Đặt \(A=\dfrac{1}{\sqrt[3]{a+7b}}+\dfrac{1}{\sqrt[3]{b+7c}}+\dfrac{1}{\sqrt[3]{c+7a}}\)

\(A=\dfrac{\sqrt[3]{64}}{\sqrt[3]{8.8\left(a+7b\right)}}+\dfrac{\sqrt[3]{64}}{\sqrt[3]{8.8\left(b+7c\right)}}+\dfrac{\sqrt[3]{64}}{\sqrt[3]{8.8\left(c+7a\right)}}\)

\(\ge\dfrac{4}{\dfrac{8+8+a+7b}{3}}+\dfrac{4}{\dfrac{8+8+b+7c}{3}}+\dfrac{4}{\dfrac{8+8+c+7a}{3}}\ge\dfrac{\left(2+2+2\right)^2}{\dfrac{8+8+a+7b+8+8+b+7c+8+8+c+7a}{3}}\)

\(=\dfrac{36.3}{8\left(a+b+c\right)+48}=\dfrac{3}{2}\)

Vậy \(A_{min}=\dfrac{3}{2}\Leftrightarrow a=b=c=1\)

a) Thay m=3 vào hệ pt, ta được:

\(\left\{{}\begin{matrix}x+3y=3\\3x+4y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+9y=9\\3x+4y=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5y=3\\x+3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{3}{5}\\x=3-3y=3-3\cdot\dfrac{3}{5}=\dfrac{6}{5}\end{matrix}\right.\)

Vậy: Khi m=3 thì hệ phương trình có nghiệm duy nhất là \(\left(x,y\right)=\left(\dfrac{6}{5};\dfrac{3}{5}\right)\)

a) Thay m=3 vào hệ phương trình, ta được:

\(\left\{{}\begin{matrix}x+3y=3\\3x+4y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+9y=9\\3x+4y=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5y=3\\x+3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{3}{5}\\x=3-3\cdot\dfrac{3}{5}=\dfrac{15}{5}-\dfrac{9}{5}=\dfrac{6}{5}\end{matrix}\right.\)

Vậy: \(\left(x,y\right)=\left(\dfrac{6}{5};\dfrac{3}{5}\right)\)

Câu 4: 

\(\dfrac{3x+5}{16}-\dfrac{3x-5}{26}=\dfrac{3x-8}{29}-\dfrac{3x+8}{13}\)

\(\Leftrightarrow\left(\dfrac{3x+5}{16}+1\right)-\left(\dfrac{3x-5}{26}+1\right)=\left(\dfrac{3x-8}{29}+1\right)-\left(\dfrac{3x-8}{13}+1\right)\)

\(\Leftrightarrow\left(3x+21\right)\left(\dfrac{1}{16}-\dfrac{1}{26}-\dfrac{1}{29}+\dfrac{1}{13}\right)=0\)

=>3x+21=0

hay x=-7

\(=>A+B-C+D=a+b-5-b-c+1-b+c+4+b-a\)

\(=-5+4=-1\)

\(N=\left|x+3\right|+\left|x+4\right|+\left|x+5\right|\)

\(\left|x+3\right|,\left|x+4\right|,\left|x+5\right|\ge0\)

\(\Rightarrow N\ge0\)

\(N=\left(x+3\right)+\left(x+4\right)+\left(x+5\right)\ge0\)

\(N=\left(x+x+x\right)+\left(3+4+5\right)\)

\(N=3x+12\)

\(\Rightarrow N=3x\ge12\)

\(\Rightarrow N=x\ge4\)

\(\Rightarrow N\ge4\)

\(\%Fe\left(FeO\right)=\dfrac{56}{72}.100\%=77,78\%\)

\(\%Fe\left(Fe_2O_3\right)=\dfrac{56.2}{160}.100\%=70\%\)

\(\%Fe\left(Fe_3O_4\right)=\dfrac{56.3}{232}.100\%=72,414\%\)

\(\%Fe\left(Fe\left(OH\right)_3\right)=\dfrac{56}{107}.100\%=52,34\%\)

\(\%Fe\left(FeCl_2\right)=\dfrac{56}{127}.100\%=44,09\%\)

\(\%Fe\left(FeSO_4.5H_2O\right)=\dfrac{56}{242}.100\%=23,14\%\)

=> FeO có hàm lượng Fe cao nhất