医治27m平方减一除以三的2m次方等于27则M等于多少
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![医治27m平方减一除以三的2m次方等于27则M等于多少](/uploads/image/f/2530433-65-3.jpg?t=%E5%8C%BB%E6%B2%BB27m%E5%B9%B3%E6%96%B9%E5%87%8F%E4%B8%80%E9%99%A4%E4%BB%A5%E4%B8%89%E7%9A%842m%E6%AC%A1%E6%96%B9%E7%AD%89%E4%BA%8E27%E5%88%99M%E7%AD%89%E4%BA%8E%E5%A4%9A%E5%B0%91)
m等于它的倒数m=1/mm^2=1m=正负1m平方加m减6,除以m减2m^2+m-6=(m+3)(m-2)(m+3)(m-2)/(m-2)=m+3除以m加3(m+3)/(m+3)=1除以m平方减3m加
原式=[m^4+4m²+4-4m²]/[m(m²-2m+2)]=[(m²+2)²-4m²]/[m(m²-2m+2)]=(m
解题思路:利用分式的运算求解。解题过程:过程请见图片。最终答案:略
(m²-2m+1)/(m²-1)÷[m-1-(m-1)/(m+1)]=(m-1)²/[(m-1)(m+1)]÷[(m+1)(m-1)-(m-1)]/(m+1)=(m-1)
(m的平方-3m+2)/(m的平方-1)=(m-1)(m-2))/[(m-1)(m+1)]=(m-2))/(m+1)再问:把m的平方-1换成m的平方-m再答:(m的平方-3m+2)/(m的平方-m)=
分子:(m+1)(m+3)(2m-m^2)=m(m+1)(m+3)(2-m)=-m(m+1)(m+3)(m-2)分母:(m^3+m^2)(m-2)(m+3)=m^2(m+1)(m-2)(m+3)进行约
(M^2-M)/(M+2/A)=(M^2-M^3)/(4-M^2)M(M-1)/(M+2/A)=M^2(1-M)/(4-M^2)1/(M+2/A)=-M/(4-M^2)-M(M+2/A)=(4-M^2
(m+2分之2m-m-2分之m)除以m的平方-4分之m=m*(2/(m+2)-1/(m-2))*(m^2-4)/m=(2/(m+2)-1/(m-2))*(m+2)(m-2)=2(m-2)-(m+2)=
4-m^2=(2+m)*(2-m)
m是方程x²-5x-1=0的根所以:m²-5m-1=0m²=5m+1m=(5±√29)/2(2m²-5m+1)/m²=(10m+2-5m+1)/(5m
判别式△=(2m+3)²-4(m-1)²=4m²+12m+9-4m²+8m-4=20m+5=5(4m+1)∵m
(m+n)÷(m²-n²)=1÷(m-n)
(2m-m²)/(m²-4m+4)=m(2-m)/(m-2)²=-m/(m-2)
原式=m+|(m-1)²|/(m-1)=m+(m-1)²/(m-1)=m+m-1=2m-1
12/(m^2-9)+2/(3-m)=[12-2(m+3)]/(m^2-9)=-2(m-3)/(m^2-9)=-2/(m+3).
答:(m^2-9)/(m^2-2m-3)=(m-3)(m+3)/[(m-3)(m+1)]=(m+3)/(m+1)=1+2/(m+1)再问:最后一步怎么得来的再答:(m+3)/(m+1)=(m+1+2)
[m/(m+3)-6/(m^2-9)]÷[2/(m-3)]=[m/(m+3)-6/(m+3)(m-3)]×[(m-3)/2]={[m(m-3)-6]/[(m+3)(m-3)]}×[(m-3)/2]=(
1/(49-m^2)÷(m^2-7m)=1/(49-m^2)(m^2-7m)=1/m(m+7)(m-7)(m-7)=1/[m(m+7)(m-7)^2]