1.已知f(x)=ax²+bx+c,f(0)=0,且f(x+1)=f(x)+x+1,试求f(x)的表达式.2.已知函数f(x),g(x)同时满足:g(x-y)=g(x)g(y)+f(x)f(y);f(-1)=-1,f(0)=0,f(1)=1,求g(1)g(2)g(3)的值. 帮我解下这两道题,步骤写详细点,我是初
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![1.已知f(x)=ax²+bx+c,f(0)=0,且f(x+1)=f(x)+x+1,试求f(x)的表达式.2.已知函数f(x),g(x)同时满足:g(x-y)=g(x)g(y)+f(x)f(y);f(-1)=-1,f(0)=0,f(1)=1,求g(1)g(2)g(3)的值. 帮我解下这两道题,步骤写详细点,我是初](/uploads/image/z/10044455-23-5.jpg?t=1.%E5%B7%B2%E7%9F%A5f%EF%BC%88x%EF%BC%89%3Dax%26%23178%3B%2Bbx%2Bc%2Cf%280%29%3D0%2C%E4%B8%94f%28x%2B1%29%3Df%28x%29%2Bx%2B1%2C%E8%AF%95%E6%B1%82f%28x%29%E7%9A%84%E8%A1%A8%E8%BE%BE%E5%BC%8F.2.%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%2Cg%28x%29%E5%90%8C%E6%97%B6%E6%BB%A1%E8%B6%B3%EF%BC%9Ag%28x-y%29%3Dg%28x%29g%28y%29%2Bf%28x%29f%28y%29%3Bf%28-1%29%3D-1%2Cf%280%29%3D0%2Cf%281%29%3D1%2C%E6%B1%82g%281%29g%282%29g%283%29%E7%9A%84%E5%80%BC.+++++++%E5%B8%AE%E6%88%91%E8%A7%A3%E4%B8%8B%E8%BF%99%E4%B8%A4%E9%81%93%E9%A2%98%2C%E6%AD%A5%E9%AA%A4%E5%86%99%E8%AF%A6%E7%BB%86%E7%82%B9%2C%E6%88%91%E6%98%AF%E5%88%9D)
1.已知f(x)=ax²+bx+c,f(0)=0,且f(x+1)=f(x)+x+1,试求f(x)的表达式.2.已知函数f(x),g(x)同时满足:g(x-y)=g(x)g(y)+f(x)f(y);f(-1)=-1,f(0)=0,f(1)=1,求g(1)g(2)g(3)的值. 帮我解下这两道题,步骤写详细点,我是初
1.已知f(x)=ax²+bx+c,f(0)=0,且f(x+1)=f(x)+x+1,试求f(x)的表达式.
2.已知函数f(x),g(x)同时满足:g(x-y)=g(x)g(y)+f(x)f(y);f(-1)=-1,f(0)=0,f(1)=1,求g(1)g(2)g(3)的值. 帮我解下这两道题,步骤写详细点,我是初三升高中,还没正式上课,尽量写得通俗易懂.
1.已知f(x)=ax²+bx+c,f(0)=0,且f(x+1)=f(x)+x+1,试求f(x)的表达式.2.已知函数f(x),g(x)同时满足:g(x-y)=g(x)g(y)+f(x)f(y);f(-1)=-1,f(0)=0,f(1)=1,求g(1)g(2)g(3)的值. 帮我解下这两道题,步骤写详细点,我是初
1.在f(x) = ax²+bx+c中取x = 0得c = f(0) = 0,故f(x) = ax²+bx.
于是f(x+1)-f(x) = a(x+1)²+b(x+1)-(ax²+bx) = a((x+1)²-x²)+b = a(2x+1)+b = 2ax+(a+b).
而由条件f(x+1)-f(x) = x+1,得x+1 = 2ax+(a+b),即(2a-1)x+(a+b-1) = 0.
因为对任意x都成立,有2a-1 = 0,a+b-1 = 0,解得a = b = 1/2.
因此f(x) = 1/2·x²+1/2·x = x(x+1)/2.
2.代入y = 0得g(x) = g(x)g(0)+f(x)f(0) = g(x)g(0).
若g(x)恒等于0,有0 = g(x-y) = g(x)g(y)+f(x)f(y) = f(x)f(y).
但代入x = y = 1得0 = 1,矛盾.因此存在a使g(a) ≠ 0.
于是由g(a) = g(a)g(0)得g(0) = 1.
代入x = y = 1得1 = g(0) = g(1)²+f(-1)² = g(1)²+1,即g(1)² = 0,故g(1) = 0.
代入x = 0,y = 1得g(-1) = g(0)g(1)+f(0)f(1) = 0.
代入y = -1得g(x+1) = g(x)g(-1)+f(x)f(-1) = -f(x),即有g(x) = -f(x-1).
代入y = 1得g(x-1) = g(x)g(1)+f(x)f(1) = f(x),即有g(x-2) = f(x-1).
于是g(x) = -g(x-2).
代入x = 2得g(2) = -g(0) = -1,而代入x = 3得g(3) = -g(1) = 0.
因此g(1) = 0,g(2) = -1,g(3) = 0.
注:实际上g(x) = cos(πx/2),f(x) = sin(πx/2)是一组满足条件的函数.
a=0.5 b=0.5 c=o,,,,把原式子代进去算就可以了