解方程:1⼀(x²+11x-8)+1⼀(x²+2x-8)+1⼀(x²-13x-8)=0

要有过程
2024-12-18 17:47:05
推荐回答(1个)
回答1:

1/(x^2+11x-8)+1/(x^2+2x-8)+1/(x^2-13x-8)=0(x^2-8+2x)(x^2-8-13x)+(x^2-8+11x)(x^2-8-13x)+(x^2-8+11x)(x^2-8+2x)=0(x^2-8)^2-11x(x^2-8)-26x^2+(x^2-8)^2-2x(x^2-8)-143x^2+(x^2-8)^2+13x(x^2-8)+22x^2=03(x^2-8)^2-147x^2=0(x^2-8)^2-49x^2=0(x^2+7x-8)(x^2-7x-8)=0(x+8)(x-1)(x-8)(x+1)=0x1=1, x2=-1, x3=8, x4=-8