若x=1,y=2,求1/(x+y)+1/(x-y)+2x/(x^2+y^2)+4x^3/(x^4+y^4)的值!把数往里带我也会啊,可是老师说要化简的…我凌乱了…各位帮帮忙
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/04 05:19:45
![若x=1,y=2,求1/(x+y)+1/(x-y)+2x/(x^2+y^2)+4x^3/(x^4+y^4)的值!把数往里带我也会啊,可是老师说要化简的…我凌乱了…各位帮帮忙](/uploads/image/z/5181368-32-8.jpg?t=%E8%8B%A5x%3D1%2Cy%3D2%2C%E6%B1%821%2F%28x%2By%29%2B1%2F%28x-y%29%2B2x%2F%28x%5E2%2By%5E2%29%2B4x%5E3%2F%28x%5E4%2By%5E4%29%E7%9A%84%E5%80%BC%21%E6%8A%8A%E6%95%B0%E5%BE%80%E9%87%8C%E5%B8%A6%E6%88%91%E4%B9%9F%E4%BC%9A%E5%95%8A%EF%BC%8C%E5%8F%AF%E6%98%AF%E8%80%81%E5%B8%88%E8%AF%B4%E8%A6%81%E5%8C%96%E7%AE%80%E7%9A%84%E2%80%A6%E6%88%91%E5%87%8C%E4%B9%B1%E4%BA%86%E2%80%A6%E5%90%84%E4%BD%8D%E5%B8%AE%E5%B8%AE%E5%BF%99)
若x=1,y=2,求1/(x+y)+1/(x-y)+2x/(x^2+y^2)+4x^3/(x^4+y^4)的值!把数往里带我也会啊,可是老师说要化简的…我凌乱了…各位帮帮忙
若x=1,y=2,求1/(x+y)+1/(x-y)+2x/(x^2+y^2)+4x^3/(x^4+y^4)的值!
把数往里带我也会啊,可是老师说要化简的…我凌乱了…各位帮帮忙
若x=1,y=2,求1/(x+y)+1/(x-y)+2x/(x^2+y^2)+4x^3/(x^4+y^4)的值!把数往里带我也会啊,可是老师说要化简的…我凌乱了…各位帮帮忙
先化简,前两个分式先通分,即1/(x+y)+1/(x-y)通分,得到2x/(x^2-y^2);
再将得到的式子与第三个分式,即2x/(x^2+y^2),相加,先提取公因式2x再通分得到:4x^3/(x^4-y^4);
再将得到的式子与第四个分式,即4x^3/(x^4+y^4),相加,先提取公因式4x^3再通分得到:8x^7/(x^8-y^8);
最后,再将x=1,y=2代入上式,得到最终结果-8/255
希望对你有用
我知道了加我Q吧。1755694803 我发不出来
将x=1,y=2,代入就行了!我算出来是8/225
接着代入数据即可。
^什么意思
带入运算 1/(1+2)+1/(1-2)+2*1/(1+4)+4*1/(1+16)= -8/255
具体步骤,一步一步推得
1/x+y+1/x-y+2x/(x^2+y^2)+4x^3/(x^4+y^4)
=x-y+x+y/(x+y)(x-y)+2x/(x^2+y^2)+4x^3/(x^4+y^4)
=2x/x^2-y^2+2x/(x^2+y^2)+4x^3/(x^4+y^4)
=4x^3/(x^2)^2-(y^2)^2+4x^3/x^4+y^4
=4x^3/x^4-y^4+4x^3/x^4+y^4
=8x^7/x^8-y^8
x=1,y=2
∴原式=-8/225
望采纳!!