Question

6m-4(mm)²| mn²=?
При m= одна целая одна четвертая
При n= одна целая одна восьмая

Помогите пожалуйста очень нужно

Answer 1

m=1(1/4), n=1(1/8)

m(6-4m^3)/mn^2=(6-4m^3)/n^2=(6-4((5/4)^3))/(9/8)^2=(6-4((125/64)))/(81/64)=(6-(125/16))/(81/64)=((96-125)/16)/(81/64)=(-29/16)/(81/64)=-(29×64/16×81)=-(29×4/81)=-116/81=-1(35/81)=-1,432

Answer 2

$6m - \frac{4(mm)^2}{mn^2} = 6m - \frac{4(m^2)^2}{mn^2} = 6m - \frac{4m^4}{m^1*n^2} = 6m- \frac{4m^3}{n^2} =\\ \\ \\ m= 1 \frac{1}{4} = \frac{5}{4} ; n= 1 \frac{1}{8} = \frac{9}{8} \\ \\ 6 * \frac{5}{4} - \frac{4 * ( \frac{5}{4})^3 }{( \frac{9}{8})^2 } = \frac{15}{2} - \frac{ \frac{125}{16} }{\frac{81}{64} } = \frac{15}{2} - \frac{125*64}{16*81} = \frac{15}{2} - \frac{500}{81} = \\ \\ = \frac{1215 - 1000}{162} = \frac{215}{162} = 1 \frac{53}{162}$