关于x的方程2x-m=3(x-1)-搜答案-智慧作业帮

3

当m的取值满足什么条件时,关于x的方程[3/x]+[6/x-1]=x+m/x(x-1)不会产生增根两边乘x(x-1)3(x-1)+6x=x+m增根即公分母为0x(x-1)=0x=0,x=1x=0代入3

把x=-2代入方程-2（m+3）-m=0(是0吧)-2m-3-m=03m=-3m=-1

1/2一般无解的时候,要求x的系数为0再问：怎么带进去不对啊再答：那就是你题目写错了，是(x-2)/(x-3)=m/(x-3)+2么？再问：是啊，我只是没加括号再答：嗯最好加上括号，要不然不知道分母到

x-(x-m)/2=(2-x)/3x-x/2+m/2=2/3-x/3x/2+x/3=2/3-m/25x=4-3mx=(4-3m)/5∵x≥0∴(4-3m)/5≥04-3m≥03m≤4m≤4/3

x1+x2=3/2x1x2=m/21.△=9-8m>=0,∴m0,∴m>0∴0

∵sinQ+cosQ=(√3+1)/2sibQcosQ=m/2∴1+2xm/2=(√3+2)/2∴m=√3/2原式=(sin²Q-cos²Q)/(sinQ+COSQ)=sinQ-c

x/(x-3)=(2x-6+m)/(x-3);因为有增根；所以分子解出来的解使分母为0；即x=32x-6+m=x解集为x=3；6-6+m+3；m=3;很高兴为您解答,skyhunter002为您答疑解

x/(x-2)-2=m^2/(x-3)(4-x)/(x-2)=m^2/(x-3)(4-x)(x-3)/(x-2)(x-3)=m^2(x-2)/(x-3)(x-2)[x^2-(7-m^2)x+12-2m

3x=2x+13x-2x=1x=1把x=1代入x-1/2=m+x得1-1/2=m+1∴m=-1/2把x=1代入(x-1)/2=m+x得(1-1)/2=m+1∴m=-1

x^3-(2m+1)x^2+(3m+2)x-m-2=(x^3-x^2)-(2mx^2-2mx)+[(m+2)x-(m+2)]=x^2(x-1)-2mx(x-1)+(m+2)(x-1)=(x-1)(x^

两边乘x-3x-2x+6=mx=6-m无解则这是增根所以x-3=0x=3所以3=6-mm=3

通分,消去分母X-2,用X=2带入的X=-1

x²-3x=m²-m-2x²-3x+9/4=m²-m+1/4(x-3/2)²=(m-1/2)²得x-3/2=m-1/2x=m+1或x-3/2

2x^2-4x-3-m=2(x-1)^2-m-5=0所以2(x-1)^2=（m+5）由0

x^2+2mx+m^2-3x-m^2/3=0x^2+(2m-3)x=-2m^2/3x^2+(2m-3)x+(2m-3)^2/4=-2m^2/3+(2m-3)^2/4[x+(2m-3)/2]^2=(4m

2x=-4m+1-2

2x²-3x+m+1=0m

|x-2|+|x-3|=m当x≤2时,x-2≤0,x-3＜0,则原式可化为：-x+2-x+3=m,x=(5-m)/2当2＜x＜3时,x-2＞0,x-3＜0,则原式可化为：x-2-x+3=m,m=1当x

x/(x-3)=2-m/(3-x)等式两边同时去分母,可得：x=2x-6+m所以x=6-m又该方程有一个正数解所以x=6-m>0,m