x-2分之1+X+2分之6=1
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![x-2分之1+X+2分之6=1](/uploads/image/f/890069-5-9.jpg?t=x-2%E5%88%86%E4%B9%8B1%2BX%2B2%E5%88%86%E4%B9%8B6%3D1)
若原题是:5/2x+3=1/3+7x移项得5/2x-7x=1/3-3合并同类项得-9/2X=-8/3两边同乘以-2/9,得x=16/27.若原题是:(5x+3)/2=(1+7x)/3去分母得3(5x+
6分之1X+3分之2X=80两边乘以65x=80*6x=96
第一题的过程如下:1/(x-2)+1/(X-6)=1/(X-7)+1/(X-1)1/(X-2)-1/(X-1)=1/(X-7)-1/(X-6)[X-1-(X-2)]/[(X-1)(X-2)]=1/[(
3/4x-2/3x=1/6,解得x=215*(1/15+2/17)*17=4715X(1/151+2/17)X17=15x17x(1/15+2/17)=15x17x1/15+15x17x2/17=17
2/3x+1/6x=155/6x=15x=18
理解为如下等式:2/x+6/(x+1)=18/[x(x+1)]2(x+1)+6x=182x+2+6x=188x=16x=2
(x-2分之1)-(x-1分之1)=(x-7分之1)-(x-6分之1)(x-2分之1)*(x-1分之1)分之1=(x-7分之1)*(x-6分之1)分之1(x-2)(x-1)=(x-7)(x-6)x=4
(1)2x/(x+1)=12x=x+1x=1(2)x/(x-2)-1/(x²-4)=1x(x+2)-1=x²-4x²+2x-1=x²-42x=-3x=-3/2(
化简等式得:1-1/(X+6)+1-1/(x+3)=1-1/(x+2)+1-1/(x+7),整理得1/(x+6)-1/(x+7)=1/(x+2)-1/(x+3);1/[(x+6)(x+7)]=1/[(
4x(/x-2)-1=3/(2-x)4x(/x-2)-3/(2-x)=14x/(x-2)+3/(x-2)=1(4x+3)/(x-2)=14x+3=x-23x=-5x=-5/3,经检验不是增根x/(x-
我来再问:再答:再答:就这样再答:谢谢采纳
(x/2)+(x/6)+(x/12)+(x/20)+(x/30)+(x/42)=x/(1*2)+x/(2*3)+x/(3*4)+x/(4*5)+x/(5*6)+x/(6*7)=x-x/2+x/2-x/
(x-3)分之x-(x^2-3x)分之(x+6)+x分之1等于几?=x²/x(x-3)-(x+6)/x(x-3)+(x-3)/x(x-3)=(x²-x-6+x-3)/x(x-3)=
1)去分母得2(x-1)+3(x+1)=62x-2+3x+3=6∴x=1经检验:x=1是增根∴方程无解2)设y=(x+1)/x,∴x/(x+1)=1/y∴原方程可化为y+5/y=6∴y平方-6y+5=
3分之2x+25%x=6分之12/3x+1/4x=1/68x+3x=211x=2x=2/11
再问:再答:
(x+1)/(x+2)-1/(x+7)=(x+2)/(x+3)-1/(X+6)(X+2-1)/(X+2)-1/(X+7)=(X+3-1)/(X+3)-1/(X+6)1-1/(X+2)-1/(X+7)=
x+2分之x+1-x+3分之x+2=x+6分之x+5-x+7分之x+61-(x+2)分之1-[1-(x+3)分之1]=1-(x+6)分之1-【1-(x+7)分之1】从而(x+3)分之1-(x+2)分之
本人一次只回答一个算了吧或者你分开提问x/(x+2)=x/(x-1)两边乘(x+2)(x-1)x²-x=x²+2x3x=0x=0经检验,x=0是方程的解