Ответы на вопрос:
4/(3m+1)+(3m-7)/(27m^3+1)=(1-6m)/(1-3m+9m^2) 4/(3m+1)+(3m-7)/(3m+1)*(9m^2-3m+1)=(1-6m)/(1-3m+9m^2) 4/(3m+1)+(3m-7)/(3m+1)*(9m^2-3m+-6m)/(1-3m+9m^2)=0 4/(3m+1)+(3m--6m)*(3m+1)/(3m+1)*(9m^2-3m+1)=0 4/(3m+1)+(3m--6m)/(9m^2-3m+1)=0 4/3m+1+3m-7-1+6m/(9m^2-3m+1)=0 4/12m-7/(9m^2-3m+1)=0 1/4*12m-7/(9m^2-3m+1)=0 12m-7/4(9m^2-3m+1)=0 12m-7/36m^2-4m+4=0
найдите а если 4
7 от а равны 40%
4
7: а=40% : 80
4
7 = 4: 7= 0,
0,57: 0,5=1,14
а= 1,14
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