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c: =>2/3x=1/10+1/2=1/10+5/10=6/10=3/5
hay \(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
d: \(\Leftrightarrow\dfrac{4}{9}:x=\dfrac{2}{3}-\dfrac{3}{5}=\dfrac{1}{15}\)
hay \(x=\dfrac{4}{9}:\dfrac{1}{15}=\dfrac{4}{9}\cdot15=\dfrac{20}{3}\)
f: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
\(\frac{1}{2}+\frac{5}{6}-\frac{3}{8}\)
\(=\frac{12}{24}+\frac{20}{24}-\frac{9}{24}\)
\(=\frac{23}{24}\)
\(\frac{10}{15}\cdot\frac{7}{4}+\frac{7}{4}\cdot\frac{9}{15}-\frac{4}{15}\cdot\frac{7}{4}\)
\(=\frac{7}{4}\cdot\left(\frac{10}{15}+\frac{9}{15}-\frac{4}{15}\right)\)
\(=\frac{7}{4}\cdot1=\frac{7}{4}\)
\(\left(\frac{4}{5}+\frac{1}{2}\right)\left(\frac{3}{13}-\frac{8}{13}\right)\)
\(=\left(\frac{8}{10}+\frac{5}{10}\right)\cdot\left(-\frac{5}{13}\right)\)
\(=\frac{13}{10}\cdot\left(-\frac{5}{13}\right)=-\frac{1}{2}\)
a, 3x-12 = 30
=> 3x = 30 + 12
=> 3x = 42
=> x = 42 : 3 = 14
Vậy x = 14
b, \(\frac{2}{3}x+\frac{1}{4}=\frac{7}{12}\)
\(\Rightarrow\frac{2}{3}x=\frac{7}{12}-\frac{1}{4}\)
\(\Rightarrow\frac{2}{3}x=\frac{7}{12}-\frac{3}{12}\)
\(\Rightarrow\frac{2}{3}x=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}\div\frac{2}{3}\Rightarrow\frac{1}{3}\cdot\frac{3}{2}=\frac{1}{2}\)
Vậy x = \(\frac{1}{2}\)
c, 2x2 = 32
=> x2 = 32 : 2
=> x2 = 16
=> x2 = 42
=> x = 4
Vậy x = 4
a, \(x:3\frac{1}{15}=1\frac{1}{2}\)
\(\Rightarrow x=1\frac{1}{2}\cdot3\frac{1}{15}\)
\(\Rightarrow x=\frac{3}{2}\cdot\frac{46}{15}=\frac{3\cdot46}{2\cdot15}=\frac{1\cdot23}{1\cdot5}=\frac{23}{5}=4\frac{3}{5}\)
\(b)x\cdot\frac{15}{28}=\frac{3}{20}\)
\(\Rightarrow x=\frac{3}{20}:\frac{15}{28}=\frac{3}{20}\cdot\frac{28}{15}=\frac{1}{5}\cdot\frac{7}{5}=\frac{7}{25}\)
Tự làm nốt câu cuối :>
a)\(11\frac{1}{4}-\left(2\frac{5}{7}+5\frac{1}{4}\right)\)
\(=\frac{45}{4}-\left(\frac{19}{7}+\frac{21}{4}\right)\)
\(=\frac{45}{4}-\left(\frac{76}{28}+\frac{147}{28}\right)\)
\(=\frac{45}{4}-\frac{223}{28}\)
\(=\frac{315}{28}-\frac{223}{28}\)
\(=\frac{23}{7}\)
b) \(\left(8\frac{5}{11}+3\frac{5}{8}\right)-3\frac{5}{11}\)
\(=\left(\frac{93}{11}+\frac{29}{8}\right)-\frac{38}{11}\)
\(=\left(\frac{744}{88}+\frac{319}{88}\right)-\frac{38}{11}\)
\(=\frac{1063}{88}-\frac{38}{11}=\frac{1063}{88}-\frac{304}{88}\)
\(=\frac{69}{8}\)
a) \(\left(\frac{4}{13}.\frac{6}{5}+\frac{4}{13}.\frac{2}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)
\(\left(\frac{4}{13}.\frac{8}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)
\(\frac{32}{65}.\left(2x+1\right)^2=\frac{10}{13}\)
\(\left(2x+1\right)^2=\frac{10}{13}\div\frac{32}{65}\)
\(\left(2x+1\right)^2=\frac{25}{16}\)
\(\Rightarrow2x+1\in\left\{\frac{5}{4};-\frac{5}{4}\right\}\)
\(\hept{\begin{cases}2x+1=\frac{5}{4}\\2x+1=-\frac{5}{4}\end{cases}\Rightarrow\hept{\begin{cases}2x=\frac{1}{4}\\2x=-\frac{9}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{8}\\x=-\frac{9}{8}\end{cases}}}\)
Vậy \(x\in\left\{\frac{1}{8};-\frac{9}{8}\right\}\)
\(x^3-\frac{9}{16}.x=0\)
\(x\left(x^2-\frac{9}{16}\right)=0\)
