1/2x=2/x+3
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4x=x+3
3x=3
x=1
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ç»æ£éªï¼x=1æ¯æ¹ç¨ç解
x/(x+1)=2x/(3x+3)+1
两边ä¹3(x+1)
3x=2x+(3x+3)
3x=5x+3
2x=-3
x=-3/2
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ç»æ£éªï¼x=-3/2æ¯æ¹ç¨ç解
2/x-1=4/x^2-1
两边ä¹(x+1)(x-1)
2(x+1)=4
2x+2=4
2x=2
x=1
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æ以åæ¹ç¨æ 解
5/x^2+x - 1/x^2-x=0
两边ä¹x(x+1)(x-1)
5(x-1)-(x+1)=0
5x-5-x-1=0
4x=6
x=3/2
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ç»æ£éªï¼x=3/2æ¯æ¹ç¨ç解
5x/(3x-4)=1/(4-3x)-2
ä¹3x-4
5x=-1-2(3x-4)=-1-6x+8
11x=7
x=7/11
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x=7/11æ¯æ¹ç¨ç解
1/(x+2) + 1/(x+7) = 1/(x+3) + 1/(x+6)
éå
(x+7+x+2)/(x+2)(x+7)=(x+6+x+3)/(x+3)(x+6)
(2x+9)/(x^2-9x+14)-(2x+9)/(x^2+9x+18)=0
(2x+9)[1/(x^2-9x+14)-1/(x^2+9x+18)]=0
å 为x^2-9x+14ä¸çäºx^2+9x+18
æ以1/(x^2-9x+14)-1/(x^2+9x+18)ä¸çäº0
æ以2x+9=0
x=-9/2
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x=-9/2æ¯æ¹ç¨ç解
7/(x^2+x)+1/(x^2-x)=6/(x^2-1)
两边åä¹x(x+1)(x-1)
7(x-1)+(x+1)=6x
8x-6=6x
2x=6
x=3
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ç»æ£éªï¼x=3æ¯æ¹ç¨ç解
åç®æ±å¼ã[X-1-ï¼8/X+1ï¼]/[X+3/X+1] å
¶ä¸X=3-æ ¹å·2
[X-1-ï¼8/X+1ï¼]/[(X+3)/(X+1)]
={[(X-1)(X+1)-8]/(X+1)}/[(X+3)/(X+1)]
=(X^2-9)/(X+3)
=(X+3)(X-3)/(X+3)
=X-3
=-æ ¹å·2
8/(4x^2-1)+(2x+3)/(1-2x)=1
8/(4x^2-1)-(2x+3)/(2x-1)=1
8/(4x^2-1)-(2x+3)(2x+1)/(2x-1)(2x+1)=1
[8-(2x+3)(2x+1)]/(4x^2-1)=1
8-(4x^2+8x+3)=(4x^2-1)
8x^2+8x-6=0
4x^2+4x-3=0
(2x+3)(2x-1)=0
x1=-3/2
x2=1/2
代å
¥æ£éªï¼x=1/2使å¾åæ¯1-2xå4x^2-1=0ãèå»
æ以åæ¹ç¨è§£:x=-3/2
(x+1)/(x+2)+(x+6)/(x+7)=(x+2)/(x+3)+(x+5)/(x+6)
1-1/(x+2)+1-1/(x+7)=1-1/(x+3)+1-1/(x+6)
-1/(x+2)-1/(x+7)=-1/(x+3)-1/(x+6)
1/(x+2)+1/(x+7)=1/(x+3)+1/(x+6)
1/(x+2)-1/(x+3)=1/(x+6)-1/(x+7)
(x+3-(x+2))/(x+2)(x+3)=(x+7-(x+6))/(x+6)(x+7)
1/(x+2)(x+3)=1/(x+6)(x+7)
(x+2)(x+3)=(x+6)(x+7)
x^2+5x+6=x^2+13x+42
8x=-36
x=-9/2
ç»æ£éªï¼x=-9/2æ¯æ¹ç¨çæ ¹ã
(2-x)/(x-3)+1/(3-x)=1
(2-x)/(x-3)-1/(x-3)=1
(2-x-1)/(x-3)=1
1-x=x-3
x=2
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