作业帮关于x的方程[x(x+2)]^2+2cos[x(x+2)]=a 且（x+1)^2+2cos(x+1)=a有解,则所有实根之和为多少
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关于x的方程[x(x+2)]^2+2cos[x(x+2)]=a 且（x+1)^2+2cos(x+1)=a有解,则所有实根之和为多少
关于x的方程[x(x+2)]^2+2cos[x(x+2)]=a 且（x+1)^2+2cos(x+1)=a有解,则所有实根之和为多少
关于x的方程[x(x+2)]^2+2cos[x(x+2)]=a 且（x+1)^2+2cos(x+1)=a有解,则所有实根之和为多少
老子不晓叠 问你个哒哒去
补充楼上，关键的一步就是韦达定理，否则解不出来
x(x+2)=x+1
x²+x-1=0
x1+x2=-b/a=-1
The answer is -4.
Let f(y)=y^2+2cos(y). Then f(x(x+2))=f(x+1) (1)
has solutions. Compute f'=2y-2sin(y)
and f''=2-2cos(y)>=0.
Since f'(0)=0, we deduce f'>=0 for y>=0.
So f is in...
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The answer is -4.
Let f(y)=y^2+2cos(y). Then f(x(x+2))=f(x+1) (1)
has solutions. Compute f'=2y-2sin(y)
and f''=2-2cos(y)>=0.
Since f'(0)=0, we deduce f'>=0 for y>=0.
So f is increasing for y>=0.
On the other hand, f is an even function, i.e., f(-y)=f(y).
So f(u)=f(v) holds if and only if either u=v or u=-v. (2)
Applying (2) to (1), we find x(x+2)+(x+1)=0 or x(x+2)-(x+1)=0.
That is, x^2+3x+1=0 or x^2+x-1=0.
It is clear that each of them has two solutions.
Moreover, by putting them together we get 4 distinct solutiosn.
So the sum of all solutions are (-3)+(-1)=-4.
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