\(\hept{\begin{cases}x=0\\x^2-\frac{9}{16}=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=\frac{9}{16}\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=\pm\frac{3}{4}\end{cases}}}\)
Vậy \(x\in\left\{0;\frac{3}{4};-\frac{3}{4}\right\}\)
\(\left|x-3\right|=2x+4\)
\(\left|x-3\right|=2x+2\cdot2\)
\(\left|x-3\right|=2\left(x+2\right)\)
\(\Rightarrow\orbr{\begin{cases}x-3=-\left[2\cdot\left(x+2\right)\right]\\x-3=2\left(x+2\right)\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x-3=-\left[2x+4\right]\\x-3=2x+2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-3=-2x-4\\x=2x+2+3\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=-2x-4+3\\x=2x+5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-2x-1\\x=2x+5\end{cases}}\) \(.....................\)
Bài 1:
a) b) c) sẽ có bạn giải cho em thôi vì nó dễ tính tay cũng đc
d) \(\frac{4}{2.5}+\frac{4}{5.8}+...+\frac{4}{23.26}\)
\(=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{23.26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{23}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\frac{6}{13}\)
\(=\frac{8}{13}\)
Bài 2:
a) b) c)
d)\(|\frac{5}{8}x+\frac{6}{7}|-\frac{4}{7}=\frac{10}{7}\)
\(\Leftrightarrow|\frac{5}{8}x+\frac{6}{7}|=2\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x+\frac{6}{7}=2\\\frac{5}{8}x+\frac{6}{7}=-2\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x=\frac{8}{7}\\\frac{5}{8}x=\frac{-20}{7}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{64}{35}\\x=\frac{-32}{7}\end{cases}}}\)
Vậy \(x\in\left\{\frac{64}{35};\frac{-32}{7}\right\}\)
Bài 1 :
a) \(\left(\frac{2}{5}-\frac{5}{8}\right):\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-9}{40}:\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-27}{44}+\frac{1}{8}\)
\(=\frac{-43}{88}\)
\(\frac{3}{5}-\frac{1}{9}-\frac{1}{2}-\frac{1}{127}+\frac{7}{18}+\frac{4}{35}+\frac{2}{7}\)=\((\frac{1}{9}-\frac{1}{2}+\frac{7}{18})+(\frac{3}{5}+\frac{2}{7}+\frac{4}{35})+\frac{1}{127}\)=\(0+1+\frac{1}{127}\)=\(1+\frac{1}{127}=\frac{128}{127}\)
a) \(\frac{1}{4}+\frac{1}{3}\div\left(2x-1\right)=-5\)
\(\Leftrightarrow\frac{1}{3}\div\left(2x-1\right)=-5-\frac{1}{4}\)
\(\Leftrightarrow\frac{1}{3}\div\left(2x-1\right)=\frac{-20}{4}-\frac{1}{4}\)
\(\Leftrightarrow\frac{1}{3}\div\left(2x-1\right)=\frac{-21}{4}\)
\(\Leftrightarrow\left(2x-1\right)=\frac{1}{3}\div\frac{-21}{4}\)
\(\Leftrightarrow\left(2x-1\right)=\frac{-4}{63}\)
\(\Leftrightarrow2x=\frac{-4}{63}+1\)
\(\Leftrightarrow2x=\frac{-4}{63}+\frac{63}{63}\)
\(\Leftrightarrow2x=\frac{59}{63}\)
\(\Leftrightarrow x=\frac{59}{63}\div2\)
\(\Leftrightarrow x=\frac{59}{126}\)
b) \(-\left(5\frac{3}{8}+x-7\frac{5}{24}\right)\div2019\frac{11}{37}=0\)
\(\Leftrightarrow-\left(5\frac{3}{8}+x-7\frac{5}{24}\right)=0.2019\frac{11}{37}\)
\(\Leftrightarrow-5\frac{3}{8}-x+7\frac{5}{24}=0\)
\(\Leftrightarrow\frac{-43}{8}-x+\frac{173}{24}=0\)
\(\Leftrightarrow\frac{-129}{24}-x+\frac{173}{24}=0\)
\(\Leftrightarrow-x+\frac{44}{24}=0\)
\(\Leftrightarrow x=\frac{44}{24}-0\)
\(\Leftrightarrow x=\frac{44}{24}=\frac{11}{6}\)
a: =>2x=-2
hay x=-1
b: =>3x-4=20
=>3x=24
hay x=8
giúp em với ạ e lm bài kiểm tra ạ