\(5(\frac{a}{b}+\frac{1}{2})=2\frac{1}{3}\)

\( \iff5.\frac{a}{b}+5.\frac{1}{2}=\frac{7}{3}\)

\(\iff5.\frac{a}{b}+\frac{5}{2}=\frac{7}{3}\)

\(\iff 5.\frac{a}{b}=\frac{7}{3}-\frac{5}{2}\)

\(\iff 5.\frac{a}{b}=\frac{-1}{6}\)

\(\iff\frac{a}{b}=\frac{-1}{6}:5\)

\(\iff\frac{a}{b}=\frac{-1}{6}.\frac{1}{5}\)

\(\iff \frac{a}{b}=\frac{-1}{30}\)

Vậy \(\frac{a}{b}=\frac{-1}{30}\)

~ Hok tốt a~

\(5\left(ab+\frac{1}{2}\right)=2\frac{1}{3}\)

\(5\left(ab+\frac{1}{2}\right)=\frac{7}{3}\)

\(ab+\frac{1}{2}=\frac{7}{3}:5\)

\(ab+\frac{1}{2}=\frac{7}{15}\)

\(ab=\frac{7}{15}-\frac{1}{2}\)

\(ab=\frac{14}{30}-\frac{15}{30}\)

\(ab=-\frac{1}{30}\)

Vậy \(ab=-\frac{1}{30}\)

hey bạn ơi đáng lẽ câu hỏi phải là: " 5 - (a/b +1/2) = 2 và 1/3" chứ

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\(A=\frac{9a^5-ab^4-18a^4b+2b^5}{3a^2b^2+ab^4-6a^2b^3-2b^5}\)

\(=\frac{a\left(9a^4-b^4\right)-2b\left(9a^4-b^4\right)}{ab^2\left(3a^2+b^2\right)-2b^3\left(3a^2+b^2\right)}\)

\(=\frac{\left(9a^4-b^4\right)\left(a-2b\right)}{\left(3a^2+b^2\right)\left(ab^2-2b^3\right)}\)

\(=\frac{\left(3a^2-b^2\right)\left(3a^2+b^2\right)\left(a-2b\right)}{\left(3a^2+b^2\right)b^2\left(a-2b\right)}\)

\(=\frac{3a^2-b^2}{b^2}\)

\(=3.\left(\frac{a}{b}\right)^2-1=3.\left(\frac{2}{3}\right)^2-1=\frac{1}{3}\)

\(15\left(2a^2-1\right)+5\left(3-\frac{1}{5a}-6a^2\right)\)

\(=30a^2-15+15-\frac{1}{a}-30a^2\)

\(=-\frac{1}{a}\)

tại \(a=2017\)=> M= \(\frac{-1}{a}=\frac{-1}{2017}\)

\(\left(x-y\right)\left(x^2+xy+y^2\right)+y^3\)

\(=x^3-y^3+y^3\)

\(=x^3\)

ại \(x=2\)=> N= \(x^3=2^3=8\)

Ta có: \(x^2-y+\frac{1}{4}=y^2-x+\frac{1}{4}=0\)

\(\Rightarrow\left(x^2-x+\frac{1}{4}\right)+\left(y^2-y+\frac{1}{4}\right)=0\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\y-\frac{1}{2}=0\end{cases}\Rightarrow}x=y=\frac{1}{2}\)

Vậy \(x=y=\frac{1}{2}\)