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\(\Rightarrow\dfrac{5}{4}-\dfrac{1}{4}x=\dfrac{3}{10}x-\dfrac{2}{5}\)
\(\Rightarrow\dfrac{5}{4}+\dfrac{2}{5}=\dfrac{3}{10}x-\dfrac{1}{4}x\)
\(\Rightarrow\dfrac{33}{20}=\dfrac{11}{20}x\)
\(\Rightarrow x=\dfrac{33}{20}\div\dfrac{11}{20}\)
\(\Rightarrow x=3\)
\(1\dfrac{1}{4}-x\dfrac{1}{4}=x\cdot30\%\cdot\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{5}{4}-x\dfrac{1}{4}=x\cdot\dfrac{3}{10}-\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{5}{4}-\dfrac{1}{4}x=\dfrac{3}{10}x-\dfrac{2}{5}\)
\(\Leftrightarrow25-5x=6x-8\)
\(\Leftrightarrow-5x-6x=-8-25\)
\(\Leftrightarrow-11x=-33\)
\(\Leftrightarrow x=3\)
Vậy x = 3
i) \(5\dfrac{8}{17}:x+\left(-\dfrac{4}{17}\right):x+3\dfrac{1}{7}:17\dfrac{1}{3}=\dfrac{4}{11}\)
\(\Rightarrow\dfrac{93}{17}:x-\dfrac{4}{17}:x+\dfrac{33}{182}=\dfrac{4}{11}\)
\(\Rightarrow\left(\dfrac{93}{17}-\dfrac{4}{17}\right):x=\dfrac{4}{11}-\dfrac{33}{182}\)
\(\Rightarrow\dfrac{89}{17}:x=\dfrac{365}{2002}\)
\(\Rightarrow x=\dfrac{89}{17}:\dfrac{365}{2002}=\dfrac{178178}{6205}\)
j) \(\dfrac{17}{2}-\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)
\(\Rightarrow\left|2x-\dfrac{3}{4}\right|=\dfrac{17}{2}-\left(-\dfrac{7}{4}\right)=\dfrac{41}{4}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{41}{4}\\2x-\dfrac{3}{4}=-\dfrac{41}{4}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}2x=11\Rightarrow x=\dfrac{11}{2}\\2x=-\dfrac{19}{2}\Rightarrow x=-\dfrac{19}{4}\end{matrix}\right.\)
k) \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Rightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{26}{25}-\dfrac{17}{25}=\dfrac{9}{25}=\left(\dfrac{3}{5}\right)^2\)\(=\left(-\dfrac{3}{5}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\Rightarrow x=\dfrac{2}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\Rightarrow x=-\dfrac{4}{5}\end{matrix}\right.\)
l) \(-1\dfrac{5}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Rightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-32}{27}-\left(-\dfrac{24}{27}\right)=-\dfrac{8}{27}=\left(-\dfrac{2}{3}\right)^3\)
\(\Rightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Rightarrow3x=-\dfrac{2}{3}+\dfrac{7}{9}=\dfrac{1}{9}\)
\(\Rightarrow x=\dfrac{1}{27}\)
j, \(\dfrac{17}{2}-\left|2x-\dfrac{3}{4}\right|=\dfrac{-7}{4}\)
\(\Rightarrow-\left|2x-\dfrac{3}{4}\right|=\dfrac{-7}{4}-\dfrac{17}{2}\)
\(\Rightarrow-\left|2x-\dfrac{3}{4}\right|=\dfrac{-41}{4}\)
\(\Rightarrow\left|2x-\dfrac{3}{4}\right|=\dfrac{41}{4}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{41}{4}\\2x-\dfrac{3}{4}=\dfrac{-41}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{-19}{4}\end{matrix}\right.\)
k, \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Rightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Rightarrow x+\dfrac{1}{5}=\pm\dfrac{3}{5}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=\dfrac{-3}{5}\end{matrix}\right.\Rightarrow}\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=\dfrac{-4}{5}\end{matrix}\right.\)
l, \(-1\dfrac{5}{27}-\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-24}{27}\)
\(\Rightarrow-\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-19}{27}\)
\(\Rightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{19}{27}\)
\(\Rightarrow3x-\dfrac{7}{9}=\dfrac{\sqrt[3]{19}}{3}\)
\(\Rightarrow3x=\dfrac{\sqrt[3]{19}}{3}+\dfrac{7}{19}\)
\(\Rightarrow...\)
Ta có: \(\dfrac{1}{2}\cdot y+\dfrac{2}{3}\cdot y=\dfrac{7}{6}\Rightarrow y\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=\dfrac{7}{6}\Rightarrow\dfrac{7}{6}y=\dfrac{7}{6}\Rightarrow y=\dfrac{7}{6}:\dfrac{7}{6}=1\)
Vậy \(D=\left\{1\right\}\)
Gọi chiều dài của ba tấm vải lần lượt là a,b,c
Theo đề, ta có: \(\dfrac{4}{7}a=\dfrac{4}{5}b=\dfrac{3}{5}c\)
hay \(\dfrac{a}{\dfrac{7}{4}}=\dfrac{b}{\dfrac{5}{4}}=\dfrac{c}{\dfrac{5}{3}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{\dfrac{7}{4}}=\dfrac{b}{\dfrac{5}{4}}=\dfrac{c}{\dfrac{5}{3}}=\dfrac{a+b+c}{\dfrac{7}{4}+\dfrac{5}{4}+\dfrac{5}{3}}=\dfrac{112}{\dfrac{14}{3}}=24\)
Do đó: a=42; b=30; c=40
thanks bạn