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, \(-1\dfrac{5}{27}-\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-24}{27}\)
\(\Rightarrow-\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-19}{27}\)
\(\Rightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{19}{27}\)
\(\Rightarrow3x-\dfrac{7}{9}=\dfrac{\sqrt[3]{19}}{3}\)
\(\Rightarrow3x=\dfrac{\sqrt[3]{19}}{3}+\dfrac{7}{19}\)
\(\Rightarrow...\)
\(a) \)\(5\dfrac{8}{17}:x+\left(\dfrac{-4}{17}\right):x+3\dfrac{1}{7}:17\dfrac{1}{3}=\dfrac{4}{11}\)
\(\dfrac{93}{17}:x+\dfrac{-4}{17}:x+\dfrac{33}{182}=\dfrac{4}{11}\)
\(\left(\dfrac{93}{17}+\dfrac{-4}{17}\right):x=\dfrac{4}{11}-\dfrac{33}{182}\)
\(\dfrac{89}{17}:x=\dfrac{365}{2002}\)
\(x=\dfrac{89}{17}:\dfrac{365}{2002}=\dfrac{178178}{6205}\)
\(b\))\(\dfrac{17}{2}-\left|2x-\dfrac{3}{4}\right|=\dfrac{-7}{4}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{17}{2}+\dfrac{7}{4}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{41}{4}\)
\(TH1:2x-\dfrac{3}{4}=\dfrac{41}{4}\Rightarrow x=\dfrac{11}{2}\)
\(TH2:2x-\dfrac{3}{4}=\dfrac{-41}{4}\Rightarrow x=\dfrac{-19}{4}\)
\(c) \)\(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\left(x+\dfrac{1}{5}\right)^2=\dfrac{26}{25}-\dfrac{17}{25}\)
\(\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\dfrac{9}{25}=\left(\dfrac{3}{5}\right)^2=\left(\dfrac{3}{-5}\right)^2=\left(\dfrac{-3}{5}\right)^2=\left(\dfrac{-3}{-5}\right)^2\)
\(Th1:x=\dfrac{2}{5}\)
\(Th2:x=\dfrac{-4}{5}\)
\(Th3:x=\dfrac{4}{5}\)(trùng kết quả nên chỉ lấy 1 phân số 4/5 thôi)
\(Th4:x=\dfrac{2}{5}\)(trùng kết quả nên chỉ lấy 1 phân số 2/5 thôi)
Câu d tương tự câu c
