Question

求这俩题解的过程

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Answer 1

变式3：令x=sint (-π/2≤t≤π/2)，那么y=(cost-2)/(2sint+3)，∴cost-2ysint=3y+2
而-√(1+4y²)≤cost-2ysint≤√(1+4y²)，∴-√(1+4y²)≤3y+2≤√(1+4y²)
∴(3y+2)²≤1+4y²，∴5y²+12y+3≤0，∴(-6-√21)/5≤y≤(-6+√21)/5
变式4：y=[2sin(x/2)cos(x/2)+sin²(x/2)+cos²(x/2)]*[cos²(x/2)-sin²(x/2)+1]
=[sin(x/2)+cos(x/2)]²*2cos²(x/2)
=2[sin(x/2)cos(x/2)+cos²(x/2)]²
=2(1/2*sinx+1/2*cosx+1/2)²
=1/2*(sinx+cosx+1)²
=1/2*[√2*sin(x+π/4)+1]²
∵-π/12≤x≤π/12，∴π/6≤x+π/4≤π/3，∴1/2≤sin(x+π/4)≤√3/2
∴(√2+2)/2≤√2*sin(x+π/4)+1≤(√6+2)/2
∴(3+2√2)/4≤1/2*[√2*sin(x+π/4)+1]²≤(5+2√6)/4
即(3+2√2)/4≤y≤(5+2√6)/4

Answer 2

变式3 x不等于 负二分之三 闭区间负一到一

Answer 3

我要傲娇的说，我不会

Answer 4

我帮你吧