(2x+1)/3=1-(x-1/5)

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/01 16:10:29

(2x+1)/3=1-(x-1/5)
(2x+1)/3=1-(x-1/5)

(2x+1)/3=1-(x-1/5)
两边同乘3得3-3x+3/5=2x+1所以计算x=13/25.应该是吧.

(1)1/4x-1/12x=1/3
x/6=1/3
x=2
(2)(x+1)/2-(x-1)/3=1
3(x+1)-2(x-1)=6
3x+3-2x+2=6
x=1
(3)2x+1/3-(x-5...

全部展开

(1)1/4x-1/12x=1/3
x/6=1/3
x=2
(2)(x+1)/2-(x-1)/3=1
3(x+1)-2(x-1)=6
3x+3-2x+2=6
x=1
(3)2x+1/3-(x-5)=3/2
2(2x+1)-6(x-5)=9
4x+2-6x+30=9
-2x=-23
x=23/2
(4)x+4/5-x+2/10=x-2/2
2(x+4)-(x+2)=5(x-2)
2x+8-x-2=5x-10
x-5x=-10-6
-4x=-16
x=4
(5)2.4-x-4/2.5=3/5x
12-2(x-4)=3x
12-2x+8=3x
-2x-3x=-20
-5x=-20
x=4
(6) x-2/3-x-1/2=x+1
2(x-2)-3(x-1)=6(x+1)
2x-4-3x+3=6x+6
-x-6x=6+1
-7x=7
x=-1
希望能解决您的问题。

收起

解九宫格题目(X为未知数)9 X 2 X 7 X 8 X XX X 4 X X 9 X 6 X1 3 X 5 X X X X 24 X X 8 5 X X 1 XX 8 9 4 X 1 6 7 XX 1 X X 3 6 X X 88 X X X X 2 X 3 6X 5 X 6 X X 9 X XX X 7 X 4 X 2 X 1 请问这道题咋做 :(x)X (x+1) X (x+2) X (x+3) =360 解下列方程!就三道!解下列方程.1、x/x-2 - 1-x²/x²-5x = 2x/x-3(x/x-2 - 1-x²不是一起的!)2、5x/x²+x-6 - 5-2x/x²-x-12=7x-10/x²-6x+8(5x/x²+x-6 - 5-2x不是一起的!)3、x-4/x-5 + x-8/x-9=x-7/x- 1.(1/X-1 )-2=X^-3X/X^-12.(X/X+1)^+5(X/X+1)+6=03.(8(X^+2X)/X^-1)+ 3(X^-1)/X^+2x x+5/x+1-3x-3/x+5=8x+28/x^2+4X-5-2(x+5)/(x+1)-(3x-3)/(x+5)=(8x+28)/(x^2+4X-5)-2 x/(x-2)-(1-x ² )/(x ² -5x+6)=(2-x)/(x-3) 解方程,急 *-----------------------------------------------*| 6 4 X | 8 X X | X X 5 || X X X | X X X | X 7 8 || X X X | X X X | X X X ||---------------+---------------+--------------- || X X X | X X X | 5 1 X || X X X | X 6 X | X X X || 8 X X | 3 5 X | 2 X X || 填九宫格帮帮忙.x x 6 x x 7 x x 98 x x x 3 x 1 x x 9 x x 6 x 5 x 3 x x x 3 x x x x 1 8x x x 9 x 1 x x x2 1 x x x x 6 x x x 6 x 7 x 3 x x 1 x x 9 x 2 x x x 47 x x 8 x x 5 x x x(x^2-1)+2x^2 *(x+1)-3x*(2x-5)=x(3x^2-4x)-28 求:(x-2)(x²-6x-9)-x(x-3)(x-5)其中x=三分之一.以及x²(x-1)-x(x²+x-1)其中x=二分之一的答案 计算(x*x+4x+5)/(x+2)-(x*x+6x+10)/(x+3)+1 5x-3(2x+1)=6x-4(5-3x)求x 5x-3(2x+1)=6x-4(5-3x)求x 2(5x-4)-3(x+6)=5(x-1)-x 求x. 麻烦解4道分式方程(20分)(x/2x-5)+(5/5-2x)=1(x+1/x*x-2x)-(1/x)=3/x-2(1-x/5-x)-(5-x/1-x)=8/x*x-6x+5(x+1/x-1)-(4/x*x-1)=1去掉括号,不管前面正负,都不变号~ 3/x-1=4/x 6/x+1=x+5/x(x+1) x(x平方-1)+2X平方(X+1)-3X(2X-5)帮个忙 已知1/√(x)+√(x)=√5,求√(x/x2+x+1)+√(x/x2-x+1)的值定义域 x>0,两边平方,x+2+1/x=5,x+1/x=3,x^2-3x+1=0,x^2+x+1-4x=0,x^2+x+1=4x,x^2-x+1-2x=0.x^2-x+1=2x,√[x/(x^2+x+1)]-√[x/(x^2-x+1)]=√[x/(4x)]-√[x/2x]=√(1/4)-√(1/2)