\(x+15\%x=\left[\left(1,09-0,29\right).1\frac{1}{4}\right]:\left[\left(18,9-16\frac{13}{20}\right).\frac{8}{9}\right]\)

\(x+15\%x=\left[0,8.1\frac{1}{4}\right]:\left[2,25.\frac{8}{9}\right]\)

\(x+15\%x=1:2\)

\(x+15\%x=\frac{1}{2}\)

\(x+15x=\frac{1}{2}\)

\(16x=\frac{1}{2}\)

\(x=\frac{1}{2}:16=\frac{1}{32}\)

Vậy x=1/32

Bài 4: 

b: Ta có: \(2x\left(x-\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\)

\(\frac{\left(1,09-0,29\right)x\frac{5}{4}}{\left(18,9-16,65\right)x\frac{8}{9}}\)=\(\frac{\frac{4}{5}x\frac{5}{4}}{\frac{9}{4}x\frac{8}{9}}\)=\(\frac{1}{2}\)\(=0,5\)

\(B=\frac{\left(1,09-0,29\right).1\frac{1}{4}}{\left(18,9-16\frac{13}{20}\right).\frac{8}{9}}=\frac{0,8.\frac{5}{4}}{\left[\left(18+0,9\right)-\left(16+\frac{13}{20}\right)\right].\frac{8}{9}}\)

\(B=\frac{\frac{4}{5}.\frac{5}{4}}{\left(18+\frac{9}{10}-16-\frac{13}{20}\right).\frac{8}{9}}\)

\(B=\frac{1}{\left[\left(18-16\right)+\left(\frac{9}{10}-\frac{13}{20}\right)\right].\frac{8}{9}}\)

\(B=\frac{1}{\left(2+\frac{1}{4}\right).\frac{8}{9}}\)

\(B=\frac{1}{\frac{9}{4}.\frac{8}{9}}=\frac{1}{2}\)

Bằng \(\frac{1}{2}\)

Ta có:

\(\dfrac{A}{B}=\dfrac{\left[0,8.7+\left(0,8\right)^2\right]\left(1,25.7-\dfrac{4}{5}.1,25\right)+31,64}{\dfrac{\left(1,09-0,29\right).\dfrac{5}{4}}{\left(18,9-16,65\right).\dfrac{9}{8}}}\)

\(\dfrac{A}{B}=\dfrac{\left[0,8\left(7+0,8\right)\right]\left[1,25\left(7-\dfrac{4}{5}\right)\right]+31,64}{\dfrac{0,8.\dfrac{5}{4}}{2,26.\dfrac{9}{8}}}\)

\(\dfrac{A}{B}=\dfrac{6,24.7,75+31,64}{\dfrac{1}{2,53125}}\)

\(\dfrac{A}{B}=\left(48,36+31,64\right).2,53125\)

\(\dfrac{A}{B}=202,5\)

Vậy.............

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