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1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}.3\)
\(\left|4-2x\right|=1\)
=>\(4-2x=\pm1\)
+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)
\(2x=4-1\) \(2x=4-\left(-1\right)\)
\(2x=3\) \(2x=4+1\)
\(x=3:2\) \(2x=5\)
\(x=1,5\) \(x=5:2\)
Vậy x=1,5 \(x=2,5\)
Vậy x=2,5
2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)
\(9:\left|x+\left(-1\right)\right|=-3\)
\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)
\(\left|x+\left(-1\right)\right|=-3\)
=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )
Vậy x = \(\varnothing\)
\(1,\\ x+\dfrac{1}{2}=-\dfrac{5}{3}\\ x=-\dfrac{5}{3}-\dfrac{1}{2}\\ x=-\dfrac{13}{6}\\ Vậyx=-\dfrac{13}{6}\)
\(2,\\ \dfrac{1}{3}-x=\dfrac{3}{5}\\ x=\dfrac{1}{3}-\dfrac{3}{5}\\ x=-\dfrac{4}{15}\\ Vậyx=-\dfrac{4}{15}\)
\(3,\\ 3-4+x=\dfrac{7}{2}\\ -1+x=\dfrac{7}{2}\\ x=\dfrac{7}{2}+1\\ x=\dfrac{9}{2}\\ Vậyx=\dfrac{9}{2}\)
\(4,\\ x-\dfrac{4}{3}=-\dfrac{7}{9}\\ x=-\dfrac{7}{9}+\dfrac{4}{3}\\ x=\dfrac{15}{27}\\ Vậyx=\dfrac{15}{27}\)
\(5,\\ x-\left(-\dfrac{7}{3}\right)=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{3}\\ x=-\dfrac{27}{18}\\ Vậyx=-\dfrac{27}{18}\)
\(6,\\ x-\dfrac{1}{5}=\dfrac{9}{10}\\ x=\dfrac{9}{10}+\dfrac{1}{5}\\ x=\dfrac{11}{10}\\ Vậyx=\dfrac{11}{10}\)
\(7,\\ x+\dfrac{5}{12}=\dfrac{3}{8}\\ x=\dfrac{3}{8}-\dfrac{5}{12}\\ x=-\dfrac{1}{24}\\ Vậyx=-\dfrac{1}{24}\)
\(8,\\ x+\dfrac{5}{4}=\dfrac{7}{6}\\ x=\dfrac{7}{6}-\dfrac{5}{4}\\ x=-\dfrac{9}{24}\\ Vậyx=-\dfrac{9}{24}\)
\(9,\\ x-\dfrac{2}{7}=\dfrac{1}{35}\\ x=\dfrac{1}{35}+\dfrac{2}{7}\\ x=\dfrac{11}{35}\\ Vậyx=\dfrac{11}{35}\\ 10,\\ x-\dfrac{1}{5}=-\dfrac{7}{10}\\ x=-\dfrac{7}{10}+\dfrac{1}{5}\\ x=-\dfrac{1}{2}\\ Vậyx=-\dfrac{1}{2}\)
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1/ \(x+\dfrac{1}{2}=\dfrac{-5}{3}\)
\(x=\dfrac{-5}{3}-\dfrac{1}{2}\)
\(x=\dfrac{-10}{6}-\dfrac{3}{6}\)
Vậy \(x=\dfrac{-13}{6}\)
2/\(\dfrac{1}{3}-x=\dfrac{3}{5}\)
\(-x=\dfrac{3}{5}-\dfrac{1}{3}\)
\(-x=\dfrac{9}{15}-\dfrac{5}{15}\)
\(-x=\dfrac{4}{15}\)
Vậy \(x=\dfrac{-4}{15}\)
3/ \(3-4+x=\dfrac{7}{2}\)
