\(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\)

Do vế trái không âm

\(\Rightarrow\hept{\begin{cases}x+\frac{3}{4}=0\\y-\frac{1}{5}=0\\x+y+z=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=-\frac{3}{4}\\y=\frac{1}{5}\\-\frac{3}{4}+\frac{1}{5}+z=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=-\frac{3}{4}\\y=\frac{1}{5}\\z=\frac{9}{20}\end{cases}}\)

Đặt \(A=\frac{3\left|a\right|+2}{\left|a\right|+5}\)

\(=\frac{3\left|a\right|+15-13}{\left|a\right|+5}=3-\frac{13}{\left|a\right|+5}\)

Để \(A\in Z\)

\(\Rightarrow13⋮\left(\left|a\right|+5\right)\)

\(\Rightarrow\left|a\right|+5\in\left(1;-1;13;-13\right)\)

\(\Rightarrow\left|a\right|=8\Rightarrow a=\hept{\begin{cases}8\\-8\end{cases}}\)

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\(-\frac{16}{3}-2x-1=0,6\)\(=\frac{3}{5}\)

\(\Rightarrow-\frac{16}{3}-2x=\frac{3}{5}+1=\frac{8}{5}\)

\(\Rightarrow2x=-\frac{16}{3}-\frac{8}{5}=-\frac{104}{15}\)

\(\Rightarrow x=-\frac{104}{15}:2=-\frac{52}{15}\)

9a = 8b => \(\frac{a}{b}=\frac{8}{9}\)

Số a là : 5 : ( 9 - 8 ) . 8 = 40

Số b là : 40 + 5 = 45

Vậy,.......

Ta có: \(9a=8b\Rightarrow\frac{a}{8}=\frac{b}{9}\)

Theo đề bài, ta có: \(\frac{a}{8}=\frac{b}{9}=\frac{b-a}{9-8}=\frac{5}{1}=5\)

\(\Rightarrow a=5\times8=40\)

\(b=5\times9=45\)