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a: \(\Leftrightarrow\left(x-\dfrac{2}{5}\right):\dfrac{3}{2}=-\dfrac{5}{4}+\dfrac{5}{2}=\dfrac{5}{4}\)
\(\Leftrightarrow x-\dfrac{2}{5}=\dfrac{5}{4}\cdot\dfrac{3}{2}=\dfrac{15}{8}\)
hay x=91/40
b: \(\Leftrightarrow\left(2.5x-3.6\right)=-1\cdot\dfrac{12}{7}=\dfrac{-12}{7}\)
=>2,5x=66/35
hay x=132/175
c: \(\Leftrightarrow\left(\dfrac{15}{4}-2x\right)=\dfrac{19}{9}:\dfrac{4}{3}=\dfrac{19}{9}\cdot\dfrac{3}{4}=\dfrac{19}{12}\)
=>2x=15/4-19/12=45/12-19/12=26/12
=>x=13/12
a: =6-6=0
b: \(=\dfrac{-5}{4}:\dfrac{2-7}{8}+\dfrac{3}{2}\cdot\dfrac{2-5}{6}\)
\(=\dfrac{-5}{4}\cdot\dfrac{8}{-5}+\dfrac{3}{2}\cdot\dfrac{-3}{6}\)
\(=2+\dfrac{-9}{12}=2-\dfrac{3}{4}=\dfrac{5}{4}\)
c: \(=2,5\cdot\left(-0,65\right)+1,5\left(-0,3-0,35\right)=-0,65\cdot4=-2,6\)
\(x\div4\frac{1}{3}=-2,5\)
\(\Leftrightarrow x\div\frac{13}{3}=\frac{-5}{2}\)
\(\Leftrightarrow x=\frac{-5}{2}.\frac{13}{3}\)
\(\Leftrightarrow x=\frac{-65}{6}\)
\(x\div\frac{-3}{5}=\frac{-10}{21}\)
\(\Leftrightarrow x=\frac{-10}{21}.\frac{-3}{5}\)
\(\Leftrightarrow x=\frac{30}{105}\)
\(\Leftrightarrow x=\frac{2}{7}\)
\(\frac{2}{3}x-\frac{1}{2}=\frac{1}{10}\)
\(\Leftrightarrow\frac{2}{3}x=\frac{1}{10}+\frac{1}{2}\)
\(\Leftrightarrow\frac{2}{3}x=\frac{6}{10}\)
\(\Leftrightarrow x=\frac{6}{10}\div\frac{2}{3}\)
\(\Leftrightarrow x=\frac{18}{20}\)
\(\Leftrightarrow x=\frac{9}{10}\)
\(\frac{1}{2}x+\frac{1}{2}=\frac{5}{2}\)
\(\Leftrightarrow\frac{1}{2}\left(x+1\right)=\frac{5}{2}\)
\(\Leftrightarrow\left(x+1\right)=\frac{5}{2}\div\frac{1}{2}\)
\(\Leftrightarrow\left(x+1\right)=5\)
\(\Leftrightarrow x=5-1\)
\(\Leftrightarrow x=4\)
1.
a) \(\frac{13}{8}:\frac{13}{4}=\frac{1}{2}=0,5=50\%\)
b) \(12,5:2,5=5=500\%\)
~ Hok tốt ~
2.
\(\left(\frac{37}{9}+\frac{13}{4}\right).\frac{9}{4}+\frac{11}{4}\)
\(=\frac{265}{36}.\frac{9}{4}+\frac{11}{4}\)
\(=\frac{265}{16}+\frac{11}{4}\)
\(=\frac{309}{16}\)
\(1+\left(\frac{9}{10}-\frac{4}{5}\right):\frac{19}{6}\)
\(=1+\frac{1}{10}:\frac{19}{6}\)
\(=1+\frac{3}{95}\)
\(=\frac{98}{95}\)
\(\left(-7+\left|13\right|\right)-\left(13-\left|-7\right|-25\right)-\left(25+\left|-10\right|-9\right)\)
\(=\left(-7+13\right)-\left(13-7-25\right)-\left(25+10-9\right)\)
\(=-7+13-13+7+25-25-10+9\)
\(=\left(-7+7\right)+\left(13-13\right)+\left(25-25\right)-10+9\)
\(=0+0+0-10+9\)
\(=-10+9\)
\(=-1\)
~ Hok tốt ~
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 + 2| = 4
\(\Leftrightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-6\end{cases}}\)
2) 3 – |2x + 1| = (-5)
\(\Leftrightarrow\left|2x+1\right|=8\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}\)
3) 12 + |3 – x| = 9
\(\Leftrightarrow\left|3-x\right|=-3\)(vô lí)
=>\(x=\varnothing\)
1) I x+2 I=4
\(\Rightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-6\end{cases}}}\)
2) \(3-|2x+1|=-5\)
\(\Leftrightarrow|2x+1|=8\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}}\)
3) \(12+|3-x|=9\)
\(\Leftrightarrow|3-x|=-3\)(vô lí vì I 3-x I \(\ge\)0)
\(\left(x-2,5\right)^2=\frac{4}{9}\)
\(\left(x-2,5\right)^2=\left(\frac{2}{3}\right)^2\)
\(x-2,5=\frac{2}{3}\)
\(x=\frac{2}{3}+2,5\)
\(x=\frac{19}{6}\)
\(\left(2x+\frac{1}{3}\right)^3=\frac{-8}{27}\)
\(\left(2x+\frac{1}{3}\right)^3=\left(\frac{-2}{3}\right)^3\)
\(2x+\frac{1}{3}=\frac{-2}{3}\)
\(2x=\frac{-2}{3}-\frac{1}{3}\)
\(2x=-1\)
\(x=\frac{-1}{2}\)
\(\left(x-2,5\right)^2=\left(\frac{2}{3}\right)^2\)
\(\Rightarrow\)\(\orbr{\begin{cases}x-2,5=\frac{2}{3}\\x-2,5=-\frac{2}{3}\end{cases}}\)
\(\Rightarrow\)\(\orbr{\begin{cases}x=\frac{2}{3}+\frac{5}{2}\\x=-\frac{2}{3}+\frac{5}{2}\end{cases}}\)
\(\Rightarrow\)\(\orbr{\begin{cases}x=\frac{4}{6}+\frac{15}{6}=\frac{19}{6}\\x=-\frac{4}{6}+\frac{15}{6}=\frac{11}{6}\end{cases}}\)