(-2)^2/m^2 +(根号3)^2/4=1 求m=?这道题怎么化简?
1个回答
(-2)^2/m^2 +(根号3)^2/4=1
=>4/m^2+3/4=1
=>4/m^2=1/4
=>m^2=16
=>m=±4
相关问题
化简求值:(-m^4/m^2)^2+(-2m)^3*m^2+(-m^2)^4/m^4,其中m=-1
化简求值:[(3m/(m+2)-m/(m-2)]/[(2m)/m^2-4)],其中m=2/3
化简4^n+2-2^2m+3/4^m-1
化简:m/(m-n)根号(m^3-2m^2n+mn^2)(m
化简2m*2m-4(m-2)(2m-3)
化简并求值(m-2)(m²+2m+4)-2m(m²-3),其中m=2
2m^2+m-1怎么化简
化简:[(4m)/(m^2-4)+(m-2)/(m+2)]/1/(m^2-4)
问您一道题:化简“根号(3-2根号2)+根号(3+2根号2)!
化简求值题,先化简,再求值.m-4/m^2-9×(1+ 14m-7 / m^2-8m+16)÷1/ m-3.其中m=5用