\(-4+x=\dfrac{7}{2}-3\)
\(-4+x=\dfrac{7}{2}-\dfrac{6}{2}\)
\(-4+x=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}+4\)
\(x=\dfrac{1}{2}+\dfrac{8}{2}\)
Vậy \(x=\dfrac{9}{2}\)
4/ \(x-\dfrac{4}{3}=\dfrac{-7}{9}\)
\(x=\dfrac{-7}{9}+\dfrac{4}{3}\)
\(x=\dfrac{-7}{9}+\dfrac{12}{9}\)
Vậy \(x=\dfrac{5}{9}\)
5/ \(x-\dfrac{-7}{2}=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}-\dfrac{7}{2}\)
\(x=\dfrac{5}{6}-\dfrac{21}{6}\)
Vậy \(x=\dfrac{-16}{6}=\dfrac{-8}{3}\)
6/ \(x-\dfrac{1}{5}=\dfrac{9}{10}\)
\(x=\dfrac{9}{10}+\dfrac{1}{5}\)
\(x=\dfrac{9}{10}+\dfrac{2}{10}\)
Vậy \(x=\dfrac{11}{10}\)
7/ \(x+\dfrac{5}{12}=\dfrac{3}{8}\)
\(x=\dfrac{3}{8}-\dfrac{5}{12}\)
\(x=\dfrac{9}{24}-\dfrac{10}{24}\)
Vậy \(x=\dfrac{-1}{24}\)
8/ \(x+\dfrac{5}{4}=\dfrac{7}{6}\)
\(x=\dfrac{7}{6}-\dfrac{5}{4}\)
\(x=\dfrac{14}{12}-\dfrac{15}{12}\)
Vậy \(x=\dfrac{-1}{12}\)
9/ \(x-\dfrac{2}{7}=\dfrac{1}{35}\)
\(x=\dfrac{1}{35}+\dfrac{2}{7}\)
\(x=\dfrac{1}{35}+\dfrac{10}{35}\)
Vậy \(x=\dfrac{11}{35}\)
10 /\(x-\dfrac{1}{5}=\dfrac{-7}{10}\)
\(x=\dfrac{-7}{10}+\dfrac{1}{5}\)
\(x=\dfrac{-7}{10}+\dfrac{2}{10}\)
Vậy \(x=\dfrac{-5}{10}=\dfrac{-1}{2}\)
1: \(\dfrac{3}{4}:\dfrac{1}{2}+x=\dfrac{2}{3}\)
=>\(x+\dfrac{3}{4}\cdot2=\dfrac{2}{3}\)
=>\(x+\dfrac{3}{2}=\dfrac{2}{3}\)
=>\(x=\dfrac{2}{3}-\dfrac{3}{2}=\dfrac{4}{6}-\dfrac{9}{6}=-\dfrac{5}{6}\)
2: \(\dfrac{7}{4}+\dfrac{1}{4}:x=2\)
=>\(\dfrac{1}{4}:x=2-\dfrac{7}{4}=\dfrac{1}{4}\)
=>\(x=\dfrac{1}{4}:\dfrac{1}{4}=1\)
3: \(\dfrac{48}{64}:\dfrac{12}{16}+0,25=\dfrac{3}{4}:\dfrac{3}{4}+0,25=1+0,25=1,25\)
4: \(\dfrac{3}{4}\cdot\dfrac{6}{9}+\dfrac{7}{12}\cdot6=\dfrac{18}{36}+\dfrac{7}{2}=\dfrac{1}{2}+\dfrac{7}{2}=\dfrac{8}{2}=4\)
5: \(5\cdot\dfrac{x}{6}-\dfrac{1}{4}=\dfrac{7}{2}\)
=>\(\dfrac{5}{6}x=\dfrac{7}{2}+\dfrac{1}{4}=\dfrac{15}{4}\)
=>\(x=\dfrac{15}{4}:\dfrac{5}{6}=\dfrac{15}{4}\cdot\dfrac{6}{5}=\dfrac{3}{2}\cdot3=\dfrac{9}{2}\)
6: \(\dfrac{3}{x+1}-\dfrac{1}{4}=\dfrac{7}{4}\)(ĐKXĐ: x<>-1)
=>\(\dfrac{3}{x+1}=\dfrac{7}{4}+\dfrac{1}{4}=\dfrac{8}{4}=2\)
=>\(x+1=\dfrac{3}{2}\)
=>\(x=\dfrac{1}{2}\left(nhận\right)\)