Tìm x, biết:
a) \(\dfrac{3}{4} + \dfrac{1}{9}:x = 0,5\)
b) \(\dfrac{3}{4} – \left( {x – \dfrac{2}{3}} \right) = 1\dfrac{1}{3}\)
c) \(\left( {\dfrac{5}{7} – x} \right).\dfrac{{11}}{{15}} = \dfrac{{ – 22}}{{45}}\)
d) \(\left( {2,5x – \dfrac{4}{7}} \right):\dfrac{8}{{21}} = – 1,5\)
Ta đổi các số thập phân về dạng phân số, rồi áp dụng quy tắc chuyển vế đổi dấu để tìm x.
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\(\begin{array}{l}a)\dfrac{3}{4} + \dfrac{1}{9}:x = 0,5\\ \Leftrightarrow \dfrac{1}{9}:x = \dfrac{1}{2} – \dfrac{3}{4}\\ \Leftrightarrow \dfrac{1}{9}:x = \dfrac{{ – 1}}{4}\\ \Leftrightarrow x = \dfrac{1}{9}:\dfrac{{ – 1}}{4}\\ \Leftrightarrow x = \dfrac{{ – 4}}{9}\end{array}\)
Vậy \(x = \dfrac{{ – 4}}{9}\)
\(\begin{array}{l}b)\dfrac{3}{4} – \left( {x – \dfrac{2}{3}} \right) = 1\dfrac{1}{3}\\ \Leftrightarrow x – \dfrac{2}{3} = \dfrac{3}{4} – \dfrac{4}{3}\\ \Leftrightarrow x – \dfrac{2}{3} = \dfrac{{ – 7}}{{12}}\\ \Leftrightarrow x = \dfrac{{ – 7}}{{12}} + \dfrac{2}{3}\\ \Leftrightarrow x = \dfrac{1}{{12}}\end{array}\)
Vậy \(x = \dfrac{{ 1}}{12}\)
\(\begin{array}{l}c)\left( {\dfrac{5}{7} – x} \right).\dfrac{{11}}{{15}} = \dfrac{{ – 22}}{{45}}\\ \Leftrightarrow \dfrac{5}{7} – x = \dfrac{{ – 22}}{{45}}:\dfrac{{11}}{{15}}\end{array}\)
\(\begin{array}{l} \Leftrightarrow \dfrac{5}{7} – x = \dfrac{{ – 2}}{3}\\ \Leftrightarrow x = \dfrac{5}{7} – \left( {\dfrac{{ – 2}}{3}} \right)\end{array}\)
\(\begin{array}{l} \Leftrightarrow x = \dfrac{5}{7} + \dfrac{2}{3}\\\Leftrightarrow x = \dfrac{15}{21} + \dfrac{14}{21}\\ \Leftrightarrow x = \dfrac{{29}}{{21}}\end{array}\)
Vậy \(x = \dfrac{{ 29}}{21}\)
\(\begin{array}{l}d)\left( {2,5x – \dfrac{4}{7}} \right):\dfrac{8}{{21}} = – 1,5\\ \Leftrightarrow \left( {\dfrac{5}{2}x – \dfrac{4}{7}} \right) = \dfrac{{ – 3}}{2}.\dfrac{8}{{21}}\\ \Leftrightarrow \dfrac{5}{2}x – \dfrac{4}{7} = \dfrac{{ – 4}}{7}\\ \Leftrightarrow \dfrac{5}{2}x = \dfrac{{ – 4}}{7} + \dfrac{4}{7}\\ \Leftrightarrow \dfrac{5}{2}x = 0\\ \Leftrightarrow x = 0\end{array}\)
Vậy \(x =0\)