\(x-\dfrac{3}{4}=\dfrac{1}{8}\\ x=\dfrac{1}{8}+\dfrac{3}{4}=\dfrac{7}{8}\)

x - \(\dfrac{3}{4}\)=\(\dfrac{1}{8}\)=>x=\(\dfrac{7}{8}\)

\(x-\dfrac{3}{4}=\dfrac{5}{12}\text{×}\dfrac{3}{10}\\ =>x-\dfrac{3}{4}=\dfrac{1}{8}\\ =>x=\dfrac{1}{8}+\dfrac{3}{4}=\dfrac{1}{8}+\dfrac{6}{8}=\dfrac{7}{8}\)

=>5căn x+2-15y=15 và 5căn x+2-2y=71/3

=>-13y=4/3 và căn x+2-3y=3

=>y=-4/39 và căn x+2=3+3y=3-12/39=105/39

=>y=-4/39 và x=887/169

27:(x-3/2)^3=(x-3/2):3

Ta có: \(\dfrac{27}{\left(x-\dfrac{3}{2}\right)^3}=\dfrac{\left(x-\dfrac{3}{2}\right)}{3}\)

\(\Rightarrow\left(x-\dfrac{3}{2}\right)^3.\left(x-\dfrac{3}{2}\right)\)=27.3

\(\Rightarrow\left(x-\dfrac{3}{2}\right)^4\)=81

\(\Rightarrow\left(x-\dfrac{3}{2}\right)^4=3^4\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=4\\x-\dfrac{3}{2}=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=4+\dfrac{3}{2}\\x=-4+\dfrac{3}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{2}+\dfrac{3}{2}\\x=\dfrac{-8}{2}+\dfrac{3}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\)

Vậy x∈\(\left\{\dfrac{11}{2};\dfrac{-5}{2}\right\}\)

\(\Leftrightarrow x^3-3x^2+3x-1-2x+3x^2-2+6x=-3\)

\(\Leftrightarrow x^3+7x-5=0\)

\(\left(x-5\right)^2=\left(18\dfrac{1}{3}:5\right).\dfrac{11}{3}\)

\(\Leftrightarrow\left(x-5\right)^2=\dfrac{55}{3}.\dfrac{1}{5}.\dfrac{11}{3}\)

\(\Leftrightarrow\left(x-5\right)^2=\dfrac{121}{9}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=\dfrac{11}{3}\\x-5=-\dfrac{11}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{26}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)

c: Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\left(y-\dfrac{1}{3}\right)^2\ge0\forall y\)

Do đó: \(\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2\ge0\forall x,y\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2-10\ge-10\forall x,y\)

Dấu '=' xảy ra khi x=-1 và \(y=\dfrac{1}{3}\)

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