x方+x分之5+x方-x分之3＝x方-1分之7-搜答案-神马作文网

5X/25x^2=1/5x,(9ab^2+6abc)/3a^2b=3ab(3b+2c)/3a^2b=(3b+2c)/a(9a^2+6ab+b^2)/(3a+b)=(3a+b)^2/(3a+b)=3a+

3/（x+1）

f(X)=1-X方分之1+X方

f(1/x)=1-x^2+1/x^2=-f(x)直接代入即可再问：能写的再具体点吗我不太明白谢谢再答：哦我把题目看错了你的题目是对的吗再问：嗯题目是对的加我Q1047763981这样说方便再答：我没有

1,去分母,等号两边同时乘以3x+33x=2x+12（3x+3）去括号移项,合并同类项,系数化为12.等号左右同乘x方-12（x+1）=4x=1,检验x=1时,方程分母为0,所以x=1是方程的增根,方

把X方—3X+1=0两边除以X得X-3+x分之一=0,x+x分之一=3,变成x方分之一+2+x方=3,变成x方分之一+x方=1,

x方-5x+1=0两边除以x得∴x-5+1/x=0x+1/x=5两边平方得x²+2+1/x²=25x²+1/x²=23

∵y分之x=5∴x=5y(4x²-5xy-6y²)分之(x²-2xy+3y²)=(100y²-25y²-6y²)分之(25y

题目表达不清楚

2-x分之x-x方+4x+4分之4-x方=2-x/x-x²+4x+4/4-x²=2-1-x²+4x+1-x²=2+4x-2x²

x四次方-6x三次方+9x二次方分之x二次方-3x=x(x-3)/x²（x²-6x+9）=(x-3)/x(x-3)²=1/x(x-3)=1/(x²-3x)如追加

x^2/(x^4+x^2+1)取倒数(x^4+x^2+1)/x^2=x^2+1+1/x^2=x^2+2+1/x^2-1=(x+1/x)^2-1=3^2-1=8(取倒数)所以x^2/(x^4+x^2+1

（x^2+2x+1）/(x^2-3x-4)=(x+1)^2/(x+1)(x-4)=(x+1)/(x-4)

把每个式子因式分解x的方+5x+4=（x+1）(x+4)x方-5x+6=(x-2)(x-3)………………

原式=[2x(x+y)-5x²]/[5x²(x+y)²]=(-3x²+2xy)/[5x²(x+y)²]=(-3x+2y)/[5x(x+y)&

=(x-4)/(x+1)(x-1)除以(x-4)(x+1)/(x+1)平方+1/(x-1)=(x-4)/(x+1)(x-1)乘以(x+1)平方/(x-4)(x+1)+1/(x-1)=(x-4)/(x+

x方-1分之x方+x化简2x(1+x)

(x-1)/(x^2+3x+2)+6/(2+x-x^2)-(10-x)/(4-x^2)=(x-1)/(x+1)(x+2)-6/(x+1)(x-2)-(x-10)/(x+2)(x-2)=[(x-1)(x

x/(x²+2x+1),(x-1)/(x²+x),1/(x²-1)式1：x/(x+1)²,式2：(x-1)/[x(x+1)],式3：1/[(x+1)(x-1)]

这个用换元法令a=x/(x²+2)则(x²+2)/x=1/aa+1/a=-10/33a²+10a+3=0(3a+1)(a+3)=0a=-1/3,a=-3x/(x²

x^2+3x+1=0x^2+1/x^2=((-3x-1)^1+1)/(-3x-1)=(9x^2+6x+2)/(-3x-1)=(-27x-9+6x+2)/(-3x-1)=7x^4+1/x^4=(x^2+