x+x/根号（x²-1）=2根号2 求x，要具体过程。

显示全部楼层 · 2013-4-13 16:50:23

x2-1>0,x>1 或 x0所以x>0所以 x>1x+x/√(x2-1)=2√2x/√(x2-1)=2√2-xx=√(x2-1)*(2√2-x)x2=(x2-1)(8-4√2x+x2)x2=8x2-4√2x^3+x^4-8+4√2x-x2x^4-4√2x^3+6x^2+4√2x-8=0x^4-√2x^3-3√2x^3+6x^2+4√2x-8=0x^3(x-√2)-3√2x^2(x-√2)+4√2(x-√2)=0...

千问 · 2013-4-13 16:50:23

你好！x2-1>0 x>1 或 x 1令 x=secu，u∈(0,π/2)则原方程即 secu + secu / tanu = 2√21/cosu + 1/sinu = 2√2(sinu+cosu)/(sinucosu) = 2√2令 t ...