C
代入f(x1×x2)=f(x1)+f(x2) 得C=0,f(x)=ln|x|f'(1)
f(4)=1得,1=ln4f'(1)
则f(x)=ln|x|/ln4(显然f(x)(0,+无穷)为增)
f(3x+1)+f(2x-6)=ln|(3x+1)(2x-6)|/ln4≤3
=>|(3x+1)(2x-6)|≤64
=>3x²-8x-35≤0
=>-7/3≤x≤5
解2:
f(4)=1
f(16)=f(4)+f(4)=2
f(64)=f(16)+f(4)=3
f((3x+1)(2x-6))=f(3x+1)+f(2x-6)≤3=f(64)
因x∈(0,+无穷)为增,
又f((3x+1)(2x-6))=f(|(3x+1)(2x-6)|)
则只需|(3x+1)(2x-6)|≤64
解得(3x²-8x-35)(3x²-8x+29)≤0
=>-7/3≤x≤5(显然3x²-8x+29＞0)
（1）pf(m/(m+1))r=-pm/(m+2)-qm/(m+1)
pf(m/(m+1))=p²m²/(m+1)²+qmp/(m+1)-p²m/(m+2)-qpm/(m+1)=-p²m/(m+2)(m+1)²0
g(0)=p²[-m/(m+2)-qm/p(m+1)]q/p≤-2(m+1)/(m+2)
显然q/p≤-2(m+1)/(m+2),与-(m+1)/(m+2)≤q/p矛盾
则当q/pf(x)=f(y)+f(x/y)
当x->y
[f(x)-f(y)]/(x-y)=[f(x/y)-f(1)]/y(x/y-1)
=>f'(x)=f'(1)/x
f(x)=ln|x|f'(1)+C
代入f(x1×x2)=f(x1)+f(x2) 得C=0,f(x)=ln|x|f'(1)
f(4)=1得,1=ln4f'(1)
则f(x)=ln|x|/ln4(显然f(x)(0,+无穷)为增)
f(3x+1)+f(2x-6)=ln|(3x+1)(2x-6)|/ln4≤3
=>|(3x+1)(2x-6)|≤64
=>3x²-8x-35≤0
=>-7/3≤x≤5
解2:
f(4)=1
f(16)=f(4)+f(4)=2
f(64)=f(16)+f(4)=3
f((3x+1)(2x-6))=f(3x+1)+f(2x-6)≤3=f(64)
因x∈(0,+无穷)为增,
又f((3x+1)(2x-6))=f(|(3x+1)(2x-6)|)
则只需|(3x+1)(2x-6)|≤64
解得(3x²-8x-35)(3x²-8x+29)≤0
=>-7/3≤x≤5(显然3x²-8x+29＞0)