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(2)f(x)的定义域为(-2,2)
令x,y满足-2<x<y<2
f(x)-f(y)=log[a,(2-x)]-log[a,(2+x)]-log[a,(2-y)]+log[a,(2+y)]
=log[a,(2-x)(2+y)/(2+x)(2-y)]
=log{a,[4-xy+2(y-x)]/[4-xy+2(x-y)]}
因为y-x>0,x-y<0,所以4-xy+2(y-x)>4-xy+2(x-y)>0
[4-xy+2(y-x)]/[4-xy+2(x-y)]>1
当0<a<1时,log{a,[4-xy+2(y-x)]/[4-xy+2(x-y)]}<0
f(x)<f(y),即函数单调递增
当a>1时,log{a,[4-xy+2(y-x)]/[4-xy+2(x-y)]}>0
f(x)>f(y),即函数单调递减
令x,y满足-2<x<y<2
f(x)-f(y)=log[a,(2-x)]-log[a,(2+x)]-log[a,(2-y)]+log[a,(2+y)]
=log[a,(2-x)(2+y)/(2+x)(2-y)]
=log{a,[4-xy+2(y-x)]/[4-xy+2(x-y)]}
因为y-x>0,x-y<0,所以4-xy+2(y-x)>4-xy+2(x-y)>0
[4-xy+2(y-x)]/[4-xy+2(x-y)]>1
当0<a<1时,log{a,[4-xy+2(y-x)]/[4-xy+2(x-y)]}<0
f(x)<f(y),即函数单调递增
当a>1时,log{a,[4-xy+2(y-x)]/[4-xy+2(x-y)]}>0
f(x)>f(y),即函数单调递减
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