用换元法解方程(1)x+1分之x平方-5x+x(x-5)分之24(x+1)+14=0(2)x+1分之2(x平方+1)+x平方+1分之6(x+1)=7(3)x平方分之x4次方+2x+1+x分之x平方+1=2
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/05 20:14:07
![用换元法解方程(1)x+1分之x平方-5x+x(x-5)分之24(x+1)+14=0(2)x+1分之2(x平方+1)+x平方+1分之6(x+1)=7(3)x平方分之x4次方+2x+1+x分之x平方+1=2](/uploads/image/z/329892-60-2.jpg?t=%E7%94%A8%E6%8D%A2%E5%85%83%E6%B3%95%E8%A7%A3%E6%96%B9%E7%A8%8B%281%29x%2B1%E5%88%86%E4%B9%8Bx%E5%B9%B3%E6%96%B9-5x%2Bx%28x-5%29%E5%88%86%E4%B9%8B24%28x%2B1%29%2B14%3D0%282%29x%2B1%E5%88%86%E4%B9%8B2%28x%E5%B9%B3%E6%96%B9%2B1%EF%BC%89%2Bx%E5%B9%B3%E6%96%B9%2B1%E5%88%86%E4%B9%8B6%EF%BC%88x%2B1%29%3D7%283%29x%E5%B9%B3%E6%96%B9%E5%88%86%E4%B9%8Bx4%E6%AC%A1%E6%96%B9%2B2x%2B1%2Bx%E5%88%86%E4%B9%8Bx%E5%B9%B3%E6%96%B9%2B1%3D2)
用换元法解方程(1)x+1分之x平方-5x+x(x-5)分之24(x+1)+14=0(2)x+1分之2(x平方+1)+x平方+1分之6(x+1)=7(3)x平方分之x4次方+2x+1+x分之x平方+1=2
用换元法解方程(1)x+1分之x平方-5x+x(x-5)分之24(x+1)+14=0(2)x+1分之2(x平方+1)+x平方+1分之6(x+1)=7
(3)x平方分之x4次方+2x+1+x分之x平方+1=2
用换元法解方程(1)x+1分之x平方-5x+x(x-5)分之24(x+1)+14=0(2)x+1分之2(x平方+1)+x平方+1分之6(x+1)=7(3)x平方分之x4次方+2x+1+x分之x平方+1=2
解方程:(1) (x²-5x)/(x+1)+24(x+1)/x(x-5)+14=0;(2).2(x²+1)/(x+1)+6(x+1)/(x²+1)=7;
(3)(x⁴+2x+1)/x²+(x²+1)/x=2 .
(1) (x²-5x)/(x+1)+24(x+1)/x(x-5)+14=0
用x(x+1)(x-5)乘方程两边得:x²(x-5)²+24(x+1)²+14x(x+1)(x-5)=0
展开化简得:x⁴+4x³-7x²-22x+24=0
x³(x-1)+5x²(x-1)-2x(x-1)-24(x-1)=0
(x-1)(x³+5x²-2x-24)=0
(x-1)[x²(x-2)+7x(x-2)+12(x-2)]=0
(x-1)(x-2)(x²+7x+12)=0
(x-1)(x-2)(x+3)(x+4)=0
故x₁=1;x₂=2;x₃-3;x₄=-4.
(2).2(x²+1)/(x+1)+6(x+1)/(x²+1)=7
去分母得 2(x²+1)²+6(x+1)²=7(x+1)(x²+1)
展开化简得 2x⁴-7x³+3x²+5x+1=0
用待定系数法可分解为 (2x² -3x-1)(x²-2x-1)=0
由2x²-3x-1=0,得x₁=(3+√17)/4; x₂=(3-√17)/4;
由x²-2x-1=0,得x₃=(2+√8)/2=1+√2; x₄=1-√2.
(3)(x⁴+2x+1)/x²+(x²+1)/x=2 .
去分母得 x⁴+2x+1+x(x²+1)=2x²
展开化简得 x⁴+x³-2x²+3x+1=0
(这个有点难,今天要休息了,明天再做.)