5分之1x=7
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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+
通分(2x+13)/(x²+13x+40)/(2x+13)/(x²+13x+42)因为x²+13x+40≠x²+13x+42所以2x+13=0x=-13/2
X+1分之X+2+X+7分之X+8=X+5分之X+6+X+3分之X+4等式两边各2个x常数和都为10所以都约去得1/x+7/x=5/x+3/x为恒等式所以x为不等于零的任意实数再问:������ڶ�Щ
X²-3X分之5X-7=X-1分之2+X-2分之3X²-3X分之5X-7=(X-1)(X-2)分之[2(X-2)+3(X-1)]X²-3X分之5X-7=(X²-
x/(1×3)+x/(3×5)+x/(5×7)+...+x/(2007×2009)=2008(x/2)(1-1/3+1/3-1/5+1/5-1/7+...+1/2007-1/2009)=2008(x/
x=-0.00369
5分之1(x-8)+4分之1x+7分之x-1=6-5分之23-x28(x-8)+35x+20(x-1)=840-28(23-x)28x-224+35x+20x-20=840-644+28x55x=44
x-4分之1=5分之3x=1/4+3/5x=17/208分之3+X=12分之11x=11/12-3/8x=13/24X-5分之2-7分之2=5分之3x=3/5+2/5+2/7x=1又2/7在右上角点击
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x+1分之x+2减去x+3分之x+4=x+5分之x+6减去X+7分之x+8方程两边分别通分,得[(x+2)(x+3)-(x+1)(x+4)]/[(x+1)(x+3)]=[(x+6)(x+7)-(x+8
x:7分之4=3分之1x=1/3×4/7x=4/218分之15:x=2分之3x=15/8÷3/2x=5/46分之1÷10分之9=5分之2:x5/27=2/5:xx=2/5÷5/27x=54/25
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/[(
35/127/124/151/18
我来再问:再答:再答:就这样再答:谢谢采纳
1*3分之x+3*5分之x+5*7分之x+...+2011*2013分之x=2012x/2(1-1/3)+x/2(1/3-1/5)+x/2(1/5-1/7)+...x/2(1/2011-1/2013)
等号前后同时分别计算减法得:(x+1)(x+3)分之2=(x+5)(x+7)分之2所以:(x+1)(x+3)=(x+5)(x+7)x^2+4x+3=x^2+12x+35x=-4
1/5(x+15)=1/2-1/3(x-7)1/5x+3=1/2-1/3x+7/31/5x+1/3x=3/6+14/6-18/68/15x=-1/6x=-1/6*15/8x=-5/16
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
1/(X-7)-1/(X-5)=1/(X-6)-1/(X-4)移项1/(X-7)+1/(X-4)=1/(X-6)+1/(X-5)通分(2x-11)/[(x-7)(x-4)]=(2x-11)/[(x-6
通分(x+5+x+8)/(x+5)(x+8)=(x+6+x+7)/(x+6)(x+7)(2x+13)/(x^2+13x+40)=(2x+13)/(x^2+13x+42)(2x+13)[1/(x^2